Theorem imdistand 451

Description: Distribution of implication with conjunction (deduction form). (Contributed by NM, 27-Aug-2004.)

Hypothesis

Ref	Expression
imdistand.1	|- ( ph -> ( ps -> ( ch -> th ) ) )

Assertion

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

Proof of Theorem imdistand

Step	Hyp	Ref	Expression
1		imdistand.1	. 2 |- ( ph -> ( ps -> ( ch -> th ) ) )
2		imdistan 448	. 2 |- ( ( ps -> ( ch -> th ) ) <-> ( ( ps /\ ch ) -> ( ps /\ th ) ) )
3	1, 2	sylib 122	1 |- ( ph -> ( ( ps /\ ch ) -> ( ps /\ th ) ) )

This proof depends on syntax axioms:   -> wi 4   /\ wa 104

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108

This proof depends on definitions:  df-bi 117

This theorem is used by:  imdistanda  452  pm5.32d  454  a2and  564  fconstfvm  5933  lbzbi  10018