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Metamath Proof Explorer

Theorem opeq2d

Description: Equality deduction for ordered pairs. (Contributed by NM, 16-Dec-2006)

		Ref	Expression
	Hypothesis	opeq1d.1	$⊢ φ \to A = B$
	Assertion	opeq2d	$⊢ φ \to ⟨C, A⟩ = ⟨C, B⟩$

Step	Hyp	Ref	Expression
1		opeq1d.1	$⊢ φ \to A = B$
2		opeq2	$⊢ A = B \to ⟨C, A⟩ = ⟨C, B⟩$
3	1 2	syl	$⊢ φ \to ⟨C, A⟩ = ⟨C, B⟩$