Theorem sbbid 1248

Description: Deduction substituting both sides of a biconditional.

Hypotheses

Ref	Expression
sbbid.1	|- (ph -> A.xph)
sbbid.2	|- (ph -> (ps <-> ch))

Assertion

Ref	Expression
sbbid	|- (ph -> ([y / x]ps <-> [y / x]ch))

Proof of Theorem sbbid

Step	Hyp	Ref	Expression
1		sbbid.1	. . 3 |- (ph -> A.xph)
2		sbbid.2	. . 3 |- (ph -> (ps <-> ch))
3	1, 2	19.21ai 1000	. 2 |- (ph -> A.x(ps <-> ch))
4		a4sbbi 1247	. 2 |- (A.x(ps <-> ch) -> ([y / x]ps <-> [y / x]ch))
5	3, 4	syl 10	1 |- (ph -> ([y / x]ps <-> [y / x]ch))

Syntax hints:   -> wi 3   <-> wb 146  A.wal 956  [wsbc 1172

This theorem is referenced by:  sbcom 1260  sbcom2 1336

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 964  ax-gen 965  ax-10 968  ax-12 970  ax-4 975  ax-5o 977  ax-6o 980  ax-9o 1125  ax-10o 1142  ax-11o 1220

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 983  df-sb 1174

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