Metamath Proof Explorer


Theorem opabssi

Description: Sufficient condition for a collection of ordered pairs to be a subclass of a relation. (Contributed by Peter Mazsa, 21-Oct-2019) (Revised by Thierry Arnoux, 18-Feb-2022)

Ref Expression
Hypothesis opabssi.1 ⊢ ( 𝜑 → ⟨ 𝑥 , 𝑦 ⟩ ∈ 𝐴 )
Assertion opabssi { ⟨ 𝑥 , 𝑦 ⟩ ∣ 𝜑 } ⊆ 𝐴

Proof

Step Hyp Ref Expression
1 opabssi.1 ⊢ ( 𝜑 → ⟨ 𝑥 , 𝑦 ⟩ ∈ 𝐴 )
2 df-opab ⊢ { ⟨ 𝑥 , 𝑦 ⟩ ∣ 𝜑 } = { 𝑧 ∣ ∃ 𝑥 ∃ 𝑦 ( 𝑧 = ⟨ 𝑥 , 𝑦 ⟩ ∧ 𝜑 ) }
3 eleq1 ⊢ ( 𝑧 = ⟨ 𝑥 , 𝑦 ⟩ → ( 𝑧 ∈ 𝐴 ↔ ⟨ 𝑥 , 𝑦 ⟩ ∈ 𝐴 ) )
4 3 biimprd ⊢ ( 𝑧 = ⟨ 𝑥 , 𝑦 ⟩ → ( ⟨ 𝑥 , 𝑦 ⟩ ∈ 𝐴 → 𝑧 ∈ 𝐴 ) )
5 4 1 impel ⊢ ( ( 𝑧 = ⟨ 𝑥 , 𝑦 ⟩ ∧ 𝜑 ) → 𝑧 ∈ 𝐴 )
6 5 exlimivv ⊢ ( ∃ 𝑥 ∃ 𝑦 ( 𝑧 = ⟨ 𝑥 , 𝑦 ⟩ ∧ 𝜑 ) → 𝑧 ∈ 𝐴 )
7 6 abssi ⊢ { 𝑧 ∣ ∃ 𝑥 ∃ 𝑦 ( 𝑧 = ⟨ 𝑥 , 𝑦 ⟩ ∧ 𝜑 ) } ⊆ 𝐴
8 2 7 eqsstri ⊢ { ⟨ 𝑥 , 𝑦 ⟩ ∣ 𝜑 } ⊆ 𝐴