Theorem axregndlem1 4954

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 1145	. . 3 |- (A.x x = z -> A.xA.x x = z)
2		hbae 1145	. . . . . 6 |- (A.x x = z -> A.zA.x x = z)
3		elirrv 4598	. . . . . . . . 9 |- -. x e. x
4		elequ1 1136	. . . . . . . . 9 |- (x = z -> (x e. x <-> z e. x))
5	3, 4	mtbii 716	. . . . . . . 8 |- (x = z -> -. z e. x)
6	5	a4s 984	. . . . . . 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 998	. . . . 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 1062	. 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 1029	. 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 954   = wceq 956   e. wcel 958  E.wex 980

This theorem is referenced by:  axregndlem2 4955  axregnd 4956

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 962  ax-gen 963  ax-8 964  ax-10 966  ax-11 967  ax-12 968  ax-13 969  ax-14 970  ax-17 971  ax-4 973  ax-5o 975  ax-6o 978  ax-9o 1123  ax-10o 1140  ax-16 1210  ax-11o 1218  ax-ext 1459  ax-sep 2703  ax-pow 2742  ax-reg 4593

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 981  df-sb 1172  df-eu 1382  df-mo 1383  df-clab 1464  df-cleq 1469  df-clel 1472  df-ne 1587  df-ral 1649  df-rex 1650  df-v 1812  df-dif 2049  df-un 2050  df-in 2051  df-ss 2053  df-nul 2281  df-pw 2402  df-sn 2412  df-pr 2413

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