Theorem sbbi 1243

Description: Equivalence inside and outside of a substitution are equivalent.

Assertion

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

Proof of Theorem sbbi

Step	Hyp	Ref	Expression
1		dfbi2 517	. . 3 |- ((ph <-> ps) <-> ((ph -> ps) /\ (ps -> ph)))
2	1	sbbii 1178	. 2 |- ([y / x](ph <-> ps) <-> [y / x]((ph -> ps) /\ (ps -> ph)))
3		sbim 1238	. . . 4 |- ([y / x](ph -> ps) <-> ([y / x]ph -> [y / x]ps))
4		sbim 1238	. . . 4 |- ([y / x](ps -> ph) <-> ([y / x]ps -> [y / x]ph))
5	3, 4	anbi12i 485	. . 3 |- (([y / x](ph -> ps) /\ [y / x](ps -> ph)) <-> (([y / x]ph -> [y / x]ps) /\ ([y / x]ps -> [y / x]ph)))
6		sban 1241	. . 3 |- ([y / x]((ph -> ps) /\ (ps -> ph)) <-> ([y / x](ph -> ps) /\ [y / x](ps -> ph)))
7		dfbi2 517	. . 3 |- (([y / x]ph <-> [y / x]ps) <-> (([y / x]ph -> [y / x]ps) /\ ([y / x]ps -> [y / x]ph)))
8	5, 6, 7	3bitr4i 183	. 2 |- ([y / x]((ph -> ps) /\ (ps -> ph)) <-> ([y / x]ph <-> [y / x]ps))
9	2, 8	bitri 173	1 |- ([y / x](ph <-> ps) <-> ([y / x]ph <-> [y / x]ps))

Syntax hints:   -> wi 3   <-> wb 146   /\ wa 223  [wsbc 1174

This theorem is referenced by:  sblbis 1244  sbrbis 1245  a4sbbi 1249  sbco 1256  equsb3lem 1333  elsb3 1335  sbal 1351  sbcbidig 1983

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 966  ax-gen 967  ax-10 970  ax-12 972  ax-4 977  ax-5o 979  ax-6o 982  ax-9o 1127  ax-10o 1144  ax-11o 1222

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 985  df-sb 1176

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