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Metamath Proof Explorer

Theorem eqcomd

Description: Deduction from commutative law for class equality. (Contributed by NM, 15-Aug-1994) Allow shortening of eqcom . (Revised by Wolf Lammen, 19-Nov-2019)

		Ref	Expression
	Hypothesis	eqcomd.1	$⊢ φ \to A = B$
	Assertion	eqcomd	$⊢ φ \to B = A$

Proof

Step	Hyp	Ref	Expression
1		eqcomd.1	$⊢ φ \to A = B$
2		eqid	$⊢ A = A$
3	1	eqeq1d	$⊢ φ \to (A = A \leftrightarrow B = A)$
4	2 3	mpbii	$⊢ φ \to B = A$