Theorem sb3an 2070

Description: Conjunction inside and outside of a substitution are equivalent. (Contributed by NM, 14-Dec-2006.)

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 936	. . 3 |- ((ph /\ ps /\ ch) <-> ((ph /\ ps) /\ ch))
2	1	sbbii 1653	. 2 |- ([y / x](ph /\ ps /\ ch) <-> [y / x]((ph /\ ps) /\ ch))
3		sban 2069	. 2 |- ([y / x]((ph /\ ps) /\ ch) <-> ([y / x](ph /\ ps) /\ [y / x]ch))
4		sban 2069	. . . 4 |- ([y / x](ph /\ ps) <-> ([y / x]ph /\ [y / x]ps))
5	4	anbi1i 676	. . 3 |- (([y / x](ph /\ ps) /\ [y / x]ch) <-> (([y / x]ph /\ [y / x]ps) /\ [y / x]ch))
6		df-3an 936	. . 3 |- (([y / x]ph /\ [y / x]ps /\ [y / x]ch) <-> (([y / x]ph /\ [y / x]ps) /\ [y / x]ch))
7	5, 6	bitr4i 243	. 2 |- (([y / x](ph /\ ps) /\ [y / x]ch) <-> ([y / x]ph /\ [y / x]ps /\ [y / x]ch))
8	2, 3, 7	3bitri 262	1 |- ([y / x](ph /\ ps /\ ch) <-> ([y / x]ph /\ [y / x]ps /\ [y / x]ch))

Syntax hints:   <-> wb 176   /\ wa 358   /\ w3a 934  [wsb 1648

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925

This theorem depends on definitions:  df-bi 177  df-an 360  df-3an 936  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649

This theorem is referenced by:  sbc3ang  3105