Metamath Proof Explorer


Theorem optocl

Description: Implicit substitution of class for ordered pair. (Contributed by NM, 5-Mar-1995) Shorten and reduce axiom usage. (Revised by TM, 29-Dec-2025)

Ref Expression
Hypotheses optocl.1 ⊢ 𝐷 = ( 𝐵 × 𝐶 )
optocl.2 ⊢ ( ⟨ 𝑥 , 𝑦 ⟩ = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
optocl.3 ⊢ ( ( 𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶 ) → 𝜑 )
Assertion optocl ( 𝐴 ∈ 𝐷 → 𝜓 )

Proof

Step Hyp Ref Expression
1 optocl.1 ⊢ 𝐷 = ( 𝐵 × 𝐶 )
2 optocl.2 ⊢ ( ⟨ 𝑥 , 𝑦 ⟩ = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
3 optocl.3 ⊢ ( ( 𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶 ) → 𝜑 )
4 elxpi ⊢ ( 𝐴 ∈ ( 𝐵 × 𝐶 ) → ∃ 𝑥 ∃ 𝑦 ( 𝐴 = ⟨ 𝑥 , 𝑦 ⟩ ∧ ( 𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶 ) ) )
5 2 eqcoms ⊢ ( 𝐴 = ⟨ 𝑥 , 𝑦 ⟩ → ( 𝜑 ↔ 𝜓 ) )
6 3 5 imbitrid ⊢ ( 𝐴 = ⟨ 𝑥 , 𝑦 ⟩ → ( ( 𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶 ) → 𝜓 ) )
7 6 imp ⊢ ( ( 𝐴 = ⟨ 𝑥 , 𝑦 ⟩ ∧ ( 𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶 ) ) → 𝜓 )
8 7 exlimivv ⊢ ( ∃ 𝑥 ∃ 𝑦 ( 𝐴 = ⟨ 𝑥 , 𝑦 ⟩ ∧ ( 𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐶 ) ) → 𝜓 )
9 4 8 syl ⊢ ( 𝐴 ∈ ( 𝐵 × 𝐶 ) → 𝜓 )
10 9 1 eleq2s ⊢ ( 𝐴 ∈ 𝐷 → 𝜓 )