Theorem sbrbif 1242

Description: Introduce right biconditional inside of a substitution.

Hypotheses

Ref	Expression
sbrbif.1	|- (ch -> A.xch)
sbrbif.2	|- ([y / x]ph <-> ps)

Assertion

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

Proof of Theorem sbrbif

Step	Hyp	Ref	Expression
1		sbrbif.2	. . 3 |- ([y / x]ph <-> ps)
2	1	sbrbis 1241	. 2 |- ([y / x](ph <-> ch) <-> (ps <-> [y / x]ch))
3		sbrbif.1	. . . 4 |- (ch -> A.xch)
4	3	sbf 1186	. . 3 |- ([y / x]ch <-> ch)
5	4	bibi2i 608	. 2 |- ((ps <-> [y / x]ch) <-> (ps <-> ch))
6	2, 5	bitr 173	1 |- ([y / x](ph <-> ch) <-> (ps <-> ch))

Syntax hints:   -> wi 3   <-> wb 146  A.wal 954  [wsbc 1170

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 962  ax-gen 963  ax-10 966  ax-12 968  ax-4 973  ax-5o 975  ax-6o 978  ax-9o 1123  ax-10o 1140  ax-11o 1218

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 981  df-sb 1172

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