P. 506 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 50501-50514   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	nfals 50501	Bound-variable hypothesis builder for "all some". (Contributed by David A. Wheeler, 12-Jul-2026.)
⊢ Ⅎ𝑥𝜑    &   ⊢ Ⅎ𝑥𝜓    ⇒   ⊢ Ⅎ𝑥∀∃𝑦(𝜑 → 𝜓)
Theorem	nfrals 50502*	Bound-variable hypothesis builder for "all some" restricted to a class. (Contributed by David A. Wheeler, 12-Jul-2026.)
⊢ Ⅎ𝑥𝐴    &   ⊢ Ⅎ𝑥𝜑    &   ⊢ Ⅎ𝑥𝜓    ⇒   ⊢ Ⅎ𝑥∀∃𝑦 ∈ 𝐴(𝜑 → 𝜓)
Theorem	cbvals 50503*	Rule used to change bound variables, using implicit substitution. (Contributed by David A. Wheeler, 12-Jul-2026.)
⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜒))    &   ⊢ (𝑥 = 𝑦 → (𝜓 ↔ 𝜃))    ⇒   ⊢ (∀∃𝑥(𝜑 → 𝜓) ↔ ∀∃𝑦(𝜒 → 𝜃))
Theorem	als-no-surprise 50504	Demonstrate that there is never a "surprise" when using the allsome quantifier, that is, it is never possible for the consequent to be both always true and always false. This uses the definition of df-als 50486; the proof itself builds on alimp-no-surprise 50479. For a contrast, see alimp-surprise 50478. (Contributed by David A. Wheeler, 27-Oct-2018.)
⊢ ¬ (∀∃𝑥(𝜑 → 𝜓) ∧ ∀∃𝑥(𝜑 → ¬ 𝜓))
Theorem	rals-no-surprise 50505	Demonstrate that there is never a "surprise" when using the allsome quantifier restricted to a class, that is, it is never possible for the consequent to be both always true and always false of the members of 𝐴 that satisfy the antecedent. This is the restricted counterpart of als-no-surprise 50504, and follows from it by dfrals2 50488. Note that this needs no assumption that 𝐴 is nonempty, because allsome requires a member of 𝐴 satisfying 𝜑, and that member would have to satisfy both 𝜓 and ¬ 𝜓. The ordinary restricted "for all" requires no such member and can be vacuously true, as shown in empty-surprise2 50481; that is the point of allsome. (Contributed by David A. Wheeler, 12-Jul-2026.)
⊢ ¬ (∀∃𝑥 ∈ 𝐴(𝜑 → 𝜓) ∧ ∀∃𝑥 ∈ 𝐴(𝜑 → ¬ 𝜓))
21.52.13  Miscellaneous Miscellaneous proofs.
Theorem	5m4e1 50506	Prove that 5 - 4 = 1. (Contributed by David A. Wheeler, 31-Jan-2017.)
⊢ (5 − 4) = 1
Theorem	2p2ne5 50507	Prove that 2 + 2 ≠ 5. In George Orwell's "1984", Part One, Chapter Seven, the protagonist Winston notes that, "In the end the Party would announce that two and two made five, and you would have to believe it." http://www.sparknotes.com/lit/1984/section4.rhtml. More generally, the phrase 2 + 2 = 5 has come to represent an obviously false dogma one may be required to believe. See the Wikipedia article for more about this: https://en.wikipedia.org/wiki/2_%2B_2_%3D_5. Unsurprisingly, we can easily prove that this claim is false. (Contributed by David A. Wheeler, 31-Jan-2017.)
⊢ (2 + 2) ≠ 5
Theorem	resolution 50508	Resolution rule. This is the primary inference rule in some automated theorem provers such as prover9. The resolution rule can be traced back to Davis and Putnam (1960). (Contributed by David A. Wheeler, 9-Feb-2017.)
⊢ (((𝜑 ∧ 𝜓) ∨ (¬ 𝜑 ∧ 𝜒)) → (𝜓 ∨ 𝜒))
Theorem	testable 50509	In classical logic all wffs are testable, that is, it is always true that (¬ 𝜑 ∨ ¬ ¬ 𝜑). This is not necessarily true in intuitionistic logic. In intuitionistic logic, if this statement is true for some 𝜑, then 𝜑 is testable. The proof is trivial because it's simply a special case of the law of the excluded middle, which is true in classical logic but not necessarily true in intuitionisic logic. (Contributed by David A. Wheeler, 5-Dec-2018.)
