Metamath Proof Explorer


Theorem opeq12d

Description: Equality deduction for ordered pairs. (Contributed by NM, 16-Dec-2006) (Proof shortened by Andrew Salmon, 29-Jun-2011)

Ref Expression
Hypotheses opeq1d.1 ( 𝜑𝐴 = 𝐵 )
opeq12d.2 ( 𝜑𝐶 = 𝐷 )
Assertion opeq12d ( 𝜑 → ⟨ 𝐴 , 𝐶 ⟩ = ⟨ 𝐵 , 𝐷 ⟩ )

Proof

Step Hyp Ref Expression
1 opeq1d.1 ( 𝜑𝐴 = 𝐵 )
2 opeq12d.2 ( 𝜑𝐶 = 𝐷 )
3 opeq12 ( ( 𝐴 = 𝐵𝐶 = 𝐷 ) → ⟨ 𝐴 , 𝐶 ⟩ = ⟨ 𝐵 , 𝐷 ⟩ )
4 1 2 3 syl2anc ( 𝜑 → ⟨ 𝐴 , 𝐶 ⟩ = ⟨ 𝐵 , 𝐷 ⟩ )