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Theorem List for Metamath Proof Explorer - 42001-42100   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremcdlemk22-3 42001* 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 42002* Part of proof of Lemma K of [Crawley] p. 118. Eliminate the (𝑅‘𝐶) ≠ (𝑅‘𝐷) requirement from cdlemk22-3 42001. (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 42003* Part of proof of Lemma K of [Crawley] p. 118. Eliminate the (𝑅‘𝑥) ≠ (𝑅‘𝐶) requirement from cdlemk23-3 42002 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 42004* Part of proof of Lemma K of [Crawley] p. 118. Eliminate the (𝑅‘𝐶) = (𝑅‘𝐷) requirement from cdlemk24-3 42003. (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 42005* 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 42006* Part of proof of Lemma K of [Crawley] p. 118. Eliminate the 𝑥 requirements from cdlemk25-3 42004. (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 42007* Part of proof of Lemma K of [Crawley] p. 118. Eliminate the 𝑃 from the conclusion of cdlemk25-3 42004. (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 42008* 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 42009* 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 42010. (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 42010* 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 42011* 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 42012* Part of proof of Lemma K of [Crawley] p. 118. cdlemk29-3 42011 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 42013* 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 42014* 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 42015* Part of proof of Lemma K of [Crawley] p. 118. Line 31, p. 119. TODO: derive more directly with r19.23 3259? (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 42016* 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 42017* 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 42018* 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 42019* 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 42020* 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 42021 Lemma for cdlemkfid3N 42025. (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 42022 Lemma for cdlemkid 42036. (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 42023 Lemma for cdlemkfid3N 42025. (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 42024* Lemma for cdlemkid 42036. (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 42025* 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 42026* Part of proof of Lemma K of [Crawley] p. 118. TODO: clean up (𝑏𝑌𝐺) stuff. 𝑉 represents 𝑌 in cdlemk31 41996. (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 42027* Convert between function and explicit forms. 𝐶 represents 𝑍 in cdlemkuu 41995. 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 42028* cdlemkyu 42027 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 42029* 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 42030* Lemma for cdlemk19y 42032. (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 42031* Part of proof of Lemma K of [Crawley] p. 118. Lemma for Eq. 5, p. 119. 𝐺, 𝐼 stand for g, h. cdlemk11ta 42029 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 42032* cdlemk19 41969 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 42033* Lemma for cdlemkid 42036. (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 42034* Lemma for cdlemkid 42036. (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 42035* Lemma for cdlemkid 42036. (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 42036* 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 42037* Substitution version of cdlemk35 42012. (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 42038* Substitution version of cdlemk35 42012. (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 42039* Substitution version of cdlemk39 42016. TODO: Can any commonality with cdlemk35s 42037 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 42040* Substitution version of cdlemk39 42016 with non-identity requirement on 𝐺 removed. TODO: Can any commonality with cdlemk35s 42037 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 42041* 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 42042* Lemma for cdlemk19x 42043. (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 42043* cdlemk19 41969 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 42044* 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 42045* 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 42046* 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 42047* 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 42048* 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 42049* 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 42050* 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 42051* 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 42052* 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 42054? (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 42053* 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 42054? (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 42054* 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 42055* Lemma for cdlemk53 42057. (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 42056* Lemma for cdlemk53 42057. (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 42057* 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 42058* 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 42059* Lemma for cdlemk55 42061. (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 42060* Lemma for cdlemk55 42061. (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 42061* 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 42062* 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 42063* 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 42064* Substitution version of cdlemk35 42012. (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 42065* Lemma for cdlemk55u 42066. (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 42066* 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 42067* Lemma for cdlemk39u 42068. (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 42068* 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 42069* cdlemk19 41969 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 42070* 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 42071* 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 42072* Use a fixed element to eliminate 𝑃 in cdlemk19u 42070. (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 42073* Use a fixed element to eliminate 𝑃 in cdlemk56 42071. (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 42074* 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 42075* Generalization of Lemma K of [Crawley] p. 118, cdlemk 42074. TODO: can this be used to shorten uses of cdlemk 42074? (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 42076 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 42077* 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 42078* 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)
 
