Theorem sbc3ang 1975

Description: Distribution of class substitution over triple conjunction.

Assertion

Ref	Expression
sbc3ang	|- (A e. B -> ([A / x](ph /\ ps /\ ch) <-> ([A / x]ph /\ [A / x]ps /\ [A / x]ch)))

Proof of Theorem sbc3ang

Step	Hyp	Ref	Expression
1		df-3an 776	. . 3 |- ((ph /\ ps /\ ch) <-> ((ph /\ ps) /\ ch))
2	1	sbcbii 1974	. 2 |- (A e. B -> ([A / x](ph /\ ps /\ ch) <-> [A / x]((ph /\ ps) /\ ch)))
3		sbcang 1967	. 2 |- (A e. B -> ([A / x]((ph /\ ps) /\ ch) <-> ([A / x](ph /\ ps) /\ [A / x]ch)))
4		sbcang 1967	. . . 4 |- (A e. B -> ([A / x](ph /\ ps) <-> ([A / x]ph /\ [A / x]ps)))
5	4	anbi1d 616	. . 3 |- (A e. B -> (([A / x](ph /\ ps) /\ [A / x]ch) <-> (([A / x]ph /\ [A / x]ps) /\ [A / x]ch)))
6		df-3an 776	. . 3 |- (([A / x]ph /\ [A / x]ps /\ [A / x]ch) <-> (([A / x]ph /\ [A / x]ps) /\ [A / x]ch))
7	5, 6	syl6bbr 537	. 2 |- (A e. B -> (([A / x](ph /\ ps) /\ [A / x]ch) <-> ([A / x]ph /\ [A / x]ps /\ [A / x]ch)))
8	2, 3, 7	3bitrd 543	1 |- (A e. B -> ([A / x](ph /\ ps /\ ch) <-> ([A / x]ph /\ [A / x]ps /\ [A / x]ch)))

Syntax hints:   -> wi 3   <-> wb 146   /\ wa 223   /\ w3a 774   e. wcel 956  [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-8 962  ax-10 964  ax-12 966  ax-17 969  ax-4 971  ax-5o 973  ax-6o 976  ax-9o 1121  ax-10o 1138  ax-11o 1216  ax-ext 1457

This theorem depends on definitions:  df-bi 147  df-an 225  df-3an 776  df-ex 979  df-sb 1170  df-clab 1462  df-cleq 1467  df-clel 1470  df-v 1808  df-sbc 1938

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