Metamath Proof Explorer


Theorem suppval1

Description: The value of the operation constructing the support of a function. (Contributed by AV, 6-Apr-2019)

Ref Expression
Assertion suppval1 ⊢ Fun ⁡ X ∧ X ∈ V ∧ Z ∈ W → X supp Z = i ∈ dom ⁡ X | X ⁡ i ≠ Z

Proof

Step Hyp Ref Expression
1 suppval ⊢ X ∈ V ∧ Z ∈ W → X supp Z = i ∈ dom ⁡ X | X i ≠ Z
2 1 3adant1 ⊢ Fun ⁡ X ∧ X ∈ V ∧ Z ∈ W → X supp Z = i ∈ dom ⁡ X | X i ≠ Z
3 funfn ⊢ Fun ⁡ X ↔ X Fn dom ⁡ X
4 3 biimpi ⊢ Fun ⁡ X → X Fn dom ⁡ X
5 4 3ad2ant1 ⊢ Fun ⁡ X ∧ X ∈ V ∧ Z ∈ W → X Fn dom ⁡ X
6 fnsnfv ⊢ X Fn dom ⁡ X ∧ i ∈ dom ⁡ X → X ⁡ i = X i
7 5 6 sylan ⊢ Fun ⁡ X ∧ X ∈ V ∧ Z ∈ W ∧ i ∈ dom ⁡ X → X ⁡ i = X i
8 7 eqcomd ⊢ Fun ⁡ X ∧ X ∈ V ∧ Z ∈ W ∧ i ∈ dom ⁡ X → X i = X ⁡ i
9 8 neeq1d ⊢ Fun ⁡ X ∧ X ∈ V ∧ Z ∈ W ∧ i ∈ dom ⁡ X → X i ≠ Z ↔ X ⁡ i ≠ Z
10 fvex ⊢ X ⁡ i ∈ V
11 sneqbg ⊢ X ⁡ i ∈ V → X ⁡ i = Z ↔ X ⁡ i = Z
12 10 11 mp1i ⊢ Fun ⁡ X ∧ X ∈ V ∧ Z ∈ W ∧ i ∈ dom ⁡ X → X ⁡ i = Z ↔ X ⁡ i = Z
13 12 necon3bid ⊢ Fun ⁡ X ∧ X ∈ V ∧ Z ∈ W ∧ i ∈ dom ⁡ X → X ⁡ i ≠ Z ↔ X ⁡ i ≠ Z
14 9 13 bitrd ⊢ Fun ⁡ X ∧ X ∈ V ∧ Z ∈ W ∧ i ∈ dom ⁡ X → X i ≠ Z ↔ X ⁡ i ≠ Z
15 14 rabbidva ⊢ Fun ⁡ X ∧ X ∈ V ∧ Z ∈ W → i ∈ dom ⁡ X | X i ≠ Z = i ∈ dom ⁡ X | X ⁡ i ≠ Z
16 2 15 eqtrd ⊢ Fun ⁡ X ∧ X ∈ V ∧ Z ∈ W → X supp Z = i ∈ dom ⁡ X | X ⁡ i ≠ Z