Theorem axc14 2519

Description: Axiom ax-c14 34698 is redundant if we assume ax-5 1991. Remark 9.6 in [Megill] p. 448 (p. 16 of the preprint), regarding axiom scheme C14'.

Note that 𝑤 is a dummy variable introduced in the proof. Its purpose is to satisfy the distinct variable requirements of dveel2 2518 and ax-5 1991. By the end of the proof it has vanished, and the final theorem has no distinct variable requirements. (Contributed by NM, 29-Jun-1995.) (Proof modification is discouraged.)

Assertion

Ref	Expression
axc14	⊢ (¬ ∀𝑧 𝑧 = 𝑥 → (¬ ∀𝑧 𝑧 = 𝑦 → (𝑥 ∈ 𝑦 → ∀𝑧 𝑥 ∈ 𝑦)))

Proof of Theorem axc14

Dummy variable 𝑤 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		hbn1 2175	. . . . 5 ⊢ (¬ ∀𝑧 𝑧 = 𝑦 → ∀𝑧 ¬ ∀𝑧 𝑧 = 𝑦)
2		dveel2 2518	. . . . 5 ⊢ (¬ ∀𝑧 𝑧 = 𝑦 → (𝑤 ∈ 𝑦 → ∀𝑧 𝑤 ∈ 𝑦))
3	1, 2	hbim1 2289	. . . 4 ⊢ ((¬ ∀𝑧 𝑧 = 𝑦 → 𝑤 ∈ 𝑦) → ∀𝑧(¬ ∀𝑧 𝑧 = 𝑦 → 𝑤 ∈ 𝑦))
4		elequ1 2152	. . . . 5 ⊢ (𝑤 = 𝑥 → (𝑤 ∈ 𝑦 ↔ 𝑥 ∈ 𝑦))
5	4	imbi2d 329	. . . 4 ⊢ (𝑤 = 𝑥 → ((¬ ∀𝑧 𝑧 = 𝑦 → 𝑤 ∈ 𝑦) ↔ (¬ ∀𝑧 𝑧 = 𝑦 → 𝑥 ∈ 𝑦)))
6	3, 5	dvelim 2487	. . 3 ⊢ (¬ ∀𝑧 𝑧 = 𝑥 → ((¬ ∀𝑧 𝑧 = 𝑦 → 𝑥 ∈ 𝑦) → ∀𝑧(¬ ∀𝑧 𝑧 = 𝑦 → 𝑥 ∈ 𝑦)))
7		nfa1 2184	. . . . 5 ⊢ Ⅎ𝑧∀𝑧 𝑧 = 𝑦
8	7	nfn 1935	. . . 4 ⊢ Ⅎ𝑧 ¬ ∀𝑧 𝑧 = 𝑦
9	8	19.21 2231	. . 3 ⊢ (∀𝑧(¬ ∀𝑧 𝑧 = 𝑦 → 𝑥 ∈ 𝑦) ↔ (¬ ∀𝑧 𝑧 = 𝑦 → ∀𝑧 𝑥 ∈ 𝑦))
10	6, 9	syl6ib 241	. 2 ⊢ (¬ ∀𝑧 𝑧 = 𝑥 → ((¬ ∀𝑧 𝑧 = 𝑦 → 𝑥 ∈ 𝑦) → (¬ ∀𝑧 𝑧 = 𝑦 → ∀𝑧 𝑥 ∈ 𝑦)))
11	10	pm2.86d 108	1 ⊢ (¬ ∀𝑧 𝑧 = 𝑥 → (¬ ∀𝑧 𝑧 = 𝑦 → (𝑥 ∈ 𝑦 → ∀𝑧 𝑥 ∈ 𝑦)))

Syntax hints:  ¬ wn 3   → wi 4  ∀wal 1629

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1870  ax-4 1885  ax-5 1991  ax-6 2057  ax-7 2093  ax-8 2147  ax-9 2154  ax-10 2174  ax-11 2190  ax-12 2203  ax-13 2408

This theorem depends on definitions:  df-bi 197  df-an 383  df-or 837  df-tru 1634  df-ex 1853  df-nf 1858

This theorem is referenced by: (None)