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