Theorem sbbi 1241

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 516	. . 3 |- ((ph <-> ps) <-> ((ph -> ps) /\ (ps -> ph)))
2	1	sbbii 1176	. 2 |- ([y / x](ph <-> ps) <-> [y / x]((ph -> ps) /\ (ps -> ph)))
3		sbim 1236	. . . 4 |- ([y / x](ph -> ps) <-> ([y / x]ph -> [y / x]ps))
4		sbim 1236	. . . 4 |- ([y / x](ps -> ph) <-> ([y / x]ps -> [y / x]ph))
5	3, 4	anbi12i 484	. . 3 |- (([y / x](ph -> ps) /\ [y / x](ps -> ph)) <-> (([y / x]ph -> [y / x]ps) /\ ([y / x]ps -> [y / x]ph)))
6		sban 1239	. . 3 |- ([y / x]((ph -> ps) /\ (ps -> ph)) <-> ([y / x](ph -> ps) /\ [y / x](ps -> ph)))
7		dfbi2 516	. . 3 |- (([y / x]ph <-> [y / x]ps) <-> (([y / x]ph -> [y / x]ps) /\ ([y / x]ps -> [y / x]ph)))
8	5, 6, 7	3bitr4 183	. 2 |- ([y / x]((ph -> ps) /\ (ps -> ph)) <-> ([y / x]ph <-> [y / x]ps))
9	2, 8	bitr 173	1 |- ([y / x](ph <-> ps) <-> ([y / x]ph <-> [y / x]ps))

Syntax hints:   -> wi 3   <-> wb 146   /\ wa 223  [wsbc 1172

This theorem is referenced by:  sblbis 1242  sbrbis 1243  a4sbbi 1247  sbco 1254  equsb3lem 1331  elsb3 1333  sbal 1349  sbcbidig 1976

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 964  ax-gen 965  ax-10 968  ax-12 970  ax-4 975  ax-5o 977  ax-6o 980  ax-9o 1125  ax-10o 1142  ax-11o 1220

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 983  df-sb 1174

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