Theoremcdleml4N 42079* 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) ∧ (𝑈 ≠ 0 ∧ 𝑉 ≠ 0 )) → ∃𝑠 ∈ 𝐸 (𝑠 ∘ 𝑈) = 𝑉)
 
Theoremcdleml5N 42080* 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) ∧ 𝑈 ≠ 0 ) → ∃𝑠 ∈ 𝐸 (𝑠 ∘ 𝑈) = 𝑉)
 
Theoremcdleml6 42081* Part of proof of Lemma L of [Crawley] p. 120. TODO: fix comment. (Contributed by NM, 11-Aug-2013.)
𝐵 = (Base‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑄 = ((oc‘𝐾)‘𝑊)    &   𝑍 = ((𝑄 ∨ (𝑅‘𝑏)) ∧ ((ℎ‘𝑄) ∨ (𝑅‘(𝑏 ∘ ◡(𝑠‘ℎ)))))    &   𝑌 = ((𝑄 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘(𝑠‘ℎ)) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑄) = 𝑌))    &   𝑈 = (𝑔 ∈ 𝑇 ↦ if((𝑠‘ℎ) = ℎ, 𝑔, 𝑋))    &   𝐸 = ((TEndo‘𝐾)‘𝑊)    &    0 = (𝑓 ∈ 𝑇 ↦ ( I ↾ 𝐵))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ ℎ ∈ 𝑇 ∧ (𝑠 ∈ 𝐸 ∧ 𝑠 ≠ 0 )) → (𝑈 ∈ 𝐸 ∧ (𝑈‘(𝑠‘ℎ)) = ℎ))
 
Theoremcdleml7 42082* Part of proof of Lemma L of [Crawley] p. 120. TODO: fix comment. (Contributed by NM, 11-Aug-2013.)
𝐵 = (Base‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑄 = ((oc‘𝐾)‘𝑊)    &   𝑍 = ((𝑄 ∨ (𝑅‘𝑏)) ∧ ((ℎ‘𝑄) ∨ (𝑅‘(𝑏 ∘ ◡(𝑠‘ℎ)))))    &   𝑌 = ((𝑄 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘(𝑠‘ℎ)) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑄) = 𝑌))    &   𝑈 = (𝑔 ∈ 𝑇 ↦ if((𝑠‘ℎ) = ℎ, 𝑔, 𝑋))    &   𝐸 = ((TEndo‘𝐾)‘𝑊)    &    0 = (𝑓 ∈ 𝑇 ↦ ( I ↾ 𝐵))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ ℎ ∈ 𝑇 ∧ (𝑠 ∈ 𝐸 ∧ 𝑠 ≠ 0 )) → ((𝑈 ∘ 𝑠)‘ℎ) = (( I ↾ 𝑇)‘ℎ))
 
Theoremcdleml8 42083* Part of proof of Lemma L of [Crawley] p. 120. TODO: fix comment. (Contributed by NM, 11-Aug-2013.)
𝐵 = (Base‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑄 = ((oc‘𝐾)‘𝑊)    &   𝑍 = ((𝑄 ∨ (𝑅‘𝑏)) ∧ ((ℎ‘𝑄) ∨ (𝑅‘(𝑏 ∘ ◡(𝑠‘ℎ)))))    &   𝑌 = ((𝑄 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘(𝑠‘ℎ)) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑄) = 𝑌))    &   𝑈 = (𝑔 ∈ 𝑇 ↦ if((𝑠‘ℎ) = ℎ, 𝑔, 𝑋))    &   𝐸 = ((TEndo‘𝐾)‘𝑊)    &    0 = (𝑓 ∈ 𝑇 ↦ ( I ↾ 𝐵))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (ℎ ∈ 𝑇 ∧ ℎ ≠ ( I ↾ 𝐵)) ∧ (𝑠 ∈ 𝐸 ∧ 𝑠 ≠ 0 )) → (𝑈 ∘ 𝑠) = ( I ↾ 𝑇))
 
