Metamath Proof Explorer


Theorem ndmaovcl

Description: The "closure" of an operation outside its domain, when the operation's value is a set in contrast to ndmovcl where it is required that the domain contains the empty set ( (/) e. S ). (Contributed by Alexander van der Vekens, 26-May-2017)

Ref Expression
Hypotheses ndmaov.1 ⊢ dom 𝐹 = ( 𝑆 × 𝑆 )
ndmaovcl.2 ⊢ ( ( 𝐴 ∈ 𝑆 ∧ 𝐵 ∈ 𝑆 ) → (( 𝐴 𝐹 𝐵 )) ∈ 𝑆 )
ndmaovcl.3 ⊢ (( 𝐴 𝐹 𝐵 )) ∈ V
Assertion ndmaovcl (( 𝐴 𝐹 𝐵 )) ∈ 𝑆

Proof

Step Hyp Ref Expression
1 ndmaov.1 ⊢ dom 𝐹 = ( 𝑆 × 𝑆 )
2 ndmaovcl.2 ⊢ ( ( 𝐴 ∈ 𝑆 ∧ 𝐵 ∈ 𝑆 ) → (( 𝐴 𝐹 𝐵 )) ∈ 𝑆 )
3 ndmaovcl.3 ⊢ (( 𝐴 𝐹 𝐵 )) ∈ V
4 opelxp ⊢ ( ⟨ 𝐴 , 𝐵 ⟩ ∈ ( 𝑆 × 𝑆 ) ↔ ( 𝐴 ∈ 𝑆 ∧ 𝐵 ∈ 𝑆 ) )
5 1 eqcomi ⊢ ( 𝑆 × 𝑆 ) = dom 𝐹
6 5 eleq2i ⊢ ( ⟨ 𝐴 , 𝐵 ⟩ ∈ ( 𝑆 × 𝑆 ) ↔ ⟨ 𝐴 , 𝐵 ⟩ ∈ dom 𝐹 )
7 ndmaov ⊢ ( ¬ ⟨ 𝐴 , 𝐵 ⟩ ∈ dom 𝐹 → (( 𝐴 𝐹 𝐵 )) = V )
8 eleq1 ⊢ ( (( 𝐴 𝐹 𝐵 )) = V → ( (( 𝐴 𝐹 𝐵 )) ∈ V ↔ V ∈ V ) )
9 8 biimpd ⊢ ( (( 𝐴 𝐹 𝐵 )) = V → ( (( 𝐴 𝐹 𝐵 )) ∈ V → V ∈ V ) )
10 vprc ⊢ ¬ V ∈ V
11 10 pm2.21i ⊢ ( V ∈ V → (( 𝐴 𝐹 𝐵 )) ∈ 𝑆 )
12 9 11 syl6com ⊢ ( (( 𝐴 𝐹 𝐵 )) ∈ V → ( (( 𝐴 𝐹 𝐵 )) = V → (( 𝐴 𝐹 𝐵 )) ∈ 𝑆 ) )
13 3 7 12 mpsyl ⊢ ( ¬ ⟨ 𝐴 , 𝐵 ⟩ ∈ dom 𝐹 → (( 𝐴 𝐹 𝐵 )) ∈ 𝑆 )
14 6 13 sylnbi ⊢ ( ¬ ⟨ 𝐴 , 𝐵 ⟩ ∈ ( 𝑆 × 𝑆 ) → (( 𝐴 𝐹 𝐵 )) ∈ 𝑆 )
15 4 14 sylnbir ⊢ ( ¬ ( 𝐴 ∈ 𝑆 ∧ 𝐵 ∈ 𝑆 ) → (( 𝐴 𝐹 𝐵 )) ∈ 𝑆 )
16 2 15 pm2.61i ⊢ (( 𝐴 𝐹 𝐵 )) ∈ 𝑆