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Theorem List for Metamath Proof Explorer - 41901-42000   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremcdlemk15-2N 41901* Part of proof of Lemma K of [Crawley] p. 118. Line 21 on p. 119. 𝑄, 𝐶 are k2, f2. (Contributed by NM, 1-Jul-2013.) (New usage is discouraged.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑄 = (𝑆‘𝐶)    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ (𝐹 ∈ 𝑇 ∧ 𝐶 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ ((𝑅‘𝐶) ≠ (𝑅‘𝐹) ∧ (𝐹 ≠ ( I ↾ 𝐵) ∧ 𝐶 ≠ ( I ↾ 𝐵)) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊))) → (𝑁‘𝑃) ≤ ((𝑃 ∨ (𝑅‘𝐹)) ∧ ((𝑄‘𝑃) ∨ (𝑅‘(𝐹 ∘ ◡𝐶)))))
 
Theoremcdlemk16-2N 41902* Part of proof of Lemma K of [Crawley] p. 118. (Contributed by NM, 1-Jul-2013.) (New usage is discouraged.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑄 = (𝑆‘𝐶)    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ (𝐹 ∈ 𝑇 ∧ 𝐶 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ ((𝑅‘𝐶) ≠ (𝑅‘𝐹) ∧ (𝐹 ≠ ( I ↾ 𝐵) ∧ 𝐶 ≠ ( I ↾ 𝐵)) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊))) → (((𝑃 ∨ (𝑅‘𝐹)) ∧ ((𝑄‘𝑃) ∨ (𝑅‘(𝐹 ∘ ◡𝐶)))) ∈ 𝐴 ∧ ¬ ((𝑃 ∨ (𝑅‘𝐹)) ∧ ((𝑄‘𝑃) ∨ (𝑅‘(𝐹 ∘ ◡𝐶)))) ≤ 𝑊))
 
Theoremcdlemk17-2N 41903* Part of proof of Lemma K of [Crawley] p. 118. Line 21 on p. 119. 𝑄, 𝐶 are k2, f2. (Contributed by NM, 1-Jul-2013.) (New usage is discouraged.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑄 = (𝑆‘𝐶)    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ (𝐹 ∈ 𝑇 ∧ 𝐶 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ ((𝑅‘𝐶) ≠ (𝑅‘𝐹) ∧ (𝐹 ≠ ( I ↾ 𝐵) ∧ 𝐶 ≠ ( I ↾ 𝐵)) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊))) → (𝑁‘𝑃) = ((𝑃 ∨ (𝑅‘𝐹)) ∧ ((𝑄‘𝑃) ∨ (𝑅‘(𝐹 ∘ ◡𝐶)))))
 
Theoremcdlemkj-2N 41904* Part of proof of Lemma K of [Crawley] p. 118. (Contributed by NM, 2-Jul-2013.) (New usage is discouraged.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑄 = (𝑆‘𝐶)    &   𝑌 = (℩𝑘 ∈ 𝑇 (𝑘‘𝑃) = ((𝑃 ∨ (𝑅‘𝐺)) ∧ ((𝑄‘𝑃) ∨ (𝑅‘(𝐺 ∘ ◡𝐶)))))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝑅‘𝐹) = (𝑅‘𝑁) ∧ 𝐺 ∈ 𝑇) ∧ (𝐹 ∈ 𝑇 ∧ 𝐶 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ (((𝑅‘𝐶) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝐶) ≠ (𝑅‘𝐺)) ∧ (𝐹 ≠ ( I ↾ 𝐵) ∧ 𝐺 ≠ ( I ↾ 𝐵) ∧ 𝐶 ≠ ( I ↾ 𝐵)) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊))) → 𝑌 ∈ 𝑇)
 
Theoremcdlemkuv-2N 41905* Part of proof of Lemma K of [Crawley] p. 118. Value of the sigma2 (p) function, given 𝑉. (Contributed by NM, 2-Jul-2013.) (New usage is discouraged.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑄 = (𝑆‘𝐶)    &   𝑉 = (𝑑 ∈ 𝑇 ↦ (℩𝑘 ∈ 𝑇 (𝑘‘𝑃) = ((𝑃 ∨ (𝑅‘𝑑)) ∧ ((𝑄‘𝑃) ∨ (𝑅‘(𝑑 ∘ ◡𝐶))))))    ⇒   (𝐺 ∈ 𝑇 → (𝑉‘𝐺) = (℩𝑘 ∈ 𝑇 (𝑘‘𝑃) = ((𝑃 ∨ (𝑅‘𝐺)) ∧ ((𝑄‘𝑃) ∨ (𝑅‘(𝐺 ∘ ◡𝐶))))))
 
Theoremcdlemkuel-2N 41906* Part of proof of Lemma K of [Crawley] p. 118. Conditions for the sigma2 (p) function to be a translation. TODO: combine cdlemkj 41885? (Contributed by NM, 2-Jul-2013.) (New usage is discouraged.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑄 = (𝑆‘𝐶)    &   𝑉 = (𝑑 ∈ 𝑇 ↦ (℩𝑘 ∈ 𝑇 (𝑘‘𝑃) = ((𝑃 ∨ (𝑅‘𝑑)) ∧ ((𝑄‘𝑃) ∨ (𝑅‘(𝑑 ∘ ◡𝐶))))))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝑅‘𝐹) = (𝑅‘𝑁) ∧ 𝐺 ∈ 𝑇) ∧ (𝐹 ∈ 𝑇 ∧ 𝐶 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ (((𝑅‘𝐶) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝐶) ≠ (𝑅‘𝐺)) ∧ (𝐹 ≠ ( I ↾ 𝐵) ∧ 𝐺 ≠ ( I ↾ 𝐵) ∧ 𝐶 ≠ ( I ↾ 𝐵)) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊))) → (𝑉‘𝐺) ∈ 𝑇)
 
Theoremcdlemkuv2-2 41907* Part of proof of Lemma K of [Crawley] p. 118. Line 16 on p. 119 for i = 2, where sigma2 (p) is 𝑉, f2 is 𝐶, and k2 is 𝑄. (Contributed by NM, 2-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑄 = (𝑆‘𝐶)    &   𝑉 = (𝑑 ∈ 𝑇 ↦ (℩𝑘 ∈ 𝑇 (𝑘‘𝑃) = ((𝑃 ∨ (𝑅‘𝑑)) ∧ ((𝑄‘𝑃) ∨ (𝑅‘(𝑑 ∘ ◡𝐶))))))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝑅‘𝐹) = (𝑅‘𝑁) ∧ 𝐺 ∈ 𝑇) ∧ (𝐹 ∈ 𝑇 ∧ 𝐶 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ (((𝑅‘𝐶) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝐶) ≠ (𝑅‘𝐺)) ∧ (𝐹 ≠ ( I ↾ 𝐵) ∧ 𝐺 ≠ ( I ↾ 𝐵) ∧ 𝐶 ≠ ( I ↾ 𝐵)) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊))) → ((𝑉‘𝐺)‘𝑃) = ((𝑃 ∨ (𝑅‘𝐺)) ∧ ((𝑄‘𝑃) ∨ (𝑅‘(𝐺 ∘ ◡𝐶)))))
 
Theoremcdlemk18-2N 41908* Part of proof of Lemma K of [Crawley] p. 118. Line 22 on p. 119. 𝑁, 𝑉, 𝑄, 𝐶 are k, sigma2 (p), k2, f2. (Contributed by NM, 2-Jul-2013.) (New usage is discouraged.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑄 = (𝑆‘𝐶)    &   𝑉 = (𝑑 ∈ 𝑇 ↦ (℩𝑘 ∈ 𝑇 (𝑘‘𝑃) = ((𝑃 ∨ (𝑅‘𝑑)) ∧ ((𝑄‘𝑃) ∨ (𝑅‘(𝑑 ∘ ◡𝐶))))))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ (𝐹 ∈ 𝑇 ∧ 𝐶 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ ((𝑅‘𝐶) ≠ (𝑅‘𝐹) ∧ (𝐹 ≠ ( I ↾ 𝐵) ∧ 𝐶 ≠ ( I ↾ 𝐵)) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊))) → (𝑁‘𝑃) = ((𝑉‘𝐹)‘𝑃))
 
Theoremcdlemk19-2N 41909* Part of proof of Lemma K of [Crawley] p. 118. Line 22 on p. 119. 𝑁, 𝑉, 𝑄, 𝐶 are k, sigma2 (p), k2, f2. (Contributed by NM, 2-Jul-2013.) (New usage is discouraged.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑄 = (𝑆‘𝐶)    &   𝑉 = (𝑑 ∈ 𝑇 ↦ (℩𝑘 ∈ 𝑇 (𝑘‘𝑃) = ((𝑃 ∨ (𝑅‘𝑑)) ∧ ((𝑄‘𝑃) ∨ (𝑅‘(𝑑 ∘ ◡𝐶))))))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ (𝐹 ∈ 𝑇 ∧ 𝐶 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ ((𝑅‘𝐶) ≠ (𝑅‘𝐹) ∧ (𝐹 ≠ ( I ↾ 𝐵) ∧ 𝐶 ≠ ( I ↾ 𝐵)) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊))) → (𝑉‘𝐹) = 𝑁)
 
Theoremcdlemk7u-2N 41910* Part of proof of Lemma K of [Crawley] p. 118. Line 5, p. 119 for the sigma2 case. (Contributed by NM, 5-Jul-2013.) (New usage is discouraged.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑄 = (𝑆‘𝐶)    &   𝑉 = (𝑑 ∈ 𝑇 ↦ (℩𝑘 ∈ 𝑇 (𝑘‘𝑃) = ((𝑃 ∨ (𝑅‘𝑑)) ∧ ((𝑄‘𝑃) ∨ (𝑅‘(𝑑 ∘ ◡𝐶))))))    &   𝑍 = (((𝐺‘𝑃) ∨ (𝑋‘𝑃)) ∧ ((𝑅‘(𝐺 ∘ ◡𝐶)) ∨ (𝑅‘(𝑋 ∘ ◡𝐶))))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ ((𝐹 ∈ 𝑇 ∧ 𝐶 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵)) ∧ (𝑋 ∈ 𝑇 ∧ 𝑋 ≠ ( I ↾ 𝐵))) ∧ (((𝑅‘𝐶) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝐺) ≠ (𝑅‘𝐶) ∧ (𝑅‘𝑋) ≠ (𝑅‘𝐶)) ∧ (𝐹 ≠ ( I ↾ 𝐵) ∧ 𝐶 ≠ ( I ↾ 𝐵)) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊))) → ((𝑉‘𝐺)‘𝑃) ≤ (((𝑉‘𝑋)‘𝑃) ∨ 𝑍))
 