Theoremcdleml9 42084* Part of proof of Lemma L of [Crawley] p. 120. TODO: fix comment. (Contributed by NM, 11-Aug-2013.)
𝐵 = (Base‘𝐾)    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑄 = ((oc‘𝐾)‘𝑊)    &   𝑍 = ((𝑄 ∨ (𝑅‘𝑏)) ∧ ((ℎ‘𝑄) ∨ (𝑅‘(𝑏 ∘ ◡(𝑠‘ℎ)))))    &   𝑌 = ((𝑄 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘(𝑠‘ℎ)) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑄) = 𝑌))    &   𝑈 = (𝑔 ∈ 𝑇 ↦ if((𝑠‘ℎ) = ℎ, 𝑔, 𝑋))    &   𝐸 = ((TEndo‘𝐾)‘𝑊)    &    0 = (𝑓 ∈ 𝑇 ↦ ( I ↾ 𝐵))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (ℎ ∈ 𝑇 ∧ ℎ ≠ ( I ↾ 𝐵)) ∧ (𝑠 ∈ 𝐸 ∧ 𝑠 ≠ 0 )) → 𝑈 ≠ 0 )
 
Theoremdva1dim 42085* Two expressions for the 1-dimensional subspaces of partial vector space A. Remark in [Crawley] p. 120 line 21, but using a non-identity translation (nonzero vector) 𝐹 whose trace is 𝑃 rather than 𝑃 itself; 𝐹 exists by cdlemf 41663. 𝐸 is the division ring base by erngdv 42093, and 𝑠‘𝐹 is the scalar product by dvavsca 42117. 𝐹 must be a non-identity translation for the expression to be a 1-dimensional subspace, although the theorem doesn't require it. (Contributed by NM, 14-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) → {𝑔 ∣ ∃𝑠 ∈ 𝐸 𝑔 = (𝑠‘𝐹)} = {𝑔 ∈ 𝑇 ∣ (𝑅‘𝑔) ≤ (𝑅‘𝐹)})
 
Theoremdvhb1dimN 42086* Two expressions for the 1-dimensional subspaces of vector space 𝐻, in the isomorphism B case where the second vector component is zero. (Contributed by NM, 23-Feb-2014.) (New usage is discouraged.)
≤ = (le‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝐸 = ((TEndo‘𝐾)‘𝑊)    &    0 = (ℎ ∈ 𝑇 ↦ ( I ↾ 𝐵))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ 𝐹 ∈ 𝑇) → {𝑔 ∈ (𝑇 × 𝐸) ∣ ∃𝑠 ∈ 𝐸 𝑔 = ⟨(𝑠‘𝐹), 0 ⟩} = {𝑔 ∈ (𝑇 × 𝐸) ∣ ((𝑅‘(1st ‘𝑔)) ≤ (𝑅‘𝐹) ∧ (2nd ‘𝑔) = 0 )})
 
Theoremerng1lem 42087 Value of the endomorphism division ring unity. (Contributed by NM, 12-Oct-2013.)
𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝐸 = ((TEndo‘𝐾)‘𝑊)    &   𝐷 = ((EDRing‘𝐾)‘𝑊)    &   ((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) → 𝐷 ∈ Ring)    ⇒   ((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) → (1r‘𝐷) = ( I ↾ 𝑇))
 
