Theorem alexeq 1409

Description: Two ways of expressing substitution of A for x in ph.

Hypothesis

Ref	Expression
alexeq.1	|- A e. V

Assertion

Ref	Expression
alexeq	|- (A.x(x = A -> ph) <-> E.x(x = A /\ ph))

Distinct variable group(s):   x,A

Proof of Theorem alexeq

Step	Hyp	Ref	Expression
1		alexeq.1	. 2 |- A e. V
2		cleq2 1110	. . . 4 |- (y = A -> (x = y <-> x = A))
3	2	imbi1d 465	. . 3 |- (y = A -> ((x = y -> ph) <-> (x = A -> ph)))
4	3	bialdv 935	. 2 |- (y = A -> (A.x(x = y -> ph) <-> A.x(x = A -> ph)))
5	2	anbi1d 469	. . 3 |- (y = A -> ((x = y /\ ph) <-> (x = A /\ ph)))
6	5	biexdv 936	. 2 |- (y = A -> (E.x(x = y /\ ph) <-> E.x(x = A /\ ph)))
7		eqs4 831	. . 3 |- (A.x(x = y -> ph) -> E.x(x = y /\ ph))
8		eq5 824	. . . . 5 |- (A.x x = y -> A.xA.x x = y)
9		hba1 698	. . . . 5 |- (A.x(x = y -> ph) -> A.xA.x(x = y -> ph))
10		ax-16 922	. . . . . 6 |- (A.x x = y -> ((x = y -> ph) -> A.x(x = y -> ph)))
11		pm3.4 266	. . . . . 6 |- ((x = y /\ ph) -> (x = y -> ph))
12	10, 11	syl5 22	. . . . 5 |- (A.x x = y -> ((x = y /\ ph) -> A.x(x = y -> ph)))
13	8, 9, 12	19.23ad 748	. . . 4 |- (A.x x = y -> (E.x(x = y /\ ph) -> A.x(x = y -> ph)))
14		eqs5 832	. . . 4 |- (-. A.x x = y -> (E.x(x = y /\ ph) -> A.x(x = y -> ph)))
15	13, 14	pm2.61i 110	. . 3 |- (E.x(x = y /\ ph) -> A.x(x = y -> ph))
16	7, 15	impbi 139	. 2 |- (A.x(x = y -> ph) <-> E.x(x = y /\ ph))
17	1, 4, 6, 16	vtoclb 1381	1 |- (A.x(x = A -> ph) <-> E.x(x = A /\ ph))

Syntax hints:   -> wi 2   <-> wb 127   /\ wa 196  A.wal 672  E.wex 678   = weq 797   = wceq 1091   e. wcel 1092  Vcvv 1348

This theorem is referenced by:  ceqex 1410

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-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-v 1349

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