Theoremcdlemk11u-2N 41911* Part of proof of Lemma K of [Crawley] p. 118. Line 17, p. 119, showing Eq. 3 (line 8, p. 119) for the sigma2 (𝑍) case. (Contributed by NM, 5-Jul-2013.) (New usage is discouraged.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑄 = (𝑆‘𝐶)    &   𝑉 = (𝑑 ∈ 𝑇 ↦ (℩𝑘 ∈ 𝑇 (𝑘‘𝑃) = ((𝑃 ∨ (𝑅‘𝑑)) ∧ ((𝑄‘𝑃) ∨ (𝑅‘(𝑑 ∘ ◡𝐶))))))    &   𝑍 = (((𝐺‘𝑃) ∨ (𝑋‘𝑃)) ∧ ((𝑅‘(𝐺 ∘ ◡𝐶)) ∨ (𝑅‘(𝑋 ∘ ◡𝐶))))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ ((𝐹 ∈ 𝑇 ∧ 𝐶 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵)) ∧ (𝑋 ∈ 𝑇 ∧ 𝑋 ≠ ( I ↾ 𝐵))) ∧ (((𝑅‘𝐶) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝐺) ≠ (𝑅‘𝐶) ∧ (𝑅‘𝑋) ≠ (𝑅‘𝐶)) ∧ (𝐹 ≠ ( I ↾ 𝐵) ∧ 𝐶 ≠ ( I ↾ 𝐵)) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊))) → ((𝑉‘𝐺)‘𝑃) ≤ (((𝑉‘𝑋)‘𝑃) ∨ (𝑅‘(𝑋 ∘ ◡𝐺))))
 
Theoremcdlemk12u-2N 41912* Part of proof of Lemma K of [Crawley] p. 118. Line 18, p. 119, showing Eq. 4 (line 10, p. 119) for the sigma2 (𝑉) case. (Contributed by NM, 5-Jul-2013.) (New usage is discouraged.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑄 = (𝑆‘𝐶)    &   𝑉 = (𝑑 ∈ 𝑇 ↦ (℩𝑘 ∈ 𝑇 (𝑘‘𝑃) = ((𝑃 ∨ (𝑅‘𝑑)) ∧ ((𝑄‘𝑃) ∨ (𝑅‘(𝑑 ∘ ◡𝐶))))))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ ((𝐹 ∈ 𝑇 ∧ 𝐶 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵)) ∧ (𝑋 ∈ 𝑇 ∧ 𝑋 ≠ ( I ↾ 𝐵))) ∧ (((𝑅‘𝐶) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝐺) ≠ (𝑅‘𝐶) ∧ (𝑅‘𝑋) ≠ (𝑅‘𝐶)) ∧ ((𝑅‘𝐺) ≠ (𝑅‘𝑋) ∧ 𝐹 ≠ ( I ↾ 𝐵) ∧ 𝐶 ≠ ( I ↾ 𝐵)) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊))) → ((𝑉‘𝐺)‘𝑃) = ((𝑃 ∨ (𝐺‘𝑃)) ∧ (((𝑉‘𝑋)‘𝑃) ∨ (𝑅‘(𝑋 ∘ ◡𝐺)))))
 
Theoremcdlemk21-2N 41913* Part of proof of Lemma K of [Crawley] p. 118. Lines 26-27, p. 119 for i=0 and j=2. (Contributed by NM, 5-Jul-2013.) (New usage is discouraged.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑄 = (𝑆‘𝐶)    &   𝑉 = (𝑑 ∈ 𝑇 ↦ (℩𝑘 ∈ 𝑇 (𝑘‘𝑃) = ((𝑃 ∨ (𝑅‘𝑑)) ∧ ((𝑄‘𝑃) ∨ (𝑅‘(𝑑 ∘ ◡𝐶))))))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ ((𝐹 ∈ 𝑇 ∧ 𝐶 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵))) ∧ (((𝑅‘𝐶) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝐺) ≠ (𝑅‘𝐶)) ∧ ((𝑅‘𝐺) ≠ (𝑅‘𝐹) ∧ 𝐹 ≠ ( I ↾ 𝐵) ∧ 𝐶 ≠ ( I ↾ 𝐵)) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊))) → ((𝑆‘𝐺)‘𝑃) = ((𝑉‘𝐺)‘𝑃))
 
Theoremcdlemk20-2N 41914* Part of proof of Lemma K of [Crawley] p. 118. Line 22, p. 119 for the i=2, j=1 case. Note typo on line 22: f should be fi. Our 𝐷, 𝐶, 𝑂, 𝑄, 𝑈, 𝑉 represent their f1, f2, k1, k2, sigma1, sigma2. (Contributed by NM, 5-Jul-2013.) (New usage is discouraged.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑄 = (𝑆‘𝐶)    &   𝑉 = (𝑑 ∈ 𝑇 ↦ (℩𝑘 ∈ 𝑇 (𝑘‘𝑃) = ((𝑃 ∨ (𝑅‘𝑑)) ∧ ((𝑄‘𝑃) ∨ (𝑅‘(𝑑 ∘ ◡𝐶))))))    &   𝑂 = (𝑆‘𝐷)    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ ((𝐹 ∈ 𝑇 ∧ 𝐶 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ (𝐷 ∈ 𝑇 ∧ 𝐷 ≠ ( I ↾ 𝐵)) ∧ (𝐶 ∈ 𝑇 ∧ 𝐶 ≠ ( I ↾ 𝐵))) ∧ (((𝑅‘𝐶) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝐷) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝐷) ≠ (𝑅‘𝐶)) ∧ (𝐹 ≠ ( I ↾ 𝐵) ∧ 𝐶 ≠ ( I ↾ 𝐵)) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊))) → ((𝑉‘𝐷)‘𝑃) = (𝑂‘𝑃))
 
Theoremcdlemk22 41915* Part of proof of Lemma K of [Crawley] p. 118. Lines 26-27, p. 119 for i=1 and j=2. (Contributed by NM, 5-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑄 = (𝑆‘𝐶)    &   𝑉 = (𝑑 ∈ 𝑇 ↦ (℩𝑘 ∈ 𝑇 (𝑘‘𝑃) = ((𝑃 ∨ (𝑅‘𝑑)) ∧ ((𝑄‘𝑃) ∨ (𝑅‘(𝑑 ∘ ◡𝐶))))))    &   𝑂 = (𝑆‘𝐷)    &   𝑈 = (𝑒 ∈ 𝑇 ↦ (℩𝑗 ∈ 𝑇 (𝑗‘𝑃) = ((𝑃 ∨ (𝑅‘𝑒)) ∧ ((𝑂‘𝑃) ∨ (𝑅‘(𝑒 ∘ ◡𝐷))))))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ 𝐹 ∈ 𝑇 ∧ 𝐷 ∈ 𝑇) ∧ ((𝑁 ∈ 𝑇 ∧ 𝐺 ∈ 𝑇 ∧ 𝐶 ∈ 𝑇) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ ((𝐹 ≠ ( I ↾ 𝐵) ∧ 𝐷 ≠ ( I ↾ 𝐵) ∧ 𝐺 ≠ ( I ↾ 𝐵)) ∧ (𝐶 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝐺) ≠ (𝑅‘𝐶) ∧ (𝑅‘𝐶) ≠ (𝑅‘𝐹)) ∧ ((𝑅‘𝐷) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝐺) ≠ (𝑅‘𝐷) ∧ (𝑅‘𝐶) ≠ (𝑅‘𝐷)))) → ((𝑈‘𝐺)‘𝑃) = ((𝑉‘𝐺)‘𝑃))
 
Theoremcdlemk30 41916* Part of proof of Lemma K of [Crawley] p. 118. TODO: fix comment. Part of attempt to simplify hypotheses. (Contributed by NM, 17-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ (𝐹 ∈ 𝑇 ∧ 𝑏 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ ((𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝐹 ≠ ( I ↾ 𝐵) ∧ 𝑏 ≠ ( I ↾ 𝐵)) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊))) → ((𝑆‘𝑏)‘𝑃) = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹)))))
 
Theoremcdlemkuu 41917* Convert between function and operation forms of 𝑌. TODO: Use operation form everywhere. (Contributed by NM, 6-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑌 = (𝑑 ∈ 𝑇, 𝑒 ∈ 𝑇 ↦ (℩𝑗 ∈ 𝑇 (𝑗‘𝑃) = ((𝑃 ∨ (𝑅‘𝑒)) ∧ (((𝑆‘𝑑)‘𝑃) ∨ (𝑅‘(𝑒 ∘ ◡𝑑))))))    &   𝑄 = (𝑆‘𝐷)    &   𝑍 = (𝑒 ∈ 𝑇 ↦ (℩𝑗 ∈ 𝑇 (𝑗‘𝑃) = ((𝑃 ∨ (𝑅‘𝑒)) ∧ ((𝑄‘𝑃) ∨ (𝑅‘(𝑒 ∘ ◡𝐷))))))    ⇒   ((𝐷 ∈ 𝑇 ∧ 𝐺 ∈ 𝑇) → (𝐷𝑌𝐺) = (𝑍‘𝐺))
 
