Theorem sbbi 1237

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		bi 514	. . 3 |- ((ph <-> ps) <-> ((ph -> ps) /\ (ps -> ph)))
2	1	sbbii 1172	. 2 |- ([y / x](ph <-> ps) <-> [y / x]((ph -> ps) /\ (ps -> ph)))
3		sbim 1232	. . . 4 |- ([y / x](ph -> ps) <-> ([y / x]ph -> [y / x]ps))
4		sbim 1232	. . . 4 |- ([y / x](ps -> ph) <-> ([y / x]ps -> [y / x]ph))
5	3, 4	anbi12i 482	. . 3 |- (([y / x](ph -> ps) /\ [y / x](ps -> ph)) <-> (([y / x]ph -> [y / x]ps) /\ ([y / x]ps -> [y / x]ph)))
6		sban 1235	. . 3 |- ([y / x]((ph -> ps) /\ (ps -> ph)) <-> ([y / x](ph -> ps) /\ [y / x](ps -> ph)))
7		bi 514	. . 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 1168

This theorem is referenced by:  sblbis 1238  sbrbis 1239  a4sbbi 1243  sbco 1250  equsb3lem 1327  elsb3 1329  sbal 1345  sbcbidig 1969

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 960  ax-gen 961  ax-10 964  ax-12 966  ax-4 971  ax-5o 973  ax-6o 976  ax-9o 1121  ax-10o 1138  ax-11o 1216

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 979  df-sb 1170

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