Theorem sbbi 2071

Description: Equivalence inside and outside of a substitution are equivalent. (Contributed by NM, 5-Aug-1993.)

Assertion

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

Proof of Theorem sbbi

Step	Hyp	Ref	Expression
1		dfbi2 609	. . 3 |- ((ph <-> ps) <-> ((ph -> ps) /\ (ps -> ph)))
2	1	sbbii 1653	. 2 |- ([y / x](ph <-> ps) <-> [y / x]((ph -> ps) /\ (ps -> ph)))
3		sbim 2065	. . . 4 |- ([y / x](ph -> ps) <-> ([y / x]ph -> [y / x]ps))
4		sbim 2065	. . . 4 |- ([y / x](ps -> ph) <-> ([y / x]ps -> [y / x]ph))
5	3, 4	anbi12i 678	. . 3 |- (([y / x](ph -> ps) /\ [y / x](ps -> ph)) <-> (([y / x]ph -> [y / x]ps) /\ ([y / x]ps -> [y / x]ph)))
6		sban 2069	. . 3 |- ([y / x]((ph -> ps) /\ (ps -> ph)) <-> ([y / x](ph -> ps) /\ [y / x](ps -> ph)))
7		dfbi2 609	. . 3 |- (([y / x]ph <-> [y / x]ps) <-> (([y / x]ph -> [y / x]ps) /\ ([y / x]ps -> [y / x]ph)))
8	5, 6, 7	3bitr4i 268	. 2 |- ([y / x]((ph -> ps) /\ (ps -> ph)) <-> ([y / x]ph <-> [y / x]ps))
9	2, 8	bitri 240	1 |- ([y / x](ph <-> ps) <-> ([y / x]ph <-> [y / x]ps))

Syntax hints:   -> wi 4   <-> wb 176   /\ wa 358  [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:  sblbis  2072  sbrbis  2073  spsbbi  2077  sbco  2083  sbidm  2085  sbal  2127  sb8eu  2222  pm13.183  2980  sbcbig  3093  sb8iota  4347