Theorem reu4 1340

Description: Restricted uniqueness using implicit substitution.

Hypothesis

Ref	Expression
reu4.1	|- (x = y -> (ph <-> ps))

Assertion

Ref	Expression
reu4	|- (E!x e. A ph <-> (E.x e. A ph /\ A.x e. A A.y e. A ((ph /\ ps) -> x = y)))

Distinct variable group(s):   x,y,A   ph,y   ps,x

Proof of Theorem reu4

Step	Hyp	Ref	Expression
1		reu2 1338	. 2 |- (E!x e. A ph <-> (E.x e. A ph /\ A.x e. A A.y e. A ((ph /\ [y / x]ph) -> x = y)))
2		ax-17 925	. . . . . . 7 |- (ps -> A.xps)
3		reu4.1	. . . . . . 7 |- (x = y -> (ph <-> ps))
4	2, 3	sbie 904	. . . . . 6 |- ([y / x]ph <-> ps)
5	4	anbi2i 367	. . . . 5 |- ((ph /\ [y / x]ph) <-> (ph /\ ps))
6	5	imbi1i 161	. . . 4 |- (((ph /\ [y / x]ph) -> x = y) <-> ((ph /\ ps) -> x = y))
7	6	bi2ral 1225	. . 3 |- (A.x e. A A.y e. A ((ph /\ [y / x]ph) -> x = y) <-> A.x e. A A.y e. A ((ph /\ ps) -> x = y))
8	7	anbi2i 367	. 2 |- ((E.x e. A ph /\ A.x e. A A.y e. A ((ph /\ [y / x]ph) -> x = y)) <-> (E.x e. A ph /\ A.x e. A A.y e. A ((ph /\ ps) -> x = y)))
9	1, 8	bitr 151	1 |- (E!x e. A ph <-> (E.x e. A ph /\ A.x e. A A.y e. A ((ph /\ ps) -> x = y)))

Syntax hints:   -> wi 2   <-> wb 127   /\ wa 196   = weq 797  [wsb 852  A.wral 1201  E.wrex 1202  E!wreu 1203

This theorem is referenced by:  wereu 2197  oawordeulem 3156  negeu 4124  receu 4215  creur 4532  creui 4533  uzwo2 4606  hlimreu 5145  pjthu 5241  pjthu2 5242

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-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853  df-eu 1009  df-cleq 1097  df-clel 1099  df-ral 1205  df-rex 1206  df-reu 1207

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