Theorem sbhypf 1985

Description: Introduce an explicit substitution into an implicit substitution hypothesis. See also csbhypf 2080. (Contributed by Raph Levien, 10-Apr-2004.)

Hypotheses

Ref	Expression
sbhypf.1	|- (ps -> A.xps)
sbhypf.2	|- (x = A -> (ph <-> ps))

Assertion

Ref	Expression
sbhypf	|- (y = A -> ([y / x]ph <-> ps))

Distinct variable groups:   x,A   x,y

Proof of Theorem sbhypf

Step	Hyp	Ref	Expression
1		visset 1859	. . 3 |- y e. V
2		eqeq1 1524	. . 3 |- (x = y -> (x = A <-> y = A))
3	1, 2	ceqsexv 1881	. 2 |- (E.x(x = y /\ x = A) <-> y = A)
4		hbs1 1371	. . . 4 |- ([y / x]ph -> A.x[y / x]ph)
5		sbhypf.1	. . . 4 |- (ps -> A.xps)
6	4, 5	hbbi 1046	. . 3 |- (([y / x]ph <-> ps) -> A.x([y / x]ph <-> ps))
7		sbequ12 1218	. . . . 5 |- (x = y -> (ph <-> [y / x]ph))
8	7	bicomd 524	. . . 4 |- (x = y -> ([y / x]ph <-> ph))
9		sbhypf.2	. . . 4 |- (x = A -> (ph <-> ps))
10	8, 9	sylan9bb 543	. . 3 |- ((x = y /\ x = A) -> ([y / x]ph <-> ps))
11	6, 10	19.23ai 1100	. 2 |- (E.x(x = y /\ x = A) -> ([y / x]ph <-> ps))
12	3, 11	sylbir 199	1 |- (y = A -> ([y / x]ph <-> ps))

Syntax hints:   -> wi 3   <-> wb 144   /\ wa 221  A.wal 990   = wceq 992  E.wex 1016  [wsbc 1207

This theorem is referenced by:  opelopabf 2899  ralxpf 3303  ac6sf 4906  nn0ind-raph 6385

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 998  ax-gen 999  ax-10 1002  ax-12 1004  ax-17 1007  ax-4 1009  ax-5o 1011  ax-6o 1014  ax-9o 1159  ax-10o 1177  ax-16 1247  ax-11o 1255  ax-ext 1500

This theorem depends on definitions:  df-bi 145  df-an 223  df-ex 1017  df-sb 1209  df-clab 1506  df-cleq 1511  df-clel 1514  df-v 1858

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