Metamath Proof Explorer


Theorem ftc1lem3

Description: Lemma for ftc1 . (Contributed by Mario Carneiro, 1-Sep-2014) (Revised by Mario Carneiro, 8-Sep-2015)

Ref Expression
Hypotheses ftc1.g G = x A B A x F t dt
ftc1.a φ A
ftc1.b φ B
ftc1.le φ A B
ftc1.s φ A B D
ftc1.d φ D
ftc1.i φ F 𝐿 1
ftc1.c φ C A B
ftc1.f φ F K CnP L C
ftc1.j J = L 𝑡
ftc1.k K = L 𝑡 D
ftc1.l L = TopOpen fld
Assertion ftc1lem3 φ F : D

Proof

Step Hyp Ref Expression
1 ftc1.g G = x A B A x F t dt
2 ftc1.a φ A
3 ftc1.b φ B
4 ftc1.le φ A B
5 ftc1.s φ A B D
6 ftc1.d φ D
7 ftc1.i φ F 𝐿 1
8 ftc1.c φ C A B
9 ftc1.f φ F K CnP L C
10 ftc1.j J = L 𝑡
11 ftc1.k K = L 𝑡 D
12 ftc1.l L = TopOpen fld
13 12 cnfldtopon L TopOn
14 ax-resscn
15 6 14 sstrdi φ D
16 resttopon L TopOn D L 𝑡 D TopOn D
17 13 15 16 sylancr φ L 𝑡 D TopOn D
18 11 17 eqeltrid φ K TopOn D
19 13 a1i φ L TopOn
20 cnpf2 K TopOn D L TopOn F K CnP L C F : D
21 18 19 9 20 syl3anc φ F : D