Theorem sylan9eq

Description: An equality transitivity deduction. (Contributed by NM, 8-May-1994) (Proof shortened by Andrew Salmon, 25-May-2011)

	Ref	Expression
Hypotheses	sylan9eq.1	|- ( ph -> A = B )
	sylan9eq.2	|- ( ps -> B = C )
Assertion	sylan9eq	|- ( ( ph /\ ps ) -> A = C )

Proof

Step	Hyp	Ref	Expression
1		sylan9eq.1	|- ( ph -> A = B )
2		sylan9eq.2	|- ( ps -> B = C )
3		eqtr	|- ( ( A = B /\ B = C ) -> A = C )
4	1 2 3	syl2an	|- ( ( ph /\ ps ) -> A = C )