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Metamath Proof Explorer

Theorem releqd

Description: Equality deduction for the relation predicate. (Contributed by NM, 8-Mar-2014)

		Ref	Expression
	Hypothesis	releqd.1	$⊢ φ \to A = B$
	Assertion	releqd	$⊢ φ \to (Rel A \leftrightarrow Rel B)$

Step	Hyp	Ref	Expression
1		releqd.1	$⊢ φ \to A = B$
2		releq	$⊢ A = B \to (Rel A \leftrightarrow Rel B)$
3	1 2	syl	$⊢ φ \to (Rel A \leftrightarrow Rel B)$