Theorem sbbid 1250

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 1002	. 2 |- (ph -> A.x(ps <-> ch))
4		a4sbbi 1249	. 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 958  [wsbc 1174

This theorem is referenced by:  sbcom 1262  sbcom2 1338

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 966  ax-gen 967  ax-10 970  ax-12 972  ax-4 977  ax-5o 979  ax-6o 982  ax-9o 1127  ax-10o 1144  ax-11o 1222

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 985  df-sb 1176

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