Metamath Proof Explorer


Theorem sylan9ss

Description: A subclass transitivity deduction. (Contributed by NM, 27-Sep-2004) (Proof shortened by Andrew Salmon, 14-Jun-2011)

Ref Expression
Hypotheses sylan9ss.1
|- ( ph -> A C_ B )
sylan9ss.2
|- ( ps -> B C_ C )
Assertion sylan9ss
|- ( ( ph /\ ps ) -> A C_ C )

Proof

Step Hyp Ref Expression
1 sylan9ss.1
 |-  ( ph -> A C_ B )
2 sylan9ss.2
 |-  ( ps -> B C_ C )
3 sstr
 |-  ( ( A C_ B /\ B C_ C ) -> A C_ C )
4 1 2 3 syl2an
 |-  ( ( ph /\ ps ) -> A C_ C )