Theorem sban 2069

Description: Conjunction inside and outside of a substitution are equivalent. (Contributed by NM, 5-Aug-1993.)

Assertion

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

Proof of Theorem sban

Step	Hyp	Ref	Expression
1		sbn 2062	. . 3 |- ([y / x] -. (ph -> -. ps) <-> -. [y / x](ph -> -. ps))
2		sbim 2065	. . . 4 |- ([y / x](ph -> -. ps) <-> ([y / x]ph -> [y / x] -. ps))
3		sbn 2062	. . . . 5 |- ([y / x] -. ps <-> -. [y / x]ps)
4	3	imbi2i 303	. . . 4 |- (([y / x]ph -> [y / x] -. ps) <-> ([y / x]ph -> -. [y / x]ps))
5	2, 4	bitri 240	. . 3 |- ([y / x](ph -> -. ps) <-> ([y / x]ph -> -. [y / x]ps))
6	1, 5	xchbinx 301	. 2 |- ([y / x] -. (ph -> -. ps) <-> -. ([y / x]ph -> -. [y / x]ps))
7		df-an 360	. . 3 |- ((ph /\ ps) <-> -. (ph -> -. ps))
8	7	sbbii 1653	. 2 |- ([y / x](ph /\ ps) <-> [y / x] -. (ph -> -. ps))
9		df-an 360	. 2 |- (([y / x]ph /\ [y / x]ps) <-> -. ([y / x]ph -> -. [y / x]ps))
10	6, 8, 9	3bitr4i 268	1 |- ([y / x](ph /\ ps) <-> ([y / x]ph /\ [y / x]ps))

Syntax hints:  -. wn 3   -> wi 4   <-> wb 176   /\ wa 358  [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-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649

This theorem is referenced by:  sb3an  2070  sbbi  2071  sbabel  2516  cbvreu  2834  sbcan  3089  sbcang  3090  rmo3  3134  inab  3523  difab  3524