Theoremcdlemk31 41918* Part of proof of Lemma K of [Crawley] p. 118. TODO: fix comment. Part of attempt to simplify hypotheses. (Contributed by NM, 17-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑌 = (𝑑 ∈ 𝑇, 𝑒 ∈ 𝑇 ↦ (℩𝑗 ∈ 𝑇 (𝑗‘𝑃) = ((𝑃 ∨ (𝑅‘𝑒)) ∧ (((𝑆‘𝑑)‘𝑃) ∨ (𝑅‘(𝑒 ∘ ◡𝑑))))))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ ((𝐹 ∈ 𝑇 ∧ 𝑏 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ 𝐺 ∈ 𝑇) ∧ (((𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐺)) ∧ (𝐹 ≠ ( I ↾ 𝐵) ∧ 𝑏 ≠ ( I ↾ 𝐵) ∧ 𝐺 ≠ ( I ↾ 𝐵)) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊))) → ((𝑏𝑌𝐺)‘𝑃) = ((𝑃 ∨ (𝑅‘𝐺)) ∧ (((𝑆‘𝑏)‘𝑃) ∨ (𝑅‘(𝐺 ∘ ◡𝑏)))))
 
Theoremcdlemk32 41919* Part of proof of Lemma K of [Crawley] p. 118. TODO: fix comment. Part of attempt to simplify hypotheses. (Contributed by NM, 17-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑌 = (𝑑 ∈ 𝑇, 𝑒 ∈ 𝑇 ↦ (℩𝑗 ∈ 𝑇 (𝑗‘𝑃) = ((𝑃 ∨ (𝑅‘𝑒)) ∧ (((𝑆‘𝑑)‘𝑃) ∨ (𝑅‘(𝑒 ∘ ◡𝑑))))))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ ((𝐹 ∈ 𝑇 ∧ 𝑏 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ 𝐺 ∈ 𝑇) ∧ (((𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐺)) ∧ (𝐹 ≠ ( I ↾ 𝐵) ∧ 𝑏 ≠ ( I ↾ 𝐵) ∧ 𝐺 ≠ ( I ↾ 𝐵)) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊))) → ((𝑏𝑌𝐺)‘𝑃) = ((𝑃 ∨ (𝑅‘𝐺)) ∧ (((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹)))) ∨ (𝑅‘(𝐺 ∘ ◡𝑏)))))
 
Theoremcdlemkuel-3 41920* Part of proof of Lemma K of [Crawley] p. 118. Conditions for the sigma2 (p) function to be a translation. TODO: combine cdlemkj 41885? (Contributed by NM, 11-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑌 = (𝑑 ∈ 𝑇, 𝑒 ∈ 𝑇 ↦ (℩𝑗 ∈ 𝑇 (𝑗‘𝑃) = ((𝑃 ∨ (𝑅‘𝑒)) ∧ (((𝑆‘𝑑)‘𝑃) ∨ (𝑅‘(𝑒 ∘ ◡𝑑))))))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝑅‘𝐹) = (𝑅‘𝑁) ∧ 𝐺 ∈ 𝑇) ∧ (𝐹 ∈ 𝑇 ∧ 𝐷 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ (((𝑅‘𝐷) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝐷) ≠ (𝑅‘𝐺)) ∧ (𝐹 ≠ ( I ↾ 𝐵) ∧ 𝐺 ≠ ( I ↾ 𝐵) ∧ 𝐷 ≠ ( I ↾ 𝐵)) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊))) → (𝐷𝑌𝐺) ∈ 𝑇)
 
Theoremcdlemkuv2-3N 41921* Part of proof of Lemma K of [Crawley] p. 118. Line 16 on p. 119 for i = 1, where sigma2 (p) is 𝑌, f1 is 𝐷, and k1 is 𝑂. (Contributed by NM, 6-Jul-2013.) (New usage is discouraged.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑌 = (𝑑 ∈ 𝑇, 𝑒 ∈ 𝑇 ↦ (℩𝑗 ∈ 𝑇 (𝑗‘𝑃) = ((𝑃 ∨ (𝑅‘𝑒)) ∧ (((𝑆‘𝑑)‘𝑃) ∨ (𝑅‘(𝑒 ∘ ◡𝑑))))))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝑅‘𝐹) = (𝑅‘𝑁) ∧ 𝐺 ∈ 𝑇) ∧ (𝐹 ∈ 𝑇 ∧ 𝐷 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ (((𝑅‘𝐷) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝐷) ≠ (𝑅‘𝐺)) ∧ (𝐹 ≠ ( I ↾ 𝐵) ∧ 𝐺 ≠ ( I ↾ 𝐵) ∧ 𝐷 ≠ ( I ↾ 𝐵)) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊))) → ((𝐷𝑌𝐺)‘𝑃) = ((𝑃 ∨ (𝑅‘𝐺)) ∧ (((𝑆‘𝐷)‘𝑃) ∨ (𝑅‘(𝐺 ∘ ◡𝐷)))))
 
Theoremcdlemk18-3N 41922* Part of proof of Lemma K of [Crawley] p. 118. Line 22 on p. 119. 𝑁, 𝑌, 𝑂, 𝐷 are k, sigma2 (p), k1, f1. (Contributed by NM, 7-Jul-2013.) (New usage is discouraged.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑌 = (𝑑 ∈ 𝑇, 𝑒 ∈ 𝑇 ↦ (℩𝑗 ∈ 𝑇 (𝑗‘𝑃) = ((𝑃 ∨ (𝑅‘𝑒)) ∧ (((𝑆‘𝑑)‘𝑃) ∨ (𝑅‘(𝑒 ∘ ◡𝑑))))))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ (𝐹 ∈ 𝑇 ∧ 𝐷 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ ((𝑅‘𝐷) ≠ (𝑅‘𝐹) ∧ (𝐹 ≠ ( I ↾ 𝐵) ∧ 𝐷 ≠ ( I ↾ 𝐵)) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊))) → ((𝐷𝑌𝐹)‘𝑃) = (𝑁‘𝑃))
 
Theoremcdlemk22-3 41923* Part of proof of Lemma K of [Crawley] p. 118. Lines 26-27, p. 119 for i=1 and j=2. (Contributed by NM, 7-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑌 = (𝑑 ∈ 𝑇, 𝑒 ∈ 𝑇 ↦ (℩𝑗 ∈ 𝑇 (𝑗‘𝑃) = ((𝑃 ∨ (𝑅‘𝑒)) ∧ (((𝑆‘𝑑)‘𝑃) ∨ (𝑅‘(𝑒 ∘ ◡𝑑))))))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ 𝐹 ∈ 𝑇 ∧ 𝐷 ∈ 𝑇) ∧ ((𝑁 ∈ 𝑇 ∧ 𝐺 ∈ 𝑇 ∧ 𝐶 ∈ 𝑇) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ ((𝐹 ≠ ( I ↾ 𝐵) ∧ 𝐷 ≠ ( I ↾ 𝐵) ∧ 𝐺 ≠ ( I ↾ 𝐵)) ∧ (𝐶 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝐺) ≠ (𝑅‘𝐶) ∧ (𝑅‘𝐶) ≠ (𝑅‘𝐹)) ∧ ((𝑅‘𝐷) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝐺) ≠ (𝑅‘𝐷) ∧ (𝑅‘𝐶) ≠ (𝑅‘𝐷)))) → ((𝐷𝑌𝐺)‘𝑃) = ((𝐶𝑌𝐺)‘𝑃))
 
Theoremcdlemk23-3 41924* Part of proof of Lemma K of [Crawley] p. 118. Eliminate the (𝑅‘𝐶) ≠ (𝑅‘𝐷) requirement from cdlemk22-3 41923. (Contributed by NM, 7-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑌 = (𝑑 ∈ 𝑇, 𝑒 ∈ 𝑇 ↦ (℩𝑗 ∈ 𝑇 (𝑗‘𝑃) = ((𝑃 ∨ (𝑅‘𝑒)) ∧ (((𝑆‘𝑑)‘𝑃) ∨ (𝑅‘(𝑒 ∘ ◡𝑑))))))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝐷 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ (𝐺 ∈ 𝑇 ∧ 𝐶 ∈ 𝑇 ∧ 𝑥 ∈ 𝑇)) ∧ ((𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ ((𝑅‘𝐹) = (𝑅‘𝑁) ∧ 𝐹 ≠ ( I ↾ 𝐵) ∧ 𝐷 ≠ ( I ↾ 𝐵)) ∧ (𝐺 ≠ ( I ↾ 𝐵) ∧ 𝐶 ≠ ( I ↾ 𝐵) ∧ 𝑥 ≠ ( I ↾ 𝐵))) ∧ (((𝑅‘𝐺) ≠ (𝑅‘𝐶) ∧ (𝑅‘𝐶) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝐷) ≠ (𝑅‘𝐹)) ∧ ((𝑅‘𝐺) ≠ (𝑅‘𝐷) ∧ (𝑅‘𝑥) ≠ (𝑅‘𝐶)) ∧ ((𝑅‘𝑥) ≠ (𝑅‘𝐷) ∧ (𝑅‘𝑥) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝐺) ≠ (𝑅‘𝑥)))) → ((𝐷𝑌𝐺)‘𝑃) = ((𝐶𝑌𝐺)‘𝑃))
 
Theoremcdlemk24-3 41925* Part of proof of Lemma K of [Crawley] p. 118. Eliminate the (𝑅‘𝑥) ≠ (𝑅‘𝐶) requirement from cdlemk23-3 41924 using (𝑅‘𝐶) = (𝑅‘𝐷). (Contributed by NM, 7-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑌 = (𝑑 ∈ 𝑇, 𝑒 ∈ 𝑇 ↦ (℩𝑗 ∈ 𝑇 (𝑗‘𝑃) = ((𝑃 ∨ (𝑅‘𝑒)) ∧ (((𝑆‘𝑑)‘𝑃) ∨ (𝑅‘(𝑒 ∘ ◡𝑑))))))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝐷 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ (𝐺 ∈ 𝑇 ∧ 𝐶 ∈ 𝑇 ∧ 𝑥 ∈ 𝑇)) ∧ ((𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ ((𝑅‘𝐹) = (𝑅‘𝑁) ∧ 𝐹 ≠ ( I ↾ 𝐵) ∧ 𝐷 ≠ ( I ↾ 𝐵)) ∧ (𝐺 ≠ ( I ↾ 𝐵) ∧ 𝐶 ≠ ( I ↾ 𝐵) ∧ 𝑥 ≠ ( I ↾ 𝐵))) ∧ (((𝑅‘𝐺) ≠ (𝑅‘𝐶) ∧ (𝑅‘𝐶) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝐷) ≠ (𝑅‘𝐹)) ∧ ((𝑅‘𝐺) ≠ (𝑅‘𝐷) ∧ (𝑅‘𝐶) = (𝑅‘𝐷)) ∧ ((𝑅‘𝑥) ≠ (𝑅‘𝐷) ∧ (𝑅‘𝑥) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝐺) ≠ (𝑅‘𝑥)))) → ((𝐷𝑌𝐺)‘𝑃) = ((𝐶𝑌𝐺)‘𝑃))
 
