Theorem sbhypf 2452

Description: Introduce an explicit substitution into an implicit substitution hypothesis. See also csbhypf 2572. (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 2295	. . 3 |- y e. _V
2		eqeq1 1890	. . 3 |- (x = y -> (x = A <-> y = A))
3	1, 2	ceqsexv 2325	. 2 |- (E.x(x = y /\ x = A) <-> y = A)
4		hbs1 1722	. . . 4 |- ([y / x]ph -> A.x[y / x]ph)
5		sbhypf.1	. . . 4 |- (ps -> A.xps)
6	4, 5	hbbi 1357	. . 3 |- (([y / x]ph <-> ps) -> A.x([y / x]ph <-> ps))
7		sbequ12 1545	. . . . 5 |- (x = y -> (ph <-> [y / x]ph))
8	7	bicomd 580	. . . 4 |- (x = y -> ([y / x]ph <-> ph))
9		sbhypf.2	. . . 4 |- (x = A -> (ph <-> ps))
10	8, 9	sylan9bb 599	. . 3 |- ((x = y /\ x = A) -> ([y / x]ph <-> ps))
11	6, 10	19.23ai 1412	. 2 |- (E.x(x = y /\ x = A) -> ([y / x]ph <-> ps))
12	3, 11	sylbir 218	1 |- (y = A -> ([y / x]ph <-> ps))

Syntax hints:   -> wi 3   <-> wb 163   /\ wa 240  A.wal 1296   = wceq 1298  E.wex 1326  [wsbc 1534

This theorem is referenced by:  opelopabf 3572  ralxpf 4043  ac6sf 5922  nn0ind-raph 7426  fdc1 15813

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 1304  ax-gen 1305  ax-10 1308  ax-12 1310  ax-17 1317  ax-4 1319  ax-5o 1321  ax-6o 1324  ax-9o 1481  ax-10o 1500  ax-16 1580  ax-11o 1588  ax-ext 1865

This theorem depends on definitions:  df-bi 164  df-an 242  df-ex 1327  df-sb 1536  df-clab 1872  df-cleq 1877  df-clel 1880  df-v 2294

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