Theorem axregndlem1 3748

Description: Lemma for the Axiom of Regularity with no distinct variable conditions.

Assertion

Ref	Expression
axregndlem1	|- (A.x x = z -> (x e. y -> E.x(x e. y /\ A.z(z e. x -> -. z e. y))))

Proof of Theorem axregndlem1

Step	Hyp	Ref	Expression
1		eq5 824	. . . . . . 7 |- (A.x x = z -> A.zA.x x = z)
2		eirrv 3449	. . . . . . . . . 10 |- -. x e. x
3		a13b 819	. . . . . . . . . 10 |- (x = z -> (x e. x <-> z e. x))
4	2, 3	mtbii 538	. . . . . . . . 9 |- (x = z -> -. z e. x)
5	4	a4s 682	. . . . . . . 8 |- (A.x x = z -> -. z e. x)
6	5	pm2.21d 74	. . . . . . 7 |- (A.x x = z -> (z e. x -> -. z e. y))
7	1, 6	19.21ai 740	. . . . . 6 |- (A.x x = z -> A.z(z e. x -> -. z e. y))
8	7	anim2i 270	. . . . 5 |- ((x e. y /\ A.x x = z) -> (x e. y /\ A.z(z e. x -> -. z e. y)))
9	8	exp 291	. . . 4 |- (x e. y -> (A.x x = z -> (x e. y /\ A.z(z e. x -> -. z e. y))))
10	9	com12 13	. . 3 |- (A.x x = z -> (x e. y -> (x e. y /\ A.z(z e. x -> -. z e. y))))
11	10	del42 841	. 2 |- (A.x x = z -> (E.x x e. y -> E.x(x e. y /\ A.z(z e. x -> -. z e. y))))
12		19.8a 712	. 2 |- (x e. y -> E.x x e. y)
13	11, 12	syl5 22	1 |- (A.x x = z -> (x e. y -> E.x(x e. y /\ A.z(z e. x -> -. z e. y))))

Syntax hints:  -. wn 1   -> wi 2   /\ wa 196  A.wal 672  E.wex 678   = weq 797   e. wel 803

This theorem is referenced by:  axregndlem2 3749  axregnd 3750

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-16 922  ax-17 925  ax-ext 1074  ax-rep 1075  ax-pow 1077  ax-reg 1078

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-ral 1205  df-rex 1206  df-v 1349  df-dif 1489  df-un 1490  df-in 1491  df-ss 1492  df-nul 1708  df-pw 1799  df-sn 1811  df-pr 1812

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