Theoremcdlemk25-3 41926* Part of proof of Lemma K of [Crawley] p. 118. Eliminate the (𝑅‘𝐶) = (𝑅‘𝐷) requirement from cdlemk24-3 41925. (Contributed by NM, 7-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑌 = (𝑑 ∈ 𝑇, 𝑒 ∈ 𝑇 ↦ (℩𝑗 ∈ 𝑇 (𝑗‘𝑃) = ((𝑃 ∨ (𝑅‘𝑒)) ∧ (((𝑆‘𝑑)‘𝑃) ∨ (𝑅‘(𝑒 ∘ ◡𝑑))))))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝐷 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ (𝐺 ∈ 𝑇 ∧ 𝐶 ∈ 𝑇 ∧ 𝑥 ∈ 𝑇)) ∧ ((𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ ((𝑅‘𝐹) = (𝑅‘𝑁) ∧ 𝐹 ≠ ( I ↾ 𝐵) ∧ 𝐷 ≠ ( I ↾ 𝐵)) ∧ (𝐺 ≠ ( I ↾ 𝐵) ∧ 𝐶 ≠ ( I ↾ 𝐵) ∧ 𝑥 ≠ ( I ↾ 𝐵))) ∧ (((𝑅‘𝐺) ≠ (𝑅‘𝐶) ∧ (𝑅‘𝐶) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝐷) ≠ (𝑅‘𝐹)) ∧ (𝑅‘𝐺) ≠ (𝑅‘𝐷) ∧ ((𝑅‘𝑥) ≠ (𝑅‘𝐷) ∧ (𝑅‘𝑥) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝐺) ≠ (𝑅‘𝑥)))) → ((𝐷𝑌𝐺)‘𝑃) = ((𝐶𝑌𝐺)‘𝑃))
 
Theoremcdlemk26b-3 41927* Part of proof of Lemma K of [Crawley] p. 118. (Contributed by NM, 14-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑌 = (𝑑 ∈ 𝑇, 𝑒 ∈ 𝑇 ↦ (℩𝑗 ∈ 𝑇 (𝑗‘𝑃) = ((𝑃 ∨ (𝑅‘𝑒)) ∧ (((𝑆‘𝑑)‘𝑃) ∨ (𝑅‘(𝑒 ∘ ◡𝑑))))))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵) ∧ 𝑁 ∈ 𝑇) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝐹) = (𝑅‘𝑁))) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊)) → ∃𝑥 ∈ 𝑇 ((𝑥 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑥) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑥) ≠ (𝑅‘𝐺)) ∧ (𝑥𝑌𝐺) ∈ 𝑇))
 
Theoremcdlemk26-3 41928* Part of proof of Lemma K of [Crawley] p. 118. Eliminate the 𝑥 requirements from cdlemk25-3 41926. (Contributed by NM, 10-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑌 = (𝑑 ∈ 𝑇, 𝑒 ∈ 𝑇 ↦ (℩𝑗 ∈ 𝑇 (𝑗‘𝑃) = ((𝑃 ∨ (𝑅‘𝑒)) ∧ (((𝑆‘𝑑)‘𝑃) ∨ (𝑅‘(𝑒 ∘ ◡𝑑))))))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝐷 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ (𝐺 ∈ 𝑇 ∧ 𝐶 ∈ 𝑇)) ∧ ((𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ ((𝑅‘𝐹) = (𝑅‘𝑁) ∧ 𝐹 ≠ ( I ↾ 𝐵) ∧ 𝐷 ≠ ( I ↾ 𝐵)) ∧ (𝐺 ≠ ( I ↾ 𝐵) ∧ 𝐶 ≠ ( I ↾ 𝐵))) ∧ (((𝑅‘𝐺) ≠ (𝑅‘𝐶) ∧ (𝑅‘𝐶) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝐷) ≠ (𝑅‘𝐹)) ∧ (𝑅‘𝐺) ≠ (𝑅‘𝐷))) → ((𝐷𝑌𝐺)‘𝑃) = ((𝐶𝑌𝐺)‘𝑃))
 
Theoremcdlemk27-3 41929* Part of proof of Lemma K of [Crawley] p. 118. Eliminate the 𝑃 from the conclusion of cdlemk25-3 41926. (Contributed by NM, 10-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑌 = (𝑑 ∈ 𝑇, 𝑒 ∈ 𝑇 ↦ (℩𝑗 ∈ 𝑇 (𝑗‘𝑃) = ((𝑃 ∨ (𝑅‘𝑒)) ∧ (((𝑆‘𝑑)‘𝑃) ∨ (𝑅‘(𝑒 ∘ ◡𝑑))))))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝐷 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ (𝐺 ∈ 𝑇 ∧ 𝐶 ∈ 𝑇)) ∧ ((𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ ((𝑅‘𝐹) = (𝑅‘𝑁) ∧ 𝐹 ≠ ( I ↾ 𝐵) ∧ 𝐷 ≠ ( I ↾ 𝐵)) ∧ (𝐺 ≠ ( I ↾ 𝐵) ∧ 𝐶 ≠ ( I ↾ 𝐵))) ∧ (((𝑅‘𝐺) ≠ (𝑅‘𝐶) ∧ (𝑅‘𝐶) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝐷) ≠ (𝑅‘𝐹)) ∧ (𝑅‘𝐺) ≠ (𝑅‘𝐷))) → (𝐷𝑌𝐺) = (𝐶𝑌𝐺))
 
Theoremcdlemk28-3 41930* Part of proof of Lemma K of [Crawley] p. 118. (Contributed by NM, 14-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑌 = (𝑑 ∈ 𝑇, 𝑒 ∈ 𝑇 ↦ (℩𝑗 ∈ 𝑇 (𝑗‘𝑃) = ((𝑃 ∨ (𝑅‘𝑒)) ∧ (((𝑆‘𝑑)‘𝑃) ∨ (𝑅‘(𝑒 ∘ ◡𝑑))))))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ ((𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵)) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵)) ∧ 𝑁 ∈ 𝑇) ∧ ((𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁))) → ∃𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐺)) → 𝑧 = (𝑏𝑌𝐺)))
 
Theoremcdlemk33N 41931* Part of proof of Lemma K of [Crawley] p. 118. TODO: fix comment. Part of attempt to simplify hypotheses. TODO: not needed, is embodied in cdlemk34 41932. (Contributed by NM, 18-Jul-2013.) (New usage is discouraged.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑌 = (𝑑 ∈ 𝑇, 𝑒 ∈ 𝑇 ↦ (℩𝑗 ∈ 𝑇 (𝑗‘𝑃) = ((𝑃 ∨ (𝑅‘𝑒)) ∧ (((𝑆‘𝑑)‘𝑃) ∨ (𝑅‘(𝑒 ∘ ◡𝑑))))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐺)) → 𝑧 = (𝑏𝑌𝐺)))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ ((𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵) ∧ 𝑁 ∈ 𝑇) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵)) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊))) → 𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐺)) → (𝑧‘𝑃) = ((𝑏𝑌𝐺)‘𝑃))))
 
Theoremcdlemk34 41932* Part of proof of Lemma K of [Crawley] p. 118. TODO: fix comment. Part of attempt to simplify hypotheses. (Contributed by NM, 18-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑌 = (𝑑 ∈ 𝑇, 𝑒 ∈ 𝑇 ↦ (℩𝑗 ∈ 𝑇 (𝑗‘𝑃) = ((𝑃 ∨ (𝑅‘𝑒)) ∧ (((𝑆‘𝑑)‘𝑃) ∨ (𝑅‘(𝑒 ∘ ◡𝑑))))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐺)) → 𝑧 = (𝑏𝑌𝐺)))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ ((𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵)) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵)) ∧ 𝑁 ∈ 𝑇) ∧ ((𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁))) → 𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐺)) → (𝑧‘𝑃) = ((𝑃 ∨ (𝑅‘𝐺)) ∧ (((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹)))) ∨ (𝑅‘(𝐺 ∘ ◡𝑏)))))))
 
Theoremcdlemk29-3 41933* Part of proof of Lemma K of [Crawley] p. 118. TODO: fix comment. (Contributed by NM, 14-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑌 = (𝑑 ∈ 𝑇, 𝑒 ∈ 𝑇 ↦ (℩𝑗 ∈ 𝑇 (𝑗‘𝑃) = ((𝑃 ∨ (𝑅‘𝑒)) ∧ (((𝑆‘𝑑)‘𝑃) ∨ (𝑅‘(𝑒 ∘ ◡𝑑))))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐺)) → 𝑧 = (𝑏𝑌𝐺)))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ ((𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵)) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵)) ∧ 𝑁 ∈ 𝑇) ∧ ((𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁))) → 𝑋 ∈ 𝑇)
 
Theoremcdlemk35 41934* Part of proof of Lemma K of [Crawley] p. 118. cdlemk29-3 41933 with shorter hypotheses. (Contributed by NM, 18-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝐺)) ∧ (𝑍 ∨ (𝑅‘(𝐺 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐺)) → (𝑧‘𝑃) = 𝑌))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ ((𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵)) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵)) ∧ 𝑁 ∈ 𝑇) ∧ ((𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁))) → 𝑋 ∈ 𝑇)
 
