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 → ( 0𝑝 ∘r ≤ 𝐹 ↔ ( 𝐴 × { 0 } ) ∘r ≤ 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 ⊢ 0𝑝 = ( ℂ × { 0 } )
10 9 fneq1i ⊢ ( 0𝑝 Fn ℂ ↔ ( ℂ × { 0 } ) Fn ℂ )
11 8 10 mpbir ⊢ 0𝑝 Fn ℂ
12 11 a1i ⊢ ( 𝜑 → 0𝑝 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 → ( 0𝑝 ‘ 𝑥 ) = 0 )
19 18 adantl ⊢ ( ( 𝜑 ∧ 𝑥 ∈ ℂ ) → ( 0𝑝 ‘ 𝑥 ) = 0 )
20 eqidd ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → ( 𝐹 ‘ 𝑥 ) = ( 𝐹 ‘ 𝑥 ) )
21 12 2 14 16 17 19 20 ofrfval ⊢ ( 𝜑 → ( 0𝑝 ∘r ≤ 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 } ) ∘r ≤ 𝐹 ↔ ∀ 𝑥 ∈ 𝐴 0 ≤ ( 𝐹 ‘ 𝑥 ) ) )
30 5 21 29 3bitr4d ⊢ ( 𝜑 → ( 0𝑝 ∘r ≤ 𝐹 ↔ ( 𝐴 × { 0 } ) ∘r ≤ 𝐹 ) )