Metamath Proof Explorer


Theorem afvpcfv0

Description: If the value of the alternative function at an argument is the universe, the function's value at this argument is the empty set. (Contributed by Alexander van der Vekens, 25-May-2017)

Ref Expression
Assertion afvpcfv0 ⊢ F ''' A = V → F ⁡ A = ∅

Proof

Step Hyp Ref Expression
1 dfafv2 ⊢ F ''' A = if F defAt A F ⁡ A V
2 1 eqeq1i ⊢ F ''' A = V ↔ if F defAt A F ⁡ A V = V
3 eqcom ⊢ if F defAt A F ⁡ A V = V ↔ V = if F defAt A F ⁡ A V
4 eqif ⊢ V = if F defAt A F ⁡ A V ↔ F defAt A ∧ V = F ⁡ A ∨ ¬ F defAt A ∧ V = V
5 3 4 bitri ⊢ if F defAt A F ⁡ A V = V ↔ F defAt A ∧ V = F ⁡ A ∨ ¬ F defAt A ∧ V = V
6 fveqvfvv ⊢ F ⁡ A = V → F ⁡ A = ∅
7 6 eqcoms ⊢ V = F ⁡ A → F ⁡ A = ∅
8 7 adantl ⊢ F defAt A ∧ V = F ⁡ A → F ⁡ A = ∅
9 fvfundmfvn0 ⊢ F ⁡ A ≠ ∅ → A ∈ dom ⁡ F ∧ Fun ⁡ F ↾ A
10 df-dfat ⊢ F defAt A ↔ A ∈ dom ⁡ F ∧ Fun ⁡ F ↾ A
11 9 10 sylibr ⊢ F ⁡ A ≠ ∅ → F defAt A
12 11 necon1bi ⊢ ¬ F defAt A → F ⁡ A = ∅
13 12 adantr ⊢ ¬ F defAt A ∧ V = V → F ⁡ A = ∅
14 8 13 jaoi ⊢ F defAt A ∧ V = F ⁡ A ∨ ¬ F defAt A ∧ V = V → F ⁡ A = ∅
15 5 14 sylbi ⊢ if F defAt A F ⁡ A V = V → F ⁡ A = ∅
16 2 15 sylbi ⊢ F ''' A = V → F ⁡ A = ∅