Theoremcdlemk36 41935* Part of proof of Lemma K of [Crawley] p. 118. TODO: fix comment. (Contributed by NM, 18-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝐺)) ∧ (𝑍 ∨ (𝑅‘(𝐺 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐺)) → (𝑧‘𝑃) = 𝑌))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵)) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵))) ∧ (𝑁 ∈ 𝑇 ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ (𝑏 ∈ 𝑇 ∧ (𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐺)))) → (𝑋‘𝑃) = 𝑌)
 
Theoremcdlemk37 41936* Part of proof of Lemma K of [Crawley] p. 118. TODO: fix comment. (Contributed by NM, 18-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝐺)) ∧ (𝑍 ∨ (𝑅‘(𝐺 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐺)) → (𝑧‘𝑃) = 𝑌))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵)) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵))) ∧ (𝑁 ∈ 𝑇 ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ (𝑏 ∈ 𝑇 ∧ (𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐺)))) → (𝑋‘𝑃) ≤ (𝑃 ∨ (𝑅‘𝐺)))
 
Theoremcdlemk38 41937* Part of proof of Lemma K of [Crawley] p. 118. Line 31, p. 119. TODO: derive more directly with r19.23 3260? (Contributed by NM, 19-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝐺)) ∧ (𝑍 ∨ (𝑅‘(𝐺 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐺)) → (𝑧‘𝑃) = 𝑌))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ ((𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵)) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵)) ∧ 𝑁 ∈ 𝑇) ∧ ((𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁))) → (𝑋‘𝑃) ≤ (𝑃 ∨ (𝑅‘𝐺)))
 
Theoremcdlemk39 41938* Part of proof of Lemma K of [Crawley] p. 118. Line 31, p. 119. Trace-preserving property of tau, represented by 𝑋. (Contributed by NM, 19-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝐺)) ∧ (𝑍 ∨ (𝑅‘(𝐺 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐺)) → (𝑧‘𝑃) = 𝑌))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ ((𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵)) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵)) ∧ 𝑁 ∈ 𝑇) ∧ ((𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁))) → (𝑅‘𝑋) ≤ (𝑅‘𝐺))
 
Theoremcdlemk40 41939* TODO: fix comment. (Contributed by NM, 31-Jul-2013.)
𝑋 = (℩𝑧 ∈ 𝑇 𝜑)    &   𝑈 = (𝑔 ∈ 𝑇 ↦ if(𝐹 = 𝑁, 𝑔, 𝑋))    ⇒   (𝐺 ∈ 𝑇 → (𝑈‘𝐺) = if(𝐹 = 𝑁, 𝐺, ⦋𝐺 / 𝑔⦌𝑋))
 
Theoremcdlemk40t 41940* TODO: fix comment. (Contributed by NM, 31-Jul-2013.)
𝑋 = (℩𝑧 ∈ 𝑇 𝜑)    &   𝑈 = (𝑔 ∈ 𝑇 ↦ if(𝐹 = 𝑁, 𝑔, 𝑋))    ⇒   ((𝐹 = 𝑁 ∧ 𝐺 ∈ 𝑇) → (𝑈‘𝐺) = 𝐺)
 
Theoremcdlemk40f 41941* TODO: fix comment. (Contributed by NM, 31-Jul-2013.)
𝑋 = (℩𝑧 ∈ 𝑇 𝜑)    &   𝑈 = (𝑔 ∈ 𝑇 ↦ if(𝐹 = 𝑁, 𝑔, 𝑋))    ⇒   ((𝐹 ≠ 𝑁 ∧ 𝐺 ∈ 𝑇) → (𝑈‘𝐺) = ⦋𝐺 / 𝑔⦌𝑋)
 
Theoremcdlemk41 41942* Part of proof of Lemma K of [Crawley] p. 118. TODO: fix comment. (Contributed by NM, 19-Jul-2013.)
𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    ⇒   (𝐺 ∈ 𝑇 → ⦋𝐺 / 𝑔⦌𝑌 = ((𝑃 ∨ (𝑅‘𝐺)) ∧ (𝑍 ∨ (𝑅‘(𝐺 ∘ ◡𝑏)))))
 
Theoremcdlemkfid1N 41943 Lemma for cdlemkfid3N 41947. (Contributed by NM, 29-Jul-2013.) (New usage is discouraged.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵) ∧ 𝐺 ∈ 𝑇) ∧ ((𝑅‘𝐺) ≠ (𝑅‘𝐹) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊))) → ((𝑃 ∨ (𝑅‘𝐺)) ∧ ((𝐹‘𝑃) ∨ (𝑅‘(𝐺 ∘ ◡𝐹)))) = (𝐺‘𝑃))
 
Theoremcdlemkid1 41944 Lemma for cdlemkid 41958. (Contributed by NM, 24-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇 ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ ((𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑏 ∈ 𝑇 ∧ 𝑏 ≠ ( I ↾ 𝐵)))) → (𝑍 ∨ (𝑅‘𝑏)) = (𝑃 ∨ (𝑅‘𝑏)))
 
Theoremcdlemkfid2N 41945 Lemma for cdlemkfid3N 41947. (Contributed by NM, 29-Jul-2013.) (New usage is discouraged.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ 𝐹 = 𝑁) ∧ (𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵) ∧ 𝑏 ∈ 𝑇) ∧ ((𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊))) → 𝑍 = (𝑏‘𝑃))
 
Theoremcdlemkid2 41946* Lemma for cdlemkid 41958. (Contributed by NM, 24-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇 ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ ((𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ 𝐺 = ( I ↾ 𝐵) ∧ (𝑏 ∈ 𝑇 ∧ 𝑏 ≠ ( I ↾ 𝐵)))) → ⦋𝐺 / 𝑔⦌𝑌 = 𝑃)
 
Theoremcdlemkfid3N 41947* TODO: is this useful or should it be deleted? (Contributed by NM, 29-Jul-2013.) (New usage is discouraged.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ 𝐹 = 𝑁) ∧ ((𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵)) ∧ 𝐺 ∈ 𝑇 ∧ (𝑏 ∈ 𝑇 ∧ 𝑏 ≠ ( I ↾ 𝐵))) ∧ ((𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐺) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊))) → ⦋𝐺 / 𝑔⦌𝑌 = (𝐺‘𝑃))
 
Theoremcdlemky 41948* Part of proof of Lemma K of [Crawley] p. 118. TODO: clean up (𝑏𝑌𝐺) stuff. 𝑉 represents 𝑌 in cdlemk31 41918. (Contributed by NM, 21-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑉 = (𝑑 ∈ 𝑇, 𝑒 ∈ 𝑇 ↦ (℩𝑗 ∈ 𝑇 (𝑗‘𝑃) = ((𝑃 ∨ (𝑅‘𝑒)) ∧ (((𝑆‘𝑑)‘𝑃) ∨ (𝑅‘(𝑒 ∘ ◡𝑑))))))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵)) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵))) ∧ (𝑁 ∈ 𝑇 ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ (𝑏 ∈ 𝑇 ∧ (𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐺)))) → ⦋𝐺 / 𝑔⦌𝑌 = ((𝑏𝑉𝐺)‘𝑃))
 
Theoremcdlemkyu 41949* Convert between function and explicit forms. 𝐶 represents 𝑍 in cdlemkuu 41917. TODO: Clean all this up. (Contributed by NM, 21-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑉 = (𝑑 ∈ 𝑇, 𝑒 ∈ 𝑇 ↦ (℩𝑗 ∈ 𝑇 (𝑗‘𝑃) = ((𝑃 ∨ (𝑅‘𝑒)) ∧ (((𝑆‘𝑑)‘𝑃) ∨ (𝑅‘(𝑒 ∘ ◡𝑑))))))    &   𝑄 = (𝑆‘𝑏)    &   𝐶 = (𝑒 ∈ 𝑇 ↦ (℩𝑗 ∈ 𝑇 (𝑗‘𝑃) = ((𝑃 ∨ (𝑅‘𝑒)) ∧ ((𝑄‘𝑃) ∨ (𝑅‘(𝑒 ∘ ◡𝑏))))))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵)) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵))) ∧ (𝑁 ∈ 𝑇 ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ (𝑏 ∈ 𝑇 ∧ (𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐺)))) → ⦋𝐺 / 𝑔⦌𝑌 = ((𝐶‘𝐺)‘𝑃))
 
Theoremcdlemkyuu 41950* cdlemkyu 41949 with some hypotheses eliminated. TODO: Clean all this up. (Contributed by NM, 21-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝐶 = (𝑒 ∈ 𝑇 ↦ (℩𝑗 ∈ 𝑇 (𝑗‘𝑃) = ((𝑃 ∨ (𝑅‘𝑒)) ∧ (((𝑆‘𝑏)‘𝑃) ∨ (𝑅‘(𝑒 ∘ ◡𝑏))))))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵)) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵))) ∧ (𝑁 ∈ 𝑇 ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ (𝑏 ∈ 𝑇 ∧ (𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐺)))) → ⦋𝐺 / 𝑔⦌𝑌 = ((𝐶‘𝐺)‘𝑃))
 
Theoremcdlemk11ta 41951* Part of proof of Lemma K of [Crawley] p. 118. Lemma for Eq. 5, p. 119. 𝐺, 𝐼 stand for g, h. TODO: fix comment. (Contributed by NM, 21-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝐶 = (𝑒 ∈ 𝑇 ↦ (℩𝑗 ∈ 𝑇 (𝑗‘𝑃) = ((𝑃 ∨ (𝑅‘𝑒)) ∧ (((𝑆‘𝑏)‘𝑃) ∨ (𝑅‘(𝑒 ∘ ◡𝑏))))))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵)) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵))) ∧ (𝑁 ∈ 𝑇 ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ (𝑏 ∈ 𝑇 ∧ (𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐺)) ∧ (𝐼 ∈ 𝑇 ∧ 𝐼 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐼)))) → ⦋𝐺 / 𝑔⦌𝑌 ≤ (⦋𝐼 / 𝑔⦌𝑌 ∨ (𝑅‘(𝐼 ∘ ◡𝐺))))
 
