Theorem axregndlem1 4934

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		hbae 1143	. . 3 |- (A.x x = z -> A.xA.x x = z)
2		hbae 1143	. . . . . 6 |- (A.x x = z -> A.zA.x x = z)
3		elirrv 4578	. . . . . . . . 9 |- -. x e. x
4		elequ1 1134	. . . . . . . . 9 |- (x = z -> (x e. x <-> z e. x))
5	3, 4	mtbii 715	. . . . . . . 8 |- (x = z -> -. z e. x)
6	5	a4s 982	. . . . . . 7 |- (A.x x = z -> -. z e. x)
7	6	pm2.21d 78	. . . . . 6 |- (A.x x = z -> (z e. x -> -. z e. y))
8	2, 7	19.21ai 996	. . . . 5 |- (A.x x = z -> A.z(z e. x -> -. z e. y))
9	8	anim2i 335	. . . 4 |- ((x e. y /\ A.x x = z) -> (x e. y /\ A.z(z e. x -> -. z e. y)))
10	9	expcom 374	. . 3 |- (A.x x = z -> (x e. y -> (x e. y /\ A.z(z e. x -> -. z e. y))))
11	1, 10	19.22d 1060	. 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 1027	. 2 |- (x e. y -> E.x x e. y)
13	11, 12	syl5 21	1 |- (A.x x = z -> (x e. y -> E.x(x e. y /\ A.z(z e. x -> -. z e. y))))

Syntax hints:  -. wn 2   -> wi 3   /\ wa 223  A.wal 952   = wceq 954   e. wcel 956  E.wex 978

This theorem is referenced by:  axregndlem2 4935  axregnd 4936

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 960  ax-gen 961  ax-8 962  ax-10 964  ax-11 965  ax-12 966  ax-13 967  ax-14 968  ax-17 969  ax-4 971  ax-5o 973  ax-6o 976  ax-9o 1121  ax-10o 1138  ax-16 1208  ax-11o 1216  ax-ext 1457  ax-sep 2698  ax-pow 2737  ax-reg 4573

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 979  df-sb 1170  df-eu 1380  df-mo 1381  df-clab 1462  df-cleq 1467  df-clel 1470  df-ne 1584  df-ral 1646  df-rex 1647  df-v 1808  df-dif 2045  df-un 2046  df-in 2047  df-ss 2049  df-nul 2277  df-pw 2398  df-sn 2408  df-pr 2409

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