⊢ (¬ 𝜑 ∨ ¬ ¬ 𝜑)
21.52.14  Theorems about algebraic numbers
Theorem	aacllem 50510*	Lemma for other theorems about 𝔸. (Contributed by Brendan Leahy, 3-Jan-2020.) (Revised by Alexander van der Vekens and David A. Wheeler, 25-Apr-2020.)
⊢ (𝜑 → 𝐴 ∈ ℂ)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    &   ⊢ ((𝜑 ∧ 𝑛 ∈ (1...𝑁)) → 𝑋 ∈ ℂ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (0...𝑁) ∧ 𝑛 ∈ (1...𝑁)) → 𝐶 ∈ ℚ)    &   ⊢ ((𝜑 ∧ 𝑘 ∈ (0...𝑁)) → (𝐴↑𝑘) = Σ𝑛 ∈ (1...𝑁)(𝐶 · 𝑋))    ⇒   ⊢ (𝜑 → 𝐴 ∈ 𝔸)
21.53  Mathbox for Kunhao Zheng
21.53.1  Weighted AM-GM inequality
Theorem	amgmwlem 50511	Weighted version of amgmlem 27120. (Contributed by Kunhao Zheng, 19-Jun-2021.)
⊢ 𝑀 = (mulGrp‘ℂfld)    &   ⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ (𝜑 → 𝐴 ≠ ∅)    &   ⊢ (𝜑 → 𝐹:𝐴⟶ℝ+)    &   ⊢ (𝜑 → 𝑊:𝐴⟶ℝ+)    &   ⊢ (𝜑 → (ℂfld Σg 𝑊) = 1)    ⇒   ⊢ (𝜑 → (𝑀 Σg (𝐹 ∘f ↑𝑐𝑊)) ≤ (ℂfld Σg (𝐹 ∘f · 𝑊)))
Theorem	amgmlemALT 50512	Alternate proof of amgmlem 27120 using amgmwlem 50511. (Contributed by Kunhao Zheng, 20-Jun-2021.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ 𝑀 = (mulGrp‘ℂfld)    &   ⊢ (𝜑 → 𝐴 ∈ Fin)    &   ⊢ (𝜑 → 𝐴 ≠ ∅)    &   ⊢ (𝜑 → 𝐹:𝐴⟶ℝ+)    ⇒   ⊢ (𝜑 → ((𝑀 Σg 𝐹)↑𝑐(1 / (♯‘𝐴))) ≤ ((ℂfld Σg 𝐹) / (♯‘𝐴)))
Theorem	amgmw2d 50513	Weighted arithmetic-geometric mean inequality for 𝑛 = 2 (compare amgm2d 44851). (Contributed by Kunhao Zheng, 20-Jun-2021.)
⊢ (𝜑 → 𝐴 ∈ ℝ+)    &   ⊢ (𝜑 → 𝑃 ∈ ℝ+)    &   ⊢ (𝜑 → 𝐵 ∈ ℝ+)    &   ⊢ (𝜑 → 𝑄 ∈ ℝ+)    &   ⊢ (𝜑 → (𝑃 + 𝑄) = 1)    ⇒   ⊢ (𝜑 → ((𝐴↑𝑐𝑃) · (𝐵↑𝑐𝑄)) ≤ ((𝐴 · 𝑃) + (𝐵 · 𝑄)))
Theorem	young2d 50514	Young's inequality for 𝑛 = 2, a direct application of amgmw2d 50513. (Contributed by Kunhao Zheng, 6-Jul-2021.)
⊢ (𝜑 → 𝐴 ∈ ℝ+)    &   ⊢ (𝜑 → 𝑃 ∈ ℝ+)    &   ⊢ (𝜑 → 𝐵 ∈ ℝ+)    &   ⊢ (𝜑 → 𝑄 ∈ ℝ+)    &   ⊢ (𝜑 → ((1 / 𝑃) + (1 / 𝑄)) = 1)    ⇒   ⊢ (𝜑 → (𝐴 · 𝐵) ≤ (((𝐴↑𝑐𝑃) / 𝑃) + ((𝐵↑𝑐𝑄) / 𝑄)))

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