Theoremcdlemk19ylem 41952* Lemma for cdlemk19y 41954. (Contributed by NM, 30-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝐶 = (𝑒 ∈ 𝑇 ↦ (℩𝑗 ∈ 𝑇 (𝑗‘𝑃) = ((𝑃 ∨ (𝑅‘𝑒)) ∧ (((𝑆‘𝑏)‘𝑃) ∨ (𝑅‘(𝑒 ∘ ◡𝑏))))))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵))) ∧ (𝑁 ∈ 𝑇 ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ (𝑏 ∈ 𝑇 ∧ (𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹)))) → ⦋𝐹 / 𝑔⦌𝑌 = (𝑁‘𝑃))
 
Theoremcdlemk11tb 41953* Part of proof of Lemma K of [Crawley] p. 118. Lemma for Eq. 5, p. 119. 𝐺, 𝐼 stand for g, h. cdlemk11ta 41951 with hypotheses removed. TODO: Can this be proved directly with no quantification? (Contributed by NM, 21-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵)) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵))) ∧ (𝑁 ∈ 𝑇 ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ (𝑏 ∈ 𝑇 ∧ (𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐺)) ∧ (𝐼 ∈ 𝑇 ∧ 𝐼 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐼)))) → ⦋𝐺 / 𝑔⦌𝑌 ≤ (⦋𝐼 / 𝑔⦌𝑌 ∨ (𝑅‘(𝐼 ∘ ◡𝐺))))
 
Theoremcdlemk19y 41954* cdlemk19 41891 with simpler hypotheses. TODO: Clean all this up. (Contributed by NM, 30-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵))) ∧ (𝑁 ∈ 𝑇 ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ (𝑏 ∈ 𝑇 ∧ (𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹)))) → ⦋𝐹 / 𝑔⦌𝑌 = (𝑁‘𝑃))
 
Theoremcdlemkid3N 41955* Lemma for cdlemkid 41958. (Contributed by NM, 25-Jul-2013.) (New usage is discouraged.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇 ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ ((𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ 𝐺 = ( I ↾ 𝐵))) → ⦋𝐺 / 𝑔⦌𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐺)) → (𝑧‘𝑃) = 𝑃)))
 
Theoremcdlemkid4 41956* Lemma for cdlemkid 41958. (Contributed by NM, 25-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇 ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ ((𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ 𝐺 = ( I ↾ 𝐵))) → ⦋𝐺 / 𝑔⦌𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐺)) → 𝑧 = ( I ↾ 𝐵))))
 
Theoremcdlemkid5 41957* Lemma for cdlemkid 41958. (Contributed by NM, 25-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇 ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ ((𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ 𝐺 = ( I ↾ 𝐵))) → ⦋𝐺 / 𝑔⦌𝑋 ∈ 𝑇)
 
Theoremcdlemkid 41958* The value of the tau function (in Lemma K of [Crawley] p. 118) on the identity relation. (Contributed by NM, 25-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇 ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ ((𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ 𝐺 = ( I ↾ 𝐵))) → ⦋𝐺 / 𝑔⦌𝑋 = ( I ↾ 𝐵))
 
Theoremcdlemk35s 41959* Substitution version of cdlemk35 41934. (Contributed by NM, 22-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ ((𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵)) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵)) ∧ 𝑁 ∈ 𝑇) ∧ ((𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁))) → ⦋𝐺 / 𝑔⦌𝑋 ∈ 𝑇)
 
Theoremcdlemk35s-id 41960* Substitution version of cdlemk35 41934. (Contributed by NM, 26-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ ((𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵)) ∧ 𝐺 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ ((𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁))) → ⦋𝐺 / 𝑔⦌𝑋 ∈ 𝑇)
 
Theoremcdlemk39s 41961* Substitution version of cdlemk39 41938. TODO: Can any commonality with cdlemk35s 41959 be exploited? (Contributed by NM, 23-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ ((𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵)) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵)) ∧ 𝑁 ∈ 𝑇) ∧ ((𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁))) → (𝑅‘⦋𝐺 / 𝑔⦌𝑋) ≤ (𝑅‘𝐺))
 
Theoremcdlemk39s-id 41962* Substitution version of cdlemk39 41938 with non-identity requirement on 𝐺 removed. TODO: Can any commonality with cdlemk35s 41959 be exploited? (Contributed by NM, 26-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ ((𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵)) ∧ 𝐺 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ ((𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁))) → (𝑅‘⦋𝐺 / 𝑔⦌𝑋) ≤ (𝑅‘𝐺))
 
Theoremcdlemk42 41963* Part of proof of Lemma K of [Crawley] p. 118. TODO: fix comment. (Contributed by NM, 20-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵)) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵))) ∧ (𝑁 ∈ 𝑇 ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ (𝑏 ∈ 𝑇 ∧ (𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐺)))) → (⦋𝐺 / 𝑔⦌𝑋‘𝑃) = ⦋𝐺 / 𝑔⦌𝑌)
 
Theoremcdlemk19xlem 41964* Lemma for cdlemk19x 41965. (Contributed by NM, 30-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ ((𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵) ∧ 𝑁 ∈ 𝑇) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊)) ∧ (𝑏 ∈ 𝑇 ∧ (𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹)))) → (⦋𝐹 / 𝑔⦌𝑋‘𝑃) = (𝑁‘𝑃))
 
Theoremcdlemk19x 41965* cdlemk19 41891 with simpler hypotheses. TODO: Clean all this up. (Contributed by NM, 30-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ (𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵) ∧ 𝑁 ∈ 𝑇) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊)) → (⦋𝐹 / 𝑔⦌𝑋‘𝑃) = (𝑁‘𝑃))
 
Theoremcdlemk42yN 41966* Part of proof of Lemma K of [Crawley] p. 118. TODO: fix comment. (Contributed by NM, 20-Jul-2013.) (New usage is discouraged.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵)) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵))) ∧ (𝑁 ∈ 𝑇 ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ (𝑏 ∈ 𝑇 ∧ (𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐺)))) → (⦋𝐺 / 𝑔⦌𝑋‘𝑃) = ((𝑃 ∨ (𝑅‘𝐺)) ∧ (𝑍 ∨ (𝑅‘(𝐺 ∘ ◡𝑏)))))
 
Theoremcdlemk11tc 41967* Part of proof of Lemma K of [Crawley] p. 118. Lemma for Eq. 5, p. 119. 𝐺, 𝐼 stand for g, h. TODO: fix comment. (Contributed by NM, 21-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵)) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵))) ∧ (𝑁 ∈ 𝑇 ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ (𝑏 ∈ 𝑇 ∧ (𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐺)) ∧ (𝐼 ∈ 𝑇 ∧ 𝐼 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐼)))) → (⦋𝐺 / 𝑔⦌𝑋‘𝑃) ≤ ((⦋𝐼 / 𝑔⦌𝑋‘𝑃) ∨ (𝑅‘(𝐼 ∘ ◡𝐺))))
 
Theoremcdlemk11t 41968* Part of proof of Lemma K of [Crawley] p. 118. Eq. 5, line 36, p. 119. 𝐺, 𝐼 stand for g, h. 𝑋 represents tau. (Contributed by NM, 21-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵)) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵))) ∧ (𝑁 ∈ 𝑇 ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ (𝐼 ∈ 𝑇 ∧ 𝐼 ≠ ( I ↾ 𝐵))) → (⦋𝐺 / 𝑔⦌𝑋‘𝑃) ≤ ((⦋𝐼 / 𝑔⦌𝑋‘𝑃) ∨ (𝑅‘(𝐼 ∘ ◡𝐺))))
 
Theoremcdlemk45 41969* Part of proof of Lemma K of [Crawley] p. 118. Line 37, p. 119. 𝐺, 𝐼 stand for g, h. 𝑋 represents tau. They do not explicitly mention the requirement (𝐺 ∘ 𝐼) ≠ ( I ↾ 𝐵). (Contributed by NM, 22-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵)) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵))) ∧ (𝑁 ∈ 𝑇 ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ (𝐼 ∈ 𝑇 ∧ 𝐼 ≠ ( I ↾ 𝐵) ∧ (𝐺 ∘ 𝐼) ≠ ( I ↾ 𝐵))) → (⦋(𝐺 ∘ 𝐼) / 𝑔⦌𝑋‘𝑃) ≤ ((⦋𝐼 / 𝑔⦌𝑋‘𝑃) ∨ (𝑅‘𝐺)))
 
Theoremcdlemk46 41970* Part of proof of Lemma K of [Crawley] p. 118. Line 38 (last line), p. 119. 𝐺, 𝐼 stand for g, h. 𝑋 represents tau. (Contributed by NM, 22-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵)) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵))) ∧ (𝑁 ∈ 𝑇 ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ (𝐼 ∈ 𝑇 ∧ 𝐼 ≠ ( I ↾ 𝐵) ∧ (𝐺 ∘ 𝐼) ≠ ( I ↾ 𝐵))) → (⦋(𝐺 ∘ 𝐼) / 𝑔⦌𝑋‘𝑃) ≤ ((⦋𝐺 / 𝑔⦌𝑋‘𝑃) ∨ (𝑅‘𝐼)))
 
Theoremcdlemk47 41971* Part of proof of Lemma K of [Crawley] p. 118. Line 2, p. 120. 𝐺, 𝐼 stand for g, h. 𝑋 represents tau. (Contributed by NM, 22-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵)) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵))) ∧ (𝑁 ∈ 𝑇 ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ (𝐼 ∈ 𝑇 ∧ 𝐼 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝐺) ≠ (𝑅‘𝐼))) → (⦋(𝐺 ∘ 𝐼) / 𝑔⦌𝑋‘𝑃) = (((⦋𝐺 / 𝑔⦌𝑋‘𝑃) ∨ (𝑅‘𝐼)) ∧ ((⦋𝐼 / 𝑔⦌𝑋‘𝑃) ∨ (𝑅‘𝐺))))
 
Theoremcdlemk48 41972* Part of proof of Lemma K of [Crawley] p. 118. Line 4, p. 120. 𝐺, 𝐼 stand for g, h. 𝑋 represents tau. (Contributed by NM, 22-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵)) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵))) ∧ (𝑁 ∈ 𝑇 ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ (𝐼 ∈ 𝑇 ∧ 𝐼 ≠ ( I ↾ 𝐵))) → ((⦋𝐺 / 𝑔⦌𝑋 ∘ ⦋𝐼 / 𝑔⦌𝑋)‘𝑃) ≤ ((⦋𝐼 / 𝑔⦌𝑋‘𝑃) ∨ (𝑅‘⦋𝐺 / 𝑔⦌𝑋)))
 
