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Metamath Proof Explorer

Theorem opeq1d

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

Ref Expression
Hypothesis opeq1d.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
Assertion opeq1d ( 𝜑 → ⟨ 𝐴 , 𝐶 ⟩ = ⟨ 𝐵 , 𝐶 ⟩ )

Proof

Step Hyp Ref Expression
1 opeq1d.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
2 opeq1 ⊢ ( 𝐴 = 𝐵 → ⟨ 𝐴 , 𝐶 ⟩ = ⟨ 𝐵 , 𝐶 ⟩ )
3 1 2 syl ⊢ ( 𝜑 → ⟨ 𝐴 , 𝐶 ⟩ = ⟨ 𝐵 , 𝐶 ⟩ )