Theorem expcomd 1330

Description: Deduction form of expcom 109. (Contributed by Alan Sare, 22-Jul-2012.)

Hypothesis

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

Assertion

Ref	Expression
expcomd	|- ( ph -> ( ch -> ( ps -> th ) ) )

Proof of Theorem expcomd

Step	Hyp	Ref	Expression
1		expcomd.1	. . 3 |- ( ph -> ( ( ps /\ ch ) -> th ) )
2	1	expd 245	. 2 |- ( ph -> ( ps -> ( ch -> th ) ) )
3	2	com23 72	1 |- ( ph -> ( ch -> ( ps -> th ) ) )

Syntax hints:   -> wi 4   /\ wa 97

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia3 101

This theorem is referenced by:  simplbi2comg  1332  2moswapdc  1990  indifdir  3193  reupick  3221  trintssm  3870  issod  4056  poxp  5853  smores2  5909  smoiun  5916  recexprlemm  6722  ltleletr  7100  fzind  8353  iccid  8794  ssfzo12bi  9081