Theoremerngdvlem1 42088* Lemma for eringring 42092. (Contributed by NM, 4-Aug-2013.)
𝐻 = (LHyp‘𝐾)    &   𝐷 = ((EDRing‘𝐾)‘𝑊)    &   𝐵 = (Base‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝐸 = ((TEndo‘𝐾)‘𝑊)    &   𝑃 = (𝑎 ∈ 𝐸, 𝑏 ∈ 𝐸 ↦ (𝑓 ∈ 𝑇 ↦ ((𝑎‘𝑓) ∘ (𝑏‘𝑓))))    &    0 = (𝑓 ∈ 𝑇 ↦ ( I ↾ 𝐵))    &   𝐼 = (𝑎 ∈ 𝐸 ↦ (𝑓 ∈ 𝑇 ↦ ◡(𝑎‘𝑓)))    ⇒   ((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) → 𝐷 ∈ Grp)
 
Theoremerngdvlem2N 42089* Lemma for eringring 42092. (Contributed by NM, 6-Aug-2013.) (New usage is discouraged.)
𝐻 = (LHyp‘𝐾)    &   𝐷 = ((EDRing‘𝐾)‘𝑊)    &   𝐵 = (Base‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝐸 = ((TEndo‘𝐾)‘𝑊)    &   𝑃 = (𝑎 ∈ 𝐸, 𝑏 ∈ 𝐸 ↦ (𝑓 ∈ 𝑇 ↦ ((𝑎‘𝑓) ∘ (𝑏‘𝑓))))    &    0 = (𝑓 ∈ 𝑇 ↦ ( I ↾ 𝐵))    &   𝐼 = (𝑎 ∈ 𝐸 ↦ (𝑓 ∈ 𝑇 ↦ ◡(𝑎‘𝑓)))    ⇒   ((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) → 𝐷 ∈ Abel)
 
Theoremerngdvlem3 42090* Lemma for eringring 42092. (Contributed by NM, 6-Aug-2013.)
𝐻 = (LHyp‘𝐾)    &   𝐷 = ((EDRing‘𝐾)‘𝑊)    &   𝐵 = (Base‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝐸 = ((TEndo‘𝐾)‘𝑊)    &   𝑃 = (𝑎 ∈ 𝐸, 𝑏 ∈ 𝐸 ↦ (𝑓 ∈ 𝑇 ↦ ((𝑎‘𝑓) ∘ (𝑏‘𝑓))))    &    0 = (𝑓 ∈ 𝑇 ↦ ( I ↾ 𝐵))    &   𝐼 = (𝑎 ∈ 𝐸 ↦ (𝑓 ∈ 𝑇 ↦ ◡(𝑎‘𝑓)))    &    + = (𝑎 ∈ 𝐸, 𝑏 ∈ 𝐸 ↦ (𝑎 ∘ 𝑏))    ⇒   ((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) → 𝐷 ∈ Ring)
 
Theoremerngdvlem4 42091* Lemma for erngdv 42093. (Contributed by NM, 11-Aug-2013.)
𝐻 = (LHyp‘𝐾)    &   𝐷 = ((EDRing‘𝐾)‘𝑊)    &   𝐵 = (Base‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝐸 = ((TEndo‘𝐾)‘𝑊)    &   𝑃 = (𝑎 ∈ 𝐸, 𝑏 ∈ 𝐸 ↦ (𝑓 ∈ 𝑇 ↦ ((𝑎‘𝑓) ∘ (𝑏‘𝑓))))    &    0 = (𝑓 ∈ 𝑇 ↦ ( I ↾ 𝐵))    &   𝐼 = (𝑎 ∈ 𝐸 ↦ (𝑓 ∈ 𝑇 ↦ ◡(𝑎‘𝑓)))    &    + = (𝑎 ∈ 𝐸, 𝑏 ∈ 𝐸 ↦ (𝑎 ∘ 𝑏))    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑄 = ((oc‘𝐾)‘𝑊)    &   𝑍 = ((𝑄 ∨ (𝑅‘𝑏)) ∧ ((ℎ‘𝑄) ∨ (𝑅‘(𝑏 ∘ ◡(𝑠‘ℎ)))))    &   𝑌 = ((𝑄 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘(𝑠‘ℎ)) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑄) = 𝑌))    &   𝑈 = (𝑔 ∈ 𝑇 ↦ if((𝑠‘ℎ) = ℎ, 𝑔, 𝑋))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (ℎ ∈ 𝑇 ∧ ℎ ≠ ( I ↾ 𝐵))) → 𝐷 ∈ DivRing)
 
