Theorem sb3an 1236

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

Assertion

Ref	Expression
sb3an	|- ([y / x](ph /\ ps /\ ch) <-> ([y / x]ph /\ [y / x]ps /\ [y / x]ch))

Proof of Theorem sb3an

Step	Hyp	Ref	Expression
1		df-3an 776	. . 3 |- ((ph /\ ps /\ ch) <-> ((ph /\ ps) /\ ch))
2	1	sbbii 1172	. 2 |- ([y / x](ph /\ ps /\ ch) <-> [y / x]((ph /\ ps) /\ ch))
3		sban 1235	. 2 |- ([y / x]((ph /\ ps) /\ ch) <-> ([y / x](ph /\ ps) /\ [y / x]ch))
4		sban 1235	. . . 4 |- ([y / x](ph /\ ps) <-> ([y / x]ph /\ [y / x]ps))
5	4	anbi1i 481	. . 3 |- (([y / x](ph /\ ps) /\ [y / x]ch) <-> (([y / x]ph /\ [y / x]ps) /\ [y / x]ch))
6		df-3an 776	. . 3 |- (([y / x]ph /\ [y / x]ps /\ [y / x]ch) <-> (([y / x]ph /\ [y / x]ps) /\ [y / x]ch))
7	5, 6	bitr4 176	. 2 |- (([y / x](ph /\ ps) /\ [y / x]ch) <-> ([y / x]ph /\ [y / x]ps /\ [y / x]ch))
8	2, 3, 7	3bitr 177	1 |- ([y / x](ph /\ ps /\ ch) <-> ([y / x]ph /\ [y / x]ps /\ [y / x]ch))

Syntax hints:   <-> wb 146   /\ wa 223   /\ w3a 774  [wsbc 1168

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-3an 776  df-ex 979  df-sb 1170

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