Metamath Proof Explorer


Theorem aoprssdm

Description: Domain of closure of an operation. In contrast to oprssdm , no additional property for S ( -. (/) e. S ) is required! (Contributed by Alexander van der Vekens, 26-May-2017)

Ref Expression
Hypothesis aoprssdm.1 ⊢ ( ( 𝑥 ∈ 𝑆 ∧ 𝑦 ∈ 𝑆 ) → (( 𝑥 𝐹 𝑦 )) ∈ 𝑆 )
Assertion aoprssdm ( 𝑆 × 𝑆 ) ⊆ dom 𝐹

Proof

Step Hyp Ref Expression
1 aoprssdm.1 ⊢ ( ( 𝑥 ∈ 𝑆 ∧ 𝑦 ∈ 𝑆 ) → (( 𝑥 𝐹 𝑦 )) ∈ 𝑆 )
2 relxp ⊢ Rel ( 𝑆 × 𝑆 )
3 opelxp ⊢ ( ⟨ 𝑥 , 𝑦 ⟩ ∈ ( 𝑆 × 𝑆 ) ↔ ( 𝑥 ∈ 𝑆 ∧ 𝑦 ∈ 𝑆 ) )
4 df-aov ⊢ (( 𝑥 𝐹 𝑦 )) = ( 𝐹 ''' ⟨ 𝑥 , 𝑦 ⟩ )
5 4 1 eqeltrrid ⊢ ( ( 𝑥 ∈ 𝑆 ∧ 𝑦 ∈ 𝑆 ) → ( 𝐹 ''' ⟨ 𝑥 , 𝑦 ⟩ ) ∈ 𝑆 )
6 afvvdm ⊢ ( ( 𝐹 ''' ⟨ 𝑥 , 𝑦 ⟩ ) ∈ 𝑆 → ⟨ 𝑥 , 𝑦 ⟩ ∈ dom 𝐹 )
7 5 6 syl ⊢ ( ( 𝑥 ∈ 𝑆 ∧ 𝑦 ∈ 𝑆 ) → ⟨ 𝑥 , 𝑦 ⟩ ∈ dom 𝐹 )
8 3 7 sylbi ⊢ ( ⟨ 𝑥 , 𝑦 ⟩ ∈ ( 𝑆 × 𝑆 ) → ⟨ 𝑥 , 𝑦 ⟩ ∈ dom 𝐹 )
9 2 8 relssi ⊢ ( 𝑆 × 𝑆 ) ⊆ dom 𝐹