Theorem sban 889

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

Assertion

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

Proof of Theorem sban

Step	Hyp	Ref	Expression
1		sbn 882	. . 3 |- ([y / x] -. (ph -> -. ps) <-> -. [y / x](ph -> -. ps))
2		sbim 886	. . . . 5 |- ([y / x](ph -> -. ps) <-> ([y / x]ph -> [y / x] -. ps))
3		sbn 882	. . . . . 6 |- ([y / x] -. ps <-> -. [y / x]ps)
4	3	imbi2i 160	. . . . 5 |- (([y / x]ph -> [y / x] -. ps) <-> ([y / x]ph -> -. [y / x]ps))
5	2, 4	bitr 151	. . . 4 |- ([y / x](ph -> -. ps) <-> ([y / x]ph -> -. [y / x]ps))
6	5	negbii 162	. . 3 |- (-. [y / x](ph -> -. ps) <-> -. ([y / x]ph -> -. [y / x]ps))
7	1, 6	bitr 151	. 2 |- ([y / x] -. (ph -> -. ps) <-> -. ([y / x]ph -> -. [y / x]ps))
8		df-an 198	. . 3 |- ((ph /\ ps) <-> -. (ph -> -. ps))
9	8	bisb 855	. 2 |- ([y / x](ph /\ ps) <-> [y / x] -. (ph -> -. ps))
10		df-an 198	. 2 |- (([y / x]ph /\ [y / x]ps) <-> -. ([y / x]ph -> -. [y / x]ps))
11	7, 9, 10	3bitr4 158	1 |- ([y / x](ph /\ ps) <-> ([y / x]ph /\ [y / x]ps))

Syntax hints:  -. wn 1   -> wi 2   <-> wb 127   /\ wa 196  [wsb 852

This theorem is referenced by:  sbbi 890  sbabel 1189  sbcan 1461  inab 1692  exss 1881  inopab 2495

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853

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