Theorem sbrbis 2073

Description: Introduce right biconditional inside of a substitution. (Contributed by NM, 18-Aug-1993.)

Hypothesis

Ref	Expression
sbrbis.1	|- ([y / x]ph <-> ps)

Assertion

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

Proof of Theorem sbrbis

Step	Hyp	Ref	Expression
1		sbbi 2071	. 2 |- ([y / x](ph <-> ch) <-> ([y / x]ph <-> [y / x]ch))
2		sbrbis.1	. . 3 |- ([y / x]ph <-> ps)
3	2	bibi1i 305	. 2 |- (([y / x]ph <-> [y / x]ch) <-> (ps <-> [y / x]ch))
4	1, 3	bitri 240	1 |- ([y / x](ph <-> ch) <-> (ps <-> [y / x]ch))

Syntax hints:   <-> wb 176  [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:  sbrbif  2074  sbabel  2515