Theoremcdlemk49 41973* Part of proof of Lemma K of [Crawley] p. 118. Line 5, p. 120. 𝐺, 𝐼 stand for g, h. 𝑋 represents tau. (Contributed by NM, 23-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵)) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵))) ∧ (𝑁 ∈ 𝑇 ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ (𝐼 ∈ 𝑇 ∧ 𝐼 ≠ ( I ↾ 𝐵))) → ((⦋𝐺 / 𝑔⦌𝑋 ∘ ⦋𝐼 / 𝑔⦌𝑋)‘𝑃) ≤ ((⦋𝐺 / 𝑔⦌𝑋‘𝑃) ∨ (𝑅‘⦋𝐼 / 𝑔⦌𝑋)))
 
Theoremcdlemk50 41974* Part of proof of Lemma K of [Crawley] p. 118. Line 6, p. 120. 𝐺, 𝐼 stand for g, h. 𝑋 represents tau. TODO: Combine into cdlemk52 41976? (Contributed by NM, 23-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵)) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵))) ∧ (𝑁 ∈ 𝑇 ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ (𝐼 ∈ 𝑇 ∧ 𝐼 ≠ ( I ↾ 𝐵))) → ((⦋𝐺 / 𝑔⦌𝑋 ∘ ⦋𝐼 / 𝑔⦌𝑋)‘𝑃) ≤ (((⦋𝐺 / 𝑔⦌𝑋‘𝑃) ∨ (𝑅‘⦋𝐼 / 𝑔⦌𝑋)) ∧ ((⦋𝐼 / 𝑔⦌𝑋‘𝑃) ∨ (𝑅‘⦋𝐺 / 𝑔⦌𝑋))))
 
Theoremcdlemk51 41975* Part of proof of Lemma K of [Crawley] p. 118. Line 6, p. 120. 𝐺, 𝐼 stand for g, h. 𝑋 represents tau. TODO: Combine into cdlemk52 41976? (Contributed by NM, 23-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵)) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵))) ∧ (𝑁 ∈ 𝑇 ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ (𝐼 ∈ 𝑇 ∧ 𝐼 ≠ ( I ↾ 𝐵))) → (((⦋𝐺 / 𝑔⦌𝑋‘𝑃) ∨ (𝑅‘⦋𝐼 / 𝑔⦌𝑋)) ∧ ((⦋𝐼 / 𝑔⦌𝑋‘𝑃) ∨ (𝑅‘⦋𝐺 / 𝑔⦌𝑋))) ≤ (((⦋𝐺 / 𝑔⦌𝑋‘𝑃) ∨ (𝑅‘𝐼)) ∧ ((⦋𝐼 / 𝑔⦌𝑋‘𝑃) ∨ (𝑅‘𝐺))))
 
Theoremcdlemk52 41976* Part of proof of Lemma K of [Crawley] p. 118. Line 6, p. 120. 𝐺, 𝐼 stand for g, h. 𝑋 represents tau. (Contributed by NM, 23-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵)) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵))) ∧ (𝑁 ∈ 𝑇 ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ (𝐼 ∈ 𝑇 ∧ 𝐼 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝐺) ≠ (𝑅‘𝐼))) → ((⦋𝐺 / 𝑔⦌𝑋 ∘ ⦋𝐼 / 𝑔⦌𝑋)‘𝑃) = (⦋(𝐺 ∘ 𝐼) / 𝑔⦌𝑋‘𝑃))
 
Theoremcdlemk53a 41977* Lemma for cdlemk53 41979. (Contributed by NM, 26-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵)) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵))) ∧ (𝑁 ∈ 𝑇 ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ (𝐼 ∈ 𝑇 ∧ 𝐼 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝐺) ≠ (𝑅‘𝐼))) → ⦋(𝐺 ∘ 𝐼) / 𝑔⦌𝑋 = (⦋𝐺 / 𝑔⦌𝑋 ∘ ⦋𝐼 / 𝑔⦌𝑋))
 
Theoremcdlemk53b 41978* Lemma for cdlemk53 41979. (Contributed by NM, 26-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ ((𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵) ∧ 𝑁 ∈ 𝑇) ∧ 𝐺 ∈ 𝑇 ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊)) ∧ (𝐼 ∈ 𝑇 ∧ 𝐼 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝐺) ≠ (𝑅‘𝐼))) → ⦋(𝐺 ∘ 𝐼) / 𝑔⦌𝑋 = (⦋𝐺 / 𝑔⦌𝑋 ∘ ⦋𝐼 / 𝑔⦌𝑋))
 
Theoremcdlemk53 41979* Part of proof of Lemma K of [Crawley] p. 118. Line 7, p. 120. 𝐺, 𝐼 stand for g, h. 𝑋 represents tau. (Contributed by NM, 26-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ ((𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵) ∧ 𝑁 ∈ 𝑇) ∧ 𝐺 ∈ 𝑇 ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊)) ∧ (𝐼 ∈ 𝑇 ∧ (𝑅‘𝐺) ≠ (𝑅‘𝐼))) → ⦋(𝐺 ∘ 𝐼) / 𝑔⦌𝑋 = (⦋𝐺 / 𝑔⦌𝑋 ∘ ⦋𝐼 / 𝑔⦌𝑋))
 
Theoremcdlemk54 41980* Part of proof of Lemma K of [Crawley] p. 118. Line 10, p. 120. 𝐺, 𝐼 stand for g, h. 𝑋 represents tau. (Contributed by NM, 26-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ ((𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵) ∧ 𝑁 ∈ 𝑇) ∧ 𝐺 ∈ 𝑇 ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊)) ∧ ((𝐼 ∈ 𝑇 ∧ (𝑅‘𝐺) = (𝑅‘𝐼)) ∧ 𝑗 ∈ 𝑇 ∧ (𝑗 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑗) ≠ (𝑅‘𝐺) ∧ (𝑅‘𝑗) ≠ (𝑅‘(𝐺 ∘ 𝐼))))) → (⦋(𝐺 ∘ 𝐼) / 𝑔⦌𝑋 ∘ ⦋𝑗 / 𝑔⦌𝑋) = ((⦋𝐺 / 𝑔⦌𝑋 ∘ ⦋𝐼 / 𝑔⦌𝑋) ∘ ⦋𝑗 / 𝑔⦌𝑋))
 
Theoremcdlemk55a 41981* Lemma for cdlemk55 41983. (Contributed by NM, 26-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ ((𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵) ∧ 𝑁 ∈ 𝑇) ∧ 𝐺 ∈ 𝑇 ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊)) ∧ ((𝐼 ∈ 𝑇 ∧ (𝑅‘𝐺) = (𝑅‘𝐼)) ∧ 𝑗 ∈ 𝑇 ∧ (𝑗 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑗) ≠ (𝑅‘𝐺) ∧ (𝑅‘𝑗) ≠ (𝑅‘(𝐺 ∘ 𝐼))))) → ⦋(𝐺 ∘ 𝐼) / 𝑔⦌𝑋 = (⦋𝐺 / 𝑔⦌𝑋 ∘ ⦋𝐼 / 𝑔⦌𝑋))
 
Theoremcdlemk55b 41982* Lemma for cdlemk55 41983. (Contributed by NM, 26-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ ((𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵) ∧ 𝑁 ∈ 𝑇) ∧ 𝐺 ∈ 𝑇 ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊)) ∧ (𝐼 ∈ 𝑇 ∧ (𝑅‘𝐺) = (𝑅‘𝐼))) → ⦋(𝐺 ∘ 𝐼) / 𝑔⦌𝑋 = (⦋𝐺 / 𝑔⦌𝑋 ∘ ⦋𝐼 / 𝑔⦌𝑋))
 
Theoremcdlemk55 41983* Part of proof of Lemma K of [Crawley] p. 118. Line 11, p. 120. 𝐺, 𝐼 stand for g, h. 𝑋 represents tau. (Contributed by NM, 26-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ ((𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵) ∧ 𝑁 ∈ 𝑇) ∧ 𝐺 ∈ 𝑇 ∧ 𝐼 ∈ 𝑇) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊)) → ⦋(𝐺 ∘ 𝐼) / 𝑔⦌𝑋 = (⦋𝐺 / 𝑔⦌𝑋 ∘ ⦋𝐼 / 𝑔⦌𝑋))
 
TheoremcdlemkyyN 41984* Part of proof of Lemma K of [Crawley] p. 118. TODO: clean up (𝑏𝑌𝐺) stuff. (Contributed by NM, 21-Jul-2013.) (New usage is discouraged.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    &   𝑆 = (𝑓 ∈ 𝑇 ↦ (℩𝑖 ∈ 𝑇 (𝑖‘𝑃) = ((𝑃 ∨ (𝑅‘𝑓)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑓 ∘ ◡𝐹))))))    &   𝑉 = (𝑑 ∈ 𝑇, 𝑒 ∈ 𝑇 ↦ (℩𝑗 ∈ 𝑇 (𝑗‘𝑃) = ((𝑃 ∨ (𝑅‘𝑒)) ∧ (((𝑆‘𝑑)‘𝑃) ∨ (𝑅‘(𝑒 ∘ ◡𝑑))))))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻 ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ ((𝐹 ∈ 𝑇 ∧ 𝐹 ≠ ( I ↾ 𝐵) ∧ 𝑁 ∈ 𝑇) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵)) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊)) ∧ (𝑏 ∈ 𝑇 ∧ (𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐺)))) → (⦋𝐺 / 𝑔⦌𝑋‘𝑃) = ((𝑏𝑉𝐺)‘𝑃))
 