Theoremeringring 42092 An endomorphism ring is a ring. TODO: fix comment. (Contributed by NM, 4-Aug-2013.)
𝐻 = (LHyp‘𝐾)    &   𝐷 = ((EDRing‘𝐾)‘𝑊)    ⇒   ((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) → 𝐷 ∈ Ring)
 
Theoremerngdv 42093 An endomorphism ring is a division ring. TODO: fix comment. (Contributed by NM, 11-Aug-2013.)
𝐻 = (LHyp‘𝐾)    &   𝐷 = ((EDRing‘𝐾)‘𝑊)    ⇒   ((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) → 𝐷 ∈ DivRing)
 
Theoremerng0g 42094* The division ring zero of an endomorphism ring. (Contributed by NM, 5-Nov-2013.) (Revised by Mario Carneiro, 23-Jun-2014.)
𝐵 = (Base‘𝐾)    &   𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝐷 = ((EDRing‘𝐾)‘𝑊)    &   𝑂 = (𝑓 ∈ 𝑇 ↦ ( I ↾ 𝐵))    &    0 = (0g‘𝐷)    ⇒   ((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) → 0 = 𝑂)
 
Theoremerng1r 42095 The division ring unity of an endomorphism ring. (Contributed by NM, 5-Nov-2013.) (Revised by Mario Carneiro, 23-Jun-2014.)
𝐻 = (LHyp‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝐷 = ((EDRing‘𝐾)‘𝑊)    &    1 = (1r‘𝐷)    ⇒   ((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) → 1 = ( I ↾ 𝑇))
 
Theoremerngdvlem1-rN 42096* Lemma for eringring 42092. (Contributed by NM, 4-Aug-2013.) (New usage is discouraged.)
𝐻 = (LHyp‘𝐾)    &   𝐷 = ((EDRingR‘𝐾)‘𝑊)    &   𝐵 = (Base‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝐸 = ((TEndo‘𝐾)‘𝑊)    &   𝑃 = (𝑎 ∈ 𝐸, 𝑏 ∈ 𝐸 ↦ (𝑓 ∈ 𝑇 ↦ ((𝑎‘𝑓) ∘ (𝑏‘𝑓))))    &   𝑂 = (𝑓 ∈ 𝑇 ↦ ( I ↾ 𝐵))    &   𝐼 = (𝑎 ∈ 𝐸 ↦ (𝑓 ∈ 𝑇 ↦ ◡(𝑎‘𝑓)))    ⇒   ((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) → 𝐷 ∈ Grp)
 
Theoremerngdvlem2-rN 42097* Lemma for eringring 42092. (Contributed by NM, 6-Aug-2013.) (New usage is discouraged.)
𝐻 = (LHyp‘𝐾)    &   𝐷 = ((EDRingR‘𝐾)‘𝑊)    &   𝐵 = (Base‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝐸 = ((TEndo‘𝐾)‘𝑊)    &   𝑃 = (𝑎 ∈ 𝐸, 𝑏 ∈ 𝐸 ↦ (𝑓 ∈ 𝑇 ↦ ((𝑎‘𝑓) ∘ (𝑏‘𝑓))))    &   𝑂 = (𝑓 ∈ 𝑇 ↦ ( I ↾ 𝐵))    &   𝐼 = (𝑎 ∈ 𝐸 ↦ (𝑓 ∈ 𝑇 ↦ ◡(𝑎‘𝑓)))    ⇒   ((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) → 𝐷 ∈ Abel)
 
