Theorem sbrbif 893

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 892	. 2 |- ([y / x](ph <-> ch) <-> (ps <-> [y / x]ch))
3		sbrbif.1	. . . 4 |- (ch -> A.xch)
4	3	sbf 870	. . 3 |- ([y / x]ch <-> ch)
5	4	bibi2i 460	. 2 |- ((ps <-> [y / x]ch) <-> (ps <-> ch))
6	2, 5	bitr 151	1 |- ([y / x](ph <-> ch) <-> (ps <-> ch))

Syntax hints:   -> wi 2   <-> wb 127  A.wal 672  [wsb 852

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853

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