Theoremcdlemk43N 41985* Part of proof of Lemma K of [Crawley] p. 118. TODO: fix comment. (Contributed by NM, 31-Jul-2013.) (New usage is discouraged.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    &   𝑈 = (𝑔 ∈ 𝑇 ↦ if(𝐹 = 𝑁, 𝑔, 𝑋))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ ((𝐹 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇 ∧ 𝐹 ≠ 𝑁) ∧ (𝐺 ∈ 𝑇 ∧ 𝐺 ≠ ( I ↾ 𝐵)) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊)) ∧ (𝑏 ∈ 𝑇 ∧ (𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐺)))) → ((𝑈‘𝐺)‘𝑃) = ⦋𝐺 / 𝑔⦌𝑌)
 
Theoremcdlemk35u 41986* Substitution version of cdlemk35 41934. (Contributed by NM, 31-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    &   𝑈 = (𝑔 ∈ 𝑇 ↦ if(𝐹 = 𝑁, 𝑔, 𝑋))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ (𝐹 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇 ∧ 𝐺 ∈ 𝑇) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊)) → (𝑈‘𝐺) ∈ 𝑇)
 
Theoremcdlemk55u1 41987* Lemma for cdlemk55u 41988. (Contributed by NM, 31-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    &   𝑈 = (𝑔 ∈ 𝑇 ↦ if(𝐹 = 𝑁, 𝑔, 𝑋))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ 𝐹 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ (((𝑅‘𝐹) = (𝑅‘𝑁) ∧ 𝐹 ≠ 𝑁) ∧ 𝐺 ∈ 𝑇 ∧ 𝐼 ∈ 𝑇) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊)) → (𝑈‘(𝐺 ∘ 𝐼)) = ((𝑈‘𝐺) ∘ (𝑈‘𝐼)))
 
Theoremcdlemk55u 41988* Part of proof of Lemma K of [Crawley] p. 118. Line 11, p. 120. 𝐺, 𝐼 stand for g, h. 𝑋 represents tau. (Contributed by NM, 31-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    &   𝑈 = (𝑔 ∈ 𝑇 ↦ if(𝐹 = 𝑁, 𝑔, 𝑋))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ 𝐹 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ ((𝑅‘𝐹) = (𝑅‘𝑁) ∧ 𝐺 ∈ 𝑇 ∧ 𝐼 ∈ 𝑇) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊)) → (𝑈‘(𝐺 ∘ 𝐼)) = ((𝑈‘𝐺) ∘ (𝑈‘𝐼)))
 
Theoremcdlemk39u1 41989* Lemma for cdlemk39u 41990. (Contributed by NM, 31-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    &   𝑈 = (𝑔 ∈ 𝑇 ↦ if(𝐹 = 𝑁, 𝑔, 𝑋))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ 𝐹 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ ((𝑅‘𝐹) = (𝑅‘𝑁) ∧ 𝐹 ≠ 𝑁 ∧ 𝐺 ∈ 𝑇) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊)) → (𝑅‘(𝑈‘𝐺)) ≤ (𝑅‘𝐺))
 
Theoremcdlemk39u 41990* Part of proof of Lemma K of [Crawley] p. 118. Line 31, p. 119. Trace-preserving property of the value of tau, represented by (𝑈‘𝐺). (Contributed by NM, 31-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    &   𝑈 = (𝑔 ∈ 𝑇 ↦ if(𝐹 = 𝑁, 𝑔, 𝑋))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ 𝐹 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ ((𝑅‘𝐹) = (𝑅‘𝑁) ∧ 𝐺 ∈ 𝑇) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊)) → (𝑅‘(𝑈‘𝐺)) ≤ (𝑅‘𝐺))
 
Theoremcdlemk19u1 41991* cdlemk19 41891 with simpler hypotheses. TODO: Clean all this up. (Contributed by NM, 31-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    &   𝑈 = (𝑔 ∈ 𝑇 ↦ if(𝐹 = 𝑁, 𝑔, 𝑋))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ (𝐹 ∈ 𝑇 ∧ 𝐹 ≠ 𝑁 ∧ 𝑁 ∈ 𝑇) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊)) → ((𝑈‘𝐹)‘𝑃) = (𝑁‘𝑃))
 
Theoremcdlemk19u 41992* Part of Lemma K of [Crawley] p. 118. Line 12, p. 120, "f (exponent) tau = k". We represent f, k, tau with 𝐹, 𝑁, 𝑈. (Contributed by NM, 31-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    &   𝑈 = (𝑔 ∈ 𝑇 ↦ if(𝐹 = 𝑁, 𝑔, 𝑋))    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) ∧ (𝐹 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊)) → (𝑈‘𝐹) = 𝑁)
 
Theoremcdlemk56 41993* Part of Lemma K of [Crawley] p. 118. Line 11, p. 120, "tau is in Delta" i.e. 𝑈 is a trace-preserving endormorphism. (Contributed by NM, 31-Jul-2013.)
𝐵 = (Base‘𝐾)    &    ≤ = (le‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    &   𝑈 = (𝑔 ∈ 𝑇 ↦ if(𝐹 = 𝑁, 𝑔, 𝑋))    &   𝐸 = ((TEndo‘𝐾)‘𝑊)    ⇒   ((((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ 𝐹 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ (𝑅‘𝐹) = (𝑅‘𝑁) ∧ (𝑃 ∈ 𝐴 ∧ ¬ 𝑃 ≤ 𝑊)) → 𝑈 ∈ 𝐸)
 
Theoremcdlemk19w 41994* Use a fixed element to eliminate 𝑃 in cdlemk19u 41992. (Contributed by NM, 1-Aug-2013.)
𝐵 = (Base‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &    ⊥ = (oc‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑃 = ( ⊥ ‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    &   𝑈 = (𝑔 ∈ 𝑇 ↦ if(𝐹 = 𝑁, 𝑔, 𝑋))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) → (𝑈‘𝐹) = 𝑁)
 
Theoremcdlemk56w 41995* Use a fixed element to eliminate 𝑃 in cdlemk56 41993. (Contributed by NM, 1-Aug-2013.)
𝐵 = (Base‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &    ⊥ = (oc‘𝐾)    &   𝐴 = (Atoms‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑃 = ( ⊥ ‘𝑊)    &   𝑍 = ((𝑃 ∨ (𝑅‘𝑏)) ∧ ((𝑁‘𝑃) ∨ (𝑅‘(𝑏 ∘ ◡𝐹))))    &   𝑌 = ((𝑃 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝐹) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑃) = 𝑌))    &   𝑈 = (𝑔 ∈ 𝑇 ↦ if(𝐹 = 𝑁, 𝑔, 𝑋))    &   𝐸 = ((TEndo‘𝐾)‘𝑊)    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) → (𝑈 ∈ 𝐸 ∧ (𝑈‘𝐹) = 𝑁))
 
Theoremcdlemk 41996* Lemma K of [Crawley] p. 118. Final result, lines 11 and 12 on p. 120: given two translations f and k with the same trace, there exists a trace-preserving endomorphism tau whose value at f is k. We use 𝐹, 𝑁, and 𝑢 to represent f, k, and tau. (Contributed by NM, 1-Aug-2013.)
𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝐸 = ((TEndo‘𝐾)‘𝑊)    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ (𝑅‘𝐹) = (𝑅‘𝑁)) → ∃𝑢 ∈ 𝐸 (𝑢‘𝐹) = 𝑁)
 
Theoremtendoex 41997* Generalization of Lemma K of [Crawley] p. 118, cdlemk 41996. TODO: can this be used to shorten uses of cdlemk 41996? (Contributed by NM, 15-Oct-2013.)
≤ = (le‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝐸 = ((TEndo‘𝐾)‘𝑊)    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝐹 ∈ 𝑇 ∧ 𝑁 ∈ 𝑇) ∧ (𝑅‘𝑁) ≤ (𝑅‘𝐹)) → ∃𝑢 ∈ 𝐸 (𝑢‘𝐹) = 𝑁)
 
Theoremcdleml1N 41998 Part of proof of Lemma L of [Crawley] p. 120. TODO: fix comment. (Contributed by NM, 1-Aug-2013.) (New usage is discouraged.)
𝐵 = (Base‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝐸 = ((TEndo‘𝐾)‘𝑊)    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝑈 ∈ 𝐸 ∧ 𝑉 ∈ 𝐸 ∧ 𝑓 ∈ 𝑇) ∧ (𝑓 ≠ ( I ↾ 𝐵) ∧ (𝑈‘𝑓) ≠ ( I ↾ 𝐵) ∧ (𝑉‘𝑓) ≠ ( I ↾ 𝐵))) → (𝑅‘(𝑈‘𝑓)) = (𝑅‘(𝑉‘𝑓)))
 
Theoremcdleml2N 41999* Part of proof of Lemma L of [Crawley] p. 120. TODO: fix comment. (Contributed by NM, 1-Aug-2013.) (New usage is discouraged.)
𝐵 = (Base‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝐸 = ((TEndo‘𝐾)‘𝑊)    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝑈 ∈ 𝐸 ∧ 𝑉 ∈ 𝐸 ∧ 𝑓 ∈ 𝑇) ∧ (𝑓 ≠ ( I ↾ 𝐵) ∧ (𝑈‘𝑓) ≠ ( I ↾ 𝐵) ∧ (𝑉‘𝑓) ≠ ( I ↾ 𝐵))) → ∃𝑠 ∈ 𝐸 (𝑠‘(𝑈‘𝑓)) = (𝑉‘𝑓))
 
Theoremcdleml3N 42000* Part of proof of Lemma L of [Crawley] p. 120. TODO: fix comment. (Contributed by NM, 1-Aug-2013.) (New usage is discouraged.)
𝐵 = (Base‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝐸 = ((TEndo‘𝐾)‘𝑊)    &    0 = (𝑔 ∈ 𝑇 ↦ ( I ↾ 𝐵))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (𝑈 ∈ 𝐸 ∧ 𝑉 ∈ 𝐸 ∧ 𝑓 ∈ 𝑇) ∧ (𝑓 ≠ ( I ↾ 𝐵) ∧ 𝑈 ≠ 0 ∧ 𝑉 ≠ 0 )) → ∃𝑠 ∈ 𝐸 (𝑠 ∘ 𝑈) = 𝑉)
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