Theorem imdistanda 15652

Description: Distribution of implication with conjunction (deduction version with conjoined antecedent).

Hypothesis

Ref	Expression
imdistanda.1	|- ((ph /\ ps) -> (ch -> th))

Assertion

Ref	Expression
imdistanda	|- (ph -> ((ps /\ ch) -> (ps /\ th)))

Proof of Theorem imdistanda

Step	Hyp	Ref	Expression
1		imdistanda.1	. . 3 |- ((ph /\ ps) -> (ch -> th))
2	1	ex 402	. 2 |- (ph -> (ps -> (ch -> th)))
3	2	imdistand 493	1 |- (ph -> ((ps /\ ch) -> (ps /\ th)))

Syntax hints:   -> wi 3   /\ wa 240

This theorem is referenced by:  isdivrng3 16112  keridl 16180  pmapjat 17314

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 164  df-an 242

Copyright terms: Public domain