Theorem opeq12d

Description: Equality deduction for ordered pairs. (Contributed by NM, 16-Dec-2006) (Proof shortened by Andrew Salmon, 29-Jun-2011)

	Ref	Expression
Hypotheses	opeq1d.1	|- ( ph -> A = B )
	opeq12d.2	|- ( ph -> C = D )
Assertion	opeq12d	|- ( ph -> <. A , C >. = <. B , D >. )

Proof

Step	Hyp	Ref	Expression
1		opeq1d.1	|- ( ph -> A = B )
2		opeq12d.2	|- ( ph -> C = D )
3		opeq12	|- ( ( A = B /\ C = D ) -> <. A , C >. = <. B , D >. )
4	1 2 3	syl2anc	|- ( ph -> <. A , C >. = <. B , D >. )