Theorem sbc3ang 1969

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 775	. . 3 |- ((ph /\ ps /\ ch) <-> ((ph /\ ps) /\ ch))
2	1	sbcbii 1968	. 2 |- (A e. B -> ([A / x](ph /\ ps /\ ch) <-> [A / x]((ph /\ ps) /\ ch)))
3		sbcang 1961	. 2 |- (A e. B -> ([A / x]((ph /\ ps) /\ ch) <-> ([A / x](ph /\ ps) /\ [A / x]ch)))
4		sbcang 1961	. . . 4 |- (A e. B -> ([A / x](ph /\ ps) <-> ([A / x]ph /\ [A / x]ps)))
5	4	anbi1d 615	. . 3 |- (A e. B -> (([A / x](ph /\ ps) /\ [A / x]ch) <-> (([A / x]ph /\ [A / x]ps) /\ [A / x]ch)))
6		df-3an 775	. . 3 |- (([A / x]ph /\ [A / x]ps /\ [A / x]ch) <-> (([A / x]ph /\ [A / x]ps) /\ [A / x]ch))
7	5, 6	syl6bbr 536	. 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 542	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 773   e. wcel 955  [wsbc 1166

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 959  ax-gen 960  ax-8 961  ax-10 963  ax-12 965  ax-17 968  ax-4 970  ax-5o 972  ax-6o 975  ax-9o 1119  ax-10o 1136  ax-11o 1213  ax-ext 1452

This theorem depends on definitions:  df-bi 147  df-an 225  df-3an 775  df-ex 978  df-sb 1168  df-clab 1457  df-cleq 1462  df-clel 1465  df-v 1803  df-sbc 1932

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