Theorem baibr 507

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 506	. 2 |- (ps -> (ph <-> ch))
3	2	bicomd 399	1 |- (ps -> (ch <-> ph))

Syntax hints:   -> wi 2   <-> wb 127   /\ wa 196

This theorem is referenced by:  exmoeu2 1040  ssnelpss 1751  canth 2945  kmlem14 3593  iscard 3659  cvexchlem 5759

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-an 198

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