Theorem inteqd

Description: Equality deduction for class intersection. (Contributed by NM, 2-Sep-2003)

	Ref	Expression
Hypothesis	inteqd.1	|- ( ph -> A = B )
Assertion	inteqd	|- ( ph -> |^| A = |^| B )

Proof

Step	Hyp	Ref	Expression
1		inteqd.1	|- ( ph -> A = B )
2		inteq	|- ( A = B -> |^| A = |^| B )
3	1 2	syl	|- ( ph -> |^| A = |^| B )