Theorem nbbn 498

Description: Move negation outside of biconditional. Compare Theorem *5.18 of [WhiteheadRussell] p. 124.

Assertion

Ref	Expression
nbbn	|- ((-. ph <-> ps) <-> -. (ph <-> ps))

Proof of Theorem nbbn

Step	Hyp	Ref	Expression
1		bicom 398	. 2 |- ((-. ph <-> ps) <-> (ps <-> -. ph))
2		bicom 398	. . . 4 |- ((ph <-> ps) <-> (ps <-> ph))
3		pm5.18 497	. . . 4 |- ((ps <-> ph) <-> -. (ps <-> -. ph))
4	2, 3	bitr 151	. . 3 |- ((ph <-> ps) <-> -. (ps <-> -. ph))
5	4	bicon2i 194	. 2 |- ((ps <-> -. ph) <-> -. (ph <-> ps))
6	1, 5	bitr 151	1 |- ((-. ph <-> ps) <-> -. (ph <-> ps))

Syntax hints:  -. wn 1   <-> wb 127

This theorem is referenced by:  xor 500  symdif2 1690  canth 2945

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198

metamath.org