Metamath Proof Explorer


Theorem ovolshftlem2

Description: Lemma for ovolshft . (Contributed by Mario Carneiro, 22-Mar-2014)

Ref Expression
Hypotheses ovolshft.1 φ A
ovolshft.2 φ C
ovolshft.3 φ B = x | x C A
ovolshft.4 M = y * | f 2 B ran . f y = sup ran seq 1 + abs f * <
Assertion ovolshftlem2 φ z * | g 2 A ran . g z = sup ran seq 1 + abs g * < M

Proof

Step Hyp Ref Expression
1 ovolshft.1 φ A
2 ovolshft.2 φ C
3 ovolshft.3 φ B = x | x C A
4 ovolshft.4 M = y * | f 2 B ran . f y = sup ran seq 1 + abs f * <
5 1 ad3antrrr φ z * g 2 A ran . g A
6 2 ad3antrrr φ z * g 2 A ran . g C
7 3 ad3antrrr φ z * g 2 A ran . g B = x | x C A
8 eqid seq 1 + abs g = seq 1 + abs g
9 2fveq3 m = n 1 st g m = 1 st g n
10 9 oveq1d m = n 1 st g m + C = 1 st g n + C
11 2fveq3 m = n 2 nd g m = 2 nd g n
12 11 oveq1d m = n 2 nd g m + C = 2 nd g n + C
13 10 12 opeq12d m = n 1 st g m + C 2 nd g m + C = 1 st g n + C 2 nd g n + C
14 13 cbvmptv m 1 st g m + C 2 nd g m + C = n 1 st g n + C 2 nd g n + C
15 simplr φ z * g 2 A ran . g g 2
16 elovolmlem g 2 g : 2
17 15 16 sylib φ z * g 2 A ran . g g : 2
18 simpr φ z * g 2 A ran . g A ran . g
19 5 6 7 4 8 14 17 18 ovolshftlem1 φ z * g 2 A ran . g sup ran seq 1 + abs g * < M
20 eleq1a sup ran seq 1 + abs g * < M z = sup ran seq 1 + abs g * < z M
21 19 20 syl φ z * g 2 A ran . g z = sup ran seq 1 + abs g * < z M
22 21 expimpd φ z * g 2 A ran . g z = sup ran seq 1 + abs g * < z M
23 22 rexlimdva φ z * g 2 A ran . g z = sup ran seq 1 + abs g * < z M
24 23 ralrimiva φ z * g 2 A ran . g z = sup ran seq 1 + abs g * < z M
25 rabss z * | g 2 A ran . g z = sup ran seq 1 + abs g * < M z * g 2 A ran . g z = sup ran seq 1 + abs g * < z M
26 24 25 sylibr φ z * | g 2 A ran . g z = sup ran seq 1 + abs g * < M