Theorem syland 277

Description: A syllogism deduction. (Contributed by NM, 15-Dec-2004.)

Hypotheses

Ref	Expression
syland.1	|- ( ph -> ( ps -> ch ) )
syland.2	|- ( ph -> ( ( ch /\ th ) -> ta ) )

Assertion

Ref	Expression
syland	|- ( ph -> ( ( ps /\ th ) -> ta ) )

Proof of Theorem syland

Step	Hyp	Ref	Expression
1		syland.1	. . 3 |- ( ph -> ( ps -> ch ) )
2		syland.2	. . . 4 |- ( ph -> ( ( ch /\ th ) -> ta ) )
3	2	expd 245	. . 3 |- ( ph -> ( ch -> ( th -> ta ) ) )
4	1, 3	syld 40	. 2 |- ( ph -> ( ps -> ( th -> ta ) ) )
5	4	impd 242	1 |- ( ph -> ( ( ps /\ th ) -> ta ) )

Syntax hints:   -> wi 4   /\ wa 97

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101

This theorem is referenced by:  sylan2d  278  syl2and  279  sylani  386  nn0seqcvgd  9880