Metamath Proof Explorer


Theorem faovcl

Description: Closure law for an operation, analogous to fovcl . (Contributed by Alexander van der Vekens, 26-May-2017)

Ref Expression
Hypothesis faovcl.1 ⊢ 𝐹 : ( 𝑅 × 𝑆 ) ⟶ 𝐶
Assertion faovcl ( ( 𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆 ) → (( 𝐴 𝐹 𝐵 )) ∈ 𝐶 )

Proof

Step Hyp Ref Expression
1 faovcl.1 ⊢ 𝐹 : ( 𝑅 × 𝑆 ) ⟶ 𝐶
2 ffnaov ⊢ ( 𝐹 : ( 𝑅 × 𝑆 ) ⟶ 𝐶 ↔ ( 𝐹 Fn ( 𝑅 × 𝑆 ) ∧ ∀ 𝑥 ∈ 𝑅 ∀ 𝑦 ∈ 𝑆 (( 𝑥 𝐹 𝑦 )) ∈ 𝐶 ) )
3 2 simprbi ⊢ ( 𝐹 : ( 𝑅 × 𝑆 ) ⟶ 𝐶 → ∀ 𝑥 ∈ 𝑅 ∀ 𝑦 ∈ 𝑆 (( 𝑥 𝐹 𝑦 )) ∈ 𝐶 )
4 1 3 ax-mp ⊢ ∀ 𝑥 ∈ 𝑅 ∀ 𝑦 ∈ 𝑆 (( 𝑥 𝐹 𝑦 )) ∈ 𝐶
5 eqidd ⊢ ( 𝑥 = 𝐴 → 𝐹 = 𝐹 )
6 id ⊢ ( 𝑥 = 𝐴 → 𝑥 = 𝐴 )
7 eqidd ⊢ ( 𝑥 = 𝐴 → 𝑦 = 𝑦 )
8 5 6 7 aoveq123d ⊢ ( 𝑥 = 𝐴 → (( 𝑥 𝐹 𝑦 )) = (( 𝐴 𝐹 𝑦 )) )
9 8 eleq1d ⊢ ( 𝑥 = 𝐴 → ( (( 𝑥 𝐹 𝑦 )) ∈ 𝐶 ↔ (( 𝐴 𝐹 𝑦 )) ∈ 𝐶 ) )
10 eqidd ⊢ ( 𝑦 = 𝐵 → 𝐹 = 𝐹 )
11 eqidd ⊢ ( 𝑦 = 𝐵 → 𝐴 = 𝐴 )
12 id ⊢ ( 𝑦 = 𝐵 → 𝑦 = 𝐵 )
13 10 11 12 aoveq123d ⊢ ( 𝑦 = 𝐵 → (( 𝐴 𝐹 𝑦 )) = (( 𝐴 𝐹 𝐵 )) )
14 13 eleq1d ⊢ ( 𝑦 = 𝐵 → ( (( 𝐴 𝐹 𝑦 )) ∈ 𝐶 ↔ (( 𝐴 𝐹 𝐵 )) ∈ 𝐶 ) )
15 9 14 rspc2v ⊢ ( ( 𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆 ) → ( ∀ 𝑥 ∈ 𝑅 ∀ 𝑦 ∈ 𝑆 (( 𝑥 𝐹 𝑦 )) ∈ 𝐶 → (( 𝐴 𝐹 𝐵 )) ∈ 𝐶 ) )
16 4 15 mpi ⊢ ( ( 𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆 ) → (( 𝐴 𝐹 𝐵 )) ∈ 𝐶 )