Metamath Proof Explorer


Theorem fdifsupp

Description: Express the support of a function F outside of B in two different ways. (Contributed by Thierry Arnoux, 5-Oct-2025)

Ref Expression
Hypotheses fdifsupp.1 ⊢ φ → A ∈ V
fdifsupp.2 ⊢ φ → Z ∈ W
fdifsupp.3 ⊢ φ → F Fn A
Assertion fdifsupp ⊢ φ → F ↾ A ∖ B supp Z = supp Z⁡ F ∖ B

Proof

Step Hyp Ref Expression
1 fdifsupp.1 ⊢ φ → A ∈ V
2 fdifsupp.2 ⊢ φ → Z ∈ W
3 fdifsupp.3 ⊢ φ → F Fn A
4 difssd ⊢ φ → A ∖ B ⊆ A
5 3 4 fnssresd ⊢ φ → F ↾ A ∖ B Fn A ∖ B
6 1 difexd ⊢ φ → A ∖ B ∈ V
7 elsuppfn ⊢ F ↾ A ∖ B Fn A ∖ B ∧ A ∖ B ∈ V ∧ Z ∈ W → x ∈ supp Z⁡ F ↾ A ∖ B ↔ x ∈ A ∖ B ∧ F ↾ A ∖ B ⁡ x ≠ Z
8 5 6 2 7 syl3anc ⊢ φ → x ∈ supp Z⁡ F ↾ A ∖ B ↔ x ∈ A ∖ B ∧ F ↾ A ∖ B ⁡ x ≠ Z
9 eldif ⊢ x ∈ A ∖ B ↔ x ∈ A ∧ ¬ x ∈ B
10 9 anbi1i ⊢ x ∈ A ∖ B ∧ F ⁡ x ≠ Z ↔ x ∈ A ∧ ¬ x ∈ B ∧ F ⁡ x ≠ Z
11 10 a1i ⊢ φ → x ∈ A ∖ B ∧ F ⁡ x ≠ Z ↔ x ∈ A ∧ ¬ x ∈ B ∧ F ⁡ x ≠ Z
12 simpr ⊢ φ ∧ x ∈ A ∖ B → x ∈ A ∖ B
13 12 fvresd ⊢ φ ∧ x ∈ A ∖ B → F ↾ A ∖ B ⁡ x = F ⁡ x
14 13 neeq1d ⊢ φ ∧ x ∈ A ∖ B → F ↾ A ∖ B ⁡ x ≠ Z ↔ F ⁡ x ≠ Z
15 14 pm5.32da ⊢ φ → x ∈ A ∖ B ∧ F ↾ A ∖ B ⁡ x ≠ Z ↔ x ∈ A ∖ B ∧ F ⁡ x ≠ Z
16 an32 ⊢ x ∈ A ∧ F ⁡ x ≠ Z ∧ ¬ x ∈ B ↔ x ∈ A ∧ ¬ x ∈ B ∧ F ⁡ x ≠ Z
17 16 a1i ⊢ φ → x ∈ A ∧ F ⁡ x ≠ Z ∧ ¬ x ∈ B ↔ x ∈ A ∧ ¬ x ∈ B ∧ F ⁡ x ≠ Z
18 11 15 17 3bitr4d ⊢ φ → x ∈ A ∖ B ∧ F ↾ A ∖ B ⁡ x ≠ Z ↔ x ∈ A ∧ F ⁡ x ≠ Z ∧ ¬ x ∈ B
19 eldif ⊢ x ∈ supp Z⁡ F ∖ B ↔ x ∈ supp Z⁡ F ∧ ¬ x ∈ B
20 1 elexd ⊢ φ → A ∈ V
21 elsuppfn ⊢ F Fn A ∧ A ∈ V ∧ Z ∈ W → x ∈ supp Z⁡ F ↔ x ∈ A ∧ F ⁡ x ≠ Z
22 3 20 2 21 syl3anc ⊢ φ → x ∈ supp Z⁡ F ↔ x ∈ A ∧ F ⁡ x ≠ Z
23 22 anbi1d ⊢ φ → x ∈ supp Z⁡ F ∧ ¬ x ∈ B ↔ x ∈ A ∧ F ⁡ x ≠ Z ∧ ¬ x ∈ B
24 19 23 bitr2id ⊢ φ → x ∈ A ∧ F ⁡ x ≠ Z ∧ ¬ x ∈ B ↔ x ∈ supp Z⁡ F ∖ B
25 8 18 24 3bitrd ⊢ φ → x ∈ supp Z⁡ F ↾ A ∖ B ↔ x ∈ supp Z⁡ F ∖ B
26 25 eqrdv ⊢ φ → F ↾ A ∖ B supp Z = supp Z⁡ F ∖ B