Metamath Proof Explorer


Theorem opabssxp

Description: An abstraction relation is a subset of a related Cartesian product. (Contributed by NM, 16-Jul-1995)

Ref Expression
Assertion opabssxp { ⟨ 𝑥 , 𝑦 ⟩ ∣ ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) ∧ 𝜑 ) } ⊆ ( 𝐴 × 𝐵 )

Proof

Step Hyp Ref Expression
1 simpl ⊢ ( ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) ∧ 𝜑 ) → ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) )
2 1 ssopab2i ⊢ { ⟨ 𝑥 , 𝑦 ⟩ ∣ ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) ∧ 𝜑 ) } ⊆ { ⟨ 𝑥 , 𝑦 ⟩ ∣ ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) }
3 df-xp ⊢ ( 𝐴 × 𝐵 ) = { ⟨ 𝑥 , 𝑦 ⟩ ∣ ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) }
4 2 3 sseqtrri ⊢ { ⟨ 𝑥 , 𝑦 ⟩ ∣ ( ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ) ∧ 𝜑 ) } ⊆ ( 𝐴 × 𝐵 )