Metamath Proof Explorer


Theorem opthg

Description: Ordered pair theorem. C and D are not required to be sets under our specific ordered pair definition. (Contributed by NM, 14-Oct-2005) (Revised by Mario Carneiro, 26-Apr-2015)

Ref Expression
Assertion opthg ( ( 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ) → ( ⟨ 𝐴 , 𝐵 ⟩ = ⟨ 𝐶 , 𝐷 ⟩ ↔ ( 𝐴 = 𝐶 ∧ 𝐵 = 𝐷 ) ) )

Proof

Step Hyp Ref Expression
1 opeq1 ⊢ ( 𝑥 = 𝐴 → ⟨ 𝑥 , 𝑦 ⟩ = ⟨ 𝐴 , 𝑦 ⟩ )
2 1 eqeq1d ⊢ ( 𝑥 = 𝐴 → ( ⟨ 𝑥 , 𝑦 ⟩ = ⟨ 𝐶 , 𝐷 ⟩ ↔ ⟨ 𝐴 , 𝑦 ⟩ = ⟨ 𝐶 , 𝐷 ⟩ ) )
3 eqeq1 ⊢ ( 𝑥 = 𝐴 → ( 𝑥 = 𝐶 ↔ 𝐴 = 𝐶 ) )
4 3 anbi1d ⊢ ( 𝑥 = 𝐴 → ( ( 𝑥 = 𝐶 ∧ 𝑦 = 𝐷 ) ↔ ( 𝐴 = 𝐶 ∧ 𝑦 = 𝐷 ) ) )
5 2 4 bibi12d ⊢ ( 𝑥 = 𝐴 → ( ( ⟨ 𝑥 , 𝑦 ⟩ = ⟨ 𝐶 , 𝐷 ⟩ ↔ ( 𝑥 = 𝐶 ∧ 𝑦 = 𝐷 ) ) ↔ ( ⟨ 𝐴 , 𝑦 ⟩ = ⟨ 𝐶 , 𝐷 ⟩ ↔ ( 𝐴 = 𝐶 ∧ 𝑦 = 𝐷 ) ) ) )
6 opeq2 ⊢ ( 𝑦 = 𝐵 → ⟨ 𝐴 , 𝑦 ⟩ = ⟨ 𝐴 , 𝐵 ⟩ )
7 6 eqeq1d ⊢ ( 𝑦 = 𝐵 → ( ⟨ 𝐴 , 𝑦 ⟩ = ⟨ 𝐶 , 𝐷 ⟩ ↔ ⟨ 𝐴 , 𝐵 ⟩ = ⟨ 𝐶 , 𝐷 ⟩ ) )
8 eqeq1 ⊢ ( 𝑦 = 𝐵 → ( 𝑦 = 𝐷 ↔ 𝐵 = 𝐷 ) )
9 8 anbi2d ⊢ ( 𝑦 = 𝐵 → ( ( 𝐴 = 𝐶 ∧ 𝑦 = 𝐷 ) ↔ ( 𝐴 = 𝐶 ∧ 𝐵 = 𝐷 ) ) )
10 7 9 bibi12d ⊢ ( 𝑦 = 𝐵 → ( ( ⟨ 𝐴 , 𝑦 ⟩ = ⟨ 𝐶 , 𝐷 ⟩ ↔ ( 𝐴 = 𝐶 ∧ 𝑦 = 𝐷 ) ) ↔ ( ⟨ 𝐴 , 𝐵 ⟩ = ⟨ 𝐶 , 𝐷 ⟩ ↔ ( 𝐴 = 𝐶 ∧ 𝐵 = 𝐷 ) ) ) )
11 vex ⊢ 𝑥 ∈ V
12 vex ⊢ 𝑦 ∈ V
13 11 12 opth ⊢ ( ⟨ 𝑥 , 𝑦 ⟩ = ⟨ 𝐶 , 𝐷 ⟩ ↔ ( 𝑥 = 𝐶 ∧ 𝑦 = 𝐷 ) )
14 5 10 13 vtocl2g ⊢ ( ( 𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ) → ( ⟨ 𝐴 , 𝐵 ⟩ = ⟨ 𝐶 , 𝐷 ⟩ ↔ ( 𝐴 = 𝐶 ∧ 𝐵 = 𝐷 ) ) )