Theorem baibr 685

Description: Move conjunction outside of biconditional.

Hypothesis

Ref	Expression
baibr.1	|- (ph <-> (ps /\ ch))

Assertion

Ref	Expression
baibr	|- (ps -> (ch <-> ph))

Proof of Theorem baibr

Step	Hyp	Ref	Expression
1		baibr.1	. . 3 |- (ph <-> (ps /\ ch))
2	1	baib 684	. 2 |- (ps -> (ph <-> ch))
3	2	bicomd 520	1 |- (ps -> (ch <-> ph))

Syntax hints:   -> wi 3   <-> wb 146   /\ wa 223

This theorem is referenced by:  pm5.44 686  exmoeu2 1413  ssnelpss 2327  reuunixfr 2902  brinxp 3228  canth 3902  kmlem14 4761  iscard 4836  islp2 7707  adjeqt 9816  lnopcnbdt 9921  cvexchlem 10251

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7

This theorem depends on definitions:  df-bi 147  df-an 225

Copyright terms: Public domain