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
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
2 0pledm.2 φ F Fn A
3 sseqin2 A A = A
4 1 3 sylib φ A = A
5 4 raleqdv φ x A 0 F x x A 0 F x
6 0cn 0
7 fnconstg 0 × 0 Fn
8 6 7 ax-mp × 0 Fn
9 df-0p 0 𝑝 = × 0
10 9 fneq1i 0 𝑝 Fn × 0 Fn
11 8 10 mpbir 0 𝑝 Fn
12 11 a1i φ 0 𝑝 Fn
13 cnex V
14 13 a1i φ V
15 ssexg A V A V
16 1 13 15 sylancl φ A V
17 eqid A = A
18 0pval x 0 𝑝 x = 0
19 18 adantl φ x 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 A 0 F x
22 fnconstg 0 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