Metamath Proof Explorer


Theorem syldd

Description: Nested syllogism deduction. Deduction associated with syld . Double deduction associated with syl . (Contributed by NM, 12-Dec-2004) (Proof shortened by Wolf Lammen, 11-May-2013)

Ref Expression
Hypotheses syldd.1
|- ( ph -> ( ps -> ( ch -> th ) ) )
syldd.2
|- ( ph -> ( ps -> ( th -> ta ) ) )
Assertion syldd
|- ( ph -> ( ps -> ( ch -> ta ) ) )

Proof

Step Hyp Ref Expression
1 syldd.1
 |-  ( ph -> ( ps -> ( ch -> th ) ) )
2 syldd.2
 |-  ( ph -> ( ps -> ( th -> ta ) ) )
3 imim2
 |-  ( ( th -> ta ) -> ( ( ch -> th ) -> ( ch -> ta ) ) )
4 2 1 3 syl6c
 |-  ( ph -> ( ps -> ( ch -> ta ) ) )