Theorem sbcgf 1469

Description: Substitution for a variable not free in a wff does not affect it.

Hypothesis

Ref	Expression
sbcgf.1	|- (ph -> A.xph)

Assertion

Ref	Expression
sbcgf	|- (A e. B -> ([A / x]ph <-> ph))

Proof of Theorem sbcgf

Step	Hyp	Ref	Expression
1		sbcco 1448	. 2 |- (A e. B -> ([A / y][y / x]ph <-> [A / x]ph))
2		eqid 810	. . . 4 |- x = x
3		sbcgf.1	. . . . . . 7 |- (ph -> A.xph)
4	3	sbf 870	. . . . . 6 |- ([y / x]ph <-> ph)
5	4	a1i 7	. . . . 5 |- (x = x -> ([y / x]ph <-> ph))
6	5	bisbcdv 1468	. . . 4 |- ((A e. B /\ x = x) -> ([A / y][y / x]ph <-> [A / y]ph))
7	2, 6	mpan2 519	. . 3 |- (A e. B -> ([A / y][y / x]ph <-> [A / y]ph))
8		sbc5g 1450	. . 3 |- (A e. B -> ([A / y]ph <-> E.y(y = A /\ ph)))
9		elex 1356	. . . . 5 |- (A e. B -> E.y y = A)
10	9	biantrurd 546	. . . 4 |- (A e. B -> (ph <-> (E.y y = A /\ ph)))
11		19.41v 963	. . . 4 |- (E.y(y = A /\ ph) <-> (E.y y = A /\ ph))
12	10, 11	syl6rbbr 417	. . 3 |- (A e. B -> (E.y(y = A /\ ph) <-> ph))
13	7, 8, 12	3bitrd 422	. 2 |- (A e. B -> ([A / y][y / x]ph <-> ph))
14	1, 13	bitr3d 408	1 |- (A e. B -> ([A / x]ph <-> ph))

Syntax hints:   -> wi 2   <-> wb 127   /\ wa 196  A.wal 672  E.wex 678   = weq 797  [wsb 852   = wceq 1091   e. wcel 1092  [wsbc 1440

This theorem is referenced by:  sbc19.21g 1470

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-clab 1093  df-cleq 1097  df-clel 1099  df-v 1349  df-sbc 1441

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