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 ( 𝜑𝐴 ⊆ ℂ )
0pledm.2 ( 𝜑𝐹 Fn 𝐴 )
Assertion 0pledm ( 𝜑 → ( 0𝑝r𝐹 ↔ ( 𝐴 × { 0 } ) ∘r𝐹 ) )

Proof

Step Hyp Ref Expression
1 0pledm.1 ( 𝜑𝐴 ⊆ ℂ )
2 0pledm.2 ( 𝜑𝐹 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 ≤ ( 𝐹𝑥 ) ↔ ∀ 𝑥𝐴 0 ≤ ( 𝐹𝑥 ) ) )
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 ( 𝜑𝐴 ∈ V )
17 eqid ( ℂ ∩ 𝐴 ) = ( ℂ ∩ 𝐴 )
18 0pval ( 𝑥 ∈ ℂ → ( 0𝑝𝑥 ) = 0 )
19 18 adantl ( ( 𝜑𝑥 ∈ ℂ ) → ( 0𝑝𝑥 ) = 0 )
20 eqidd ( ( 𝜑𝑥𝐴 ) → ( 𝐹𝑥 ) = ( 𝐹𝑥 ) )
21 12 2 14 16 17 19 20 ofrfval ( 𝜑 → ( 0𝑝r𝐹 ↔ ∀ 𝑥 ∈ ( ℂ ∩ 𝐴 ) 0 ≤ ( 𝐹𝑥 ) ) )
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 ( 𝐴𝐴 ) = 𝐴
26 c0ex 0 ∈ V
27 26 fvconst2 ( 𝑥𝐴 → ( ( 𝐴 × { 0 } ) ‘ 𝑥 ) = 0 )
28 27 adantl ( ( 𝜑𝑥𝐴 ) → ( ( 𝐴 × { 0 } ) ‘ 𝑥 ) = 0 )
29 24 2 16 16 25 28 20 ofrfval ( 𝜑 → ( ( 𝐴 × { 0 } ) ∘r𝐹 ↔ ∀ 𝑥𝐴 0 ≤ ( 𝐹𝑥 ) ) )
30 5 21 29 3bitr4d ( 𝜑 → ( 0𝑝r𝐹 ↔ ( 𝐴 × { 0 } ) ∘r𝐹 ) )