Theorem sbrbif 2074

Description: Introduce right biconditional inside of a substitution. (Contributed by NM, 18-Aug-1993.) (Revised by Mario Carneiro, 4-Oct-2016.)

Hypotheses

Ref	Expression
sbrbif.1	|- F/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 2073	. 2 |- ([y / x](ph <-> ch) <-> (ps <-> [y / x]ch))
3		sbrbif.1	. . . 4 |- F/xch
4	3	sbf 2026	. . 3 |- ([y / x]ch <-> ch)
5	4	bibi2i 304	. 2 |- ((ps <-> [y / x]ch) <-> (ps <-> ch))
6	2, 5	bitri 240	1 |- ([y / x](ph <-> ch) <-> (ps <-> ch))

Syntax hints:   <-> wb 176   F/wnf 1544  [wsb 1648

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925

This theorem depends on definitions:  df-bi 177  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649

This theorem is referenced by: (None)