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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 → 𝐴 = 𝐵 )
	Assertion	eqcomd	⊢ ( 𝜑 → 𝐵 = 𝐴 )

Proof

Step	Hyp	Ref	Expression
1		eqcomd.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
2		eqid	⊢ 𝐴 = 𝐴
3	1	eqeq1d	⊢ ( 𝜑 → ( 𝐴 = 𝐴 ↔ 𝐵 = 𝐴 ) )
4	2 3	mpbii	⊢ ( 𝜑 → 𝐵 = 𝐴 )