Theorem ax15 1006

Description: Axiom ax-15 806 is redundant if we assume ax-17 925. Remark 9.6 in [Megill] p. 448 (p. 16 of the preprint), regarding axiom scheme C14'.

Assertion

Ref	Expression
ax15	⊢ (¬ ∀z z = x → (¬ ∀z z = y → (x ∈ y → ∀z x ∈ y)))

Proof of Theorem ax15

Step	Hyp	Ref	Expression
1		hbn1 708	. . . . 5 ⊢ (¬ ∀z z = y → ∀z ¬ ∀z z = y)
2		ddeel2 1004	. . . . 5 ⊢ (¬ ∀z z = y → (w ∈ y → ∀z w ∈ y))
3	1, 2	hbim1 781	. . . 4 ⊢ ((¬ ∀z z = y → w ∈ y) → ∀z(¬ ∀z z = y → w ∈ y))
4		a13b 819	. . . . 5 ⊢ (w = x → (w ∈ y ↔ x ∈ y))
5	4	imbi2d 464	. . . 4 ⊢ (w = x → ((¬ ∀z z = y → w ∈ y) ↔ (¬ ∀z z = y → x ∈ y)))
6	3, 5	ddelim 1000	. . 3 ⊢ (¬ ∀z z = x → ((¬ ∀z z = y → x ∈ y) → ∀z(¬ ∀z z = y → x ∈ y)))
7	1	19.21 738	. . 3 ⊢ (∀z(¬ ∀z z = y → x ∈ y) ↔ (¬ ∀z z = y → ∀z x ∈ y))
8	6, 7	syl6ib 185	. 2 ⊢ (¬ ∀z z = x → ((¬ ∀z z = y → x ∈ y) → (¬ ∀z z = y → ∀z x ∈ y)))
9	8	pm2.86d 65	1 ⊢ (¬ ∀z z = x → (¬ ∀z z = y → (x ∈ y → ∀z x ∈ y)))

Syntax hints:  ¬ wn 1   → wi 2  ∀wal 672   = weq 797   ∈ wel 803

This theorem is referenced by:  axrepnd 3740  axpowndlem4 3746  axregndlem2 3749  axinfndlem1 3751  axinfnd 3752  axacndlem4 3756  axacndlem5 3757  axacnd 3758

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-13 804  ax-14 805  ax-17 925

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853

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