Metamath Proof Explorer


Theorem 0pledm

Description: Adjust the domain of the left argument to match the right, which works better in our theorems. (Contributed by Mario Carneiro, 28-Jul-2014)

Ref Expression
Hypotheses 0pledm.1 ⊢ φ → A ⊆ ℂ
0pledm.2 ⊢ φ → F Fn A
Assertion 0pledm ⊢ φ → 0 𝑝 ≤ f F ↔ A × 0 ≤ f F

Proof

Step Hyp Ref Expression
1 0pledm.1 ⊢ φ → A ⊆ ℂ
2 0pledm.2 ⊢ φ → F Fn A
3 sseqin2 ⊢ A ⊆ ℂ ↔ ℂ ∩ A = A
4 1 3 sylib ⊢ φ → ℂ ∩ A = A
5 4 raleqdv ⊢ φ → ∀ x ∈ ℂ ∩ A 0 ≤ F ⁡ x ↔ ∀ x ∈ A 0 ≤ F ⁡ x
6 0cn ⊢ 0 ∈ ℂ
7 fnconstg ⊢ 0 ∈ ℂ → ℂ × 0 Fn ℂ
8 6 7 ax-mp ⊢ ℂ × 0 Fn ℂ
9 df-0p ⊢ 0 𝑝 = ℂ × 0
10 9 fneq1i ⊢ 0 𝑝 Fn ℂ ↔ ℂ × 0 Fn ℂ
11 8 10 mpbir ⊢ 0 𝑝 Fn ℂ
12 11 a1i ⊢ φ → 0 𝑝 Fn ℂ
13 cnex ⊢ ℂ ∈ V
14 13 a1i ⊢ φ → ℂ ∈ V
15 ssexg ⊢ A ⊆ ℂ ∧ ℂ ∈ V → A ∈ V
16 1 13 15 sylancl ⊢ φ → A ∈ V
17 eqid ⊢ ℂ ∩ A = ℂ ∩ A
18 0pval ⊢ x ∈ ℂ → 0 𝑝 ⁡ x = 0
19 18 adantl ⊢ φ ∧ x ∈ ℂ → 0 𝑝 ⁡ x = 0
20 eqidd ⊢ φ ∧ x ∈ A → F ⁡ x = F ⁡ x
21 12 2 14 16 17 19 20 ofrfval ⊢ φ → 0 𝑝 ≤ f F ↔ ∀ x ∈ ℂ ∩ A 0 ≤ F ⁡ x
22 fnconstg ⊢ 0 ∈ ℂ → A × 0 Fn A
23 6 22 ax-mp ⊢ A × 0 Fn A
24 23 a1i ⊢ φ → A × 0 Fn A
25 inidm ⊢ A ∩ A = A
26 c0ex ⊢ 0 ∈ V
27 26 fvconst2 ⊢ x ∈ A → A × 0 ⁡ x = 0
28 27 adantl ⊢ φ ∧ x ∈ A → A × 0 ⁡ x = 0
29 24 2 16 16 25 28 20 ofrfval ⊢ φ → A × 0 ≤ f F ↔ ∀ x ∈ A 0 ≤ F ⁡ x
30 5 21 29 3bitr4d ⊢ φ → 0 𝑝 ≤ f F ↔ A × 0 ≤ f F