Theoremerngdvlem3-rN 42098* Lemma for eringring 42092. (Contributed by NM, 6-Aug-2013.) (New usage is discouraged.)
𝐻 = (LHyp‘𝐾)    &   𝐷 = ((EDRingR‘𝐾)‘𝑊)    &   𝐵 = (Base‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝐸 = ((TEndo‘𝐾)‘𝑊)    &   𝑃 = (𝑎 ∈ 𝐸, 𝑏 ∈ 𝐸 ↦ (𝑓 ∈ 𝑇 ↦ ((𝑎‘𝑓) ∘ (𝑏‘𝑓))))    &   𝑂 = (𝑓 ∈ 𝑇 ↦ ( I ↾ 𝐵))    &   𝐼 = (𝑎 ∈ 𝐸 ↦ (𝑓 ∈ 𝑇 ↦ ◡(𝑎‘𝑓)))    &   𝑀 = (𝑎 ∈ 𝐸, 𝑏 ∈ 𝐸 ↦ (𝑏 ∘ 𝑎))    ⇒   ((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) → 𝐷 ∈ Ring)
 
Theoremerngdvlem4-rN 42099* Lemma for erngdv 42093. (Contributed by NM, 11-Aug-2013.) (New usage is discouraged.)
𝐻 = (LHyp‘𝐾)    &   𝐷 = ((EDRingR‘𝐾)‘𝑊)    &   𝐵 = (Base‘𝐾)    &   𝑇 = ((LTrn‘𝐾)‘𝑊)    &   𝐸 = ((TEndo‘𝐾)‘𝑊)    &   𝑃 = (𝑎 ∈ 𝐸, 𝑏 ∈ 𝐸 ↦ (𝑓 ∈ 𝑇 ↦ ((𝑎‘𝑓) ∘ (𝑏‘𝑓))))    &   𝑂 = (𝑓 ∈ 𝑇 ↦ ( I ↾ 𝐵))    &   𝐼 = (𝑎 ∈ 𝐸 ↦ (𝑓 ∈ 𝑇 ↦ ◡(𝑎‘𝑓)))    &   𝑀 = (𝑎 ∈ 𝐸, 𝑏 ∈ 𝐸 ↦ (𝑏 ∘ 𝑎))    &    ∨ = (join‘𝐾)    &    ∧ = (meet‘𝐾)    &   𝑅 = ((trL‘𝐾)‘𝑊)    &   𝑄 = ((oc‘𝐾)‘𝑊)    &   𝑍 = ((𝑄 ∨ (𝑅‘𝑏)) ∧ ((ℎ‘𝑄) ∨ (𝑅‘(𝑏 ∘ ◡(𝑠‘ℎ)))))    &   𝑌 = ((𝑄 ∨ (𝑅‘𝑔)) ∧ (𝑍 ∨ (𝑅‘(𝑔 ∘ ◡𝑏))))    &   𝑋 = (℩𝑧 ∈ 𝑇 ∀𝑏 ∈ 𝑇 ((𝑏 ≠ ( I ↾ 𝐵) ∧ (𝑅‘𝑏) ≠ (𝑅‘(𝑠‘ℎ)) ∧ (𝑅‘𝑏) ≠ (𝑅‘𝑔)) → (𝑧‘𝑄) = 𝑌))    &   𝑈 = (𝑔 ∈ 𝑇 ↦ if((𝑠‘ℎ) = ℎ, 𝑔, 𝑋))    ⇒   (((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) ∧ (ℎ ∈ 𝑇 ∧ ℎ ≠ ( I ↾ 𝐵))) → 𝐷 ∈ DivRing)
 
Theoremerngring-rN 42100 An endomorphism ring is a ring. TODO: fix comment. (Contributed by NM, 4-Aug-2013.) (New usage is discouraged.)
𝐻 = (LHyp‘𝐾)    &   𝐷 = ((EDRingR‘𝐾)‘𝑊)    ⇒   ((𝐾 ∈ HL ∧ 𝑊 ∈ 𝐻) → 𝐷 ∈ Ring)
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