Metamath Proof Explorer


Theorem seq1st

Description: A sequence whose iteration function ignores the second argument is only affected by the first point of the initial value function. (Contributed by Mario Carneiro, 11-Feb-2015)

Ref Expression
Hypotheses algrf.1 Z = M
algrf.2 R = seq M F 1 st Z × A
Assertion seq1st M A V R = seq M F 1 st M A

Proof

Step Hyp Ref Expression
1 algrf.1 Z = M
2 algrf.2 R = seq M F 1 st Z × A
3 seqfn M seq M F 1 st Z × A Fn M
4 3 adantr M A V seq M F 1 st Z × A Fn M
5 seqfn M seq M F 1 st M A Fn M
6 5 adantr M A V seq M F 1 st M A Fn M
7 fveq2 y = M seq M F 1 st Z × A y = seq M F 1 st Z × A M
8 fveq2 y = M seq M F 1 st M A y = seq M F 1 st M A M
9 7 8 eqeq12d y = M seq M F 1 st Z × A y = seq M F 1 st M A y seq M F 1 st Z × A M = seq M F 1 st M A M
10 9 imbi2d y = M A V seq M F 1 st Z × A y = seq M F 1 st M A y A V seq M F 1 st Z × A M = seq M F 1 st M A M
11 fveq2 y = x seq M F 1 st Z × A y = seq M F 1 st Z × A x
12 fveq2 y = x seq M F 1 st M A y = seq M F 1 st M A x
13 11 12 eqeq12d y = x seq M F 1 st Z × A y = seq M F 1 st M A y seq M F 1 st Z × A x = seq M F 1 st M A x
14 13 imbi2d y = x A V seq M F 1 st Z × A y = seq M F 1 st M A y A V seq M F 1 st Z × A x = seq M F 1 st M A x
15 fveq2 y = x + 1 seq M F 1 st Z × A y = seq M F 1 st Z × A x + 1
16 fveq2 y = x + 1 seq M F 1 st M A y = seq M F 1 st M A x + 1
17 15 16 eqeq12d y = x + 1 seq M F 1 st Z × A y = seq M F 1 st M A y seq M F 1 st Z × A x + 1 = seq M F 1 st M A x + 1
18 17 imbi2d y = x + 1 A V seq M F 1 st Z × A y = seq M F 1 st M A y A V seq M F 1 st Z × A x + 1 = seq M F 1 st M A x + 1
19 seq1 M seq M F 1 st Z × A M = Z × A M
20 19 adantr M A V seq M F 1 st Z × A M = Z × A M
21 seq1 M seq M F 1 st M A M = M A M
22 21 adantr M A V seq M F 1 st M A M = M A M
23 id A V A V
24 uzid M M M
25 24 1 eleqtrrdi M M Z
26 fvconst2g A V M Z Z × A M = A
27 23 25 26 syl2anr M A V Z × A M = A
28 fvsng M A V M A M = A
29 27 28 eqtr4d M A V Z × A M = M A M
30 22 29 eqtr4d M A V seq M F 1 st M A M = Z × A M
31 20 30 eqtr4d M A V seq M F 1 st Z × A M = seq M F 1 st M A M
32 31 ex M A V seq M F 1 st Z × A M = seq M F 1 st M A M
33 fveq2 seq M F 1 st Z × A x = seq M F 1 st M A x F seq M F 1 st Z × A x = F seq M F 1 st M A x
34 seqp1 x M seq M F 1 st Z × A x + 1 = seq M F 1 st Z × A x F 1 st Z × A x + 1
35 fvex seq M F 1 st Z × A x V
36 fvex Z × A x + 1 V
37 35 36 algrflem seq M F 1 st Z × A x F 1 st Z × A x + 1 = F seq M F 1 st Z × A x
38 34 37 eqtrdi x M seq M F 1 st Z × A x + 1 = F seq M F 1 st Z × A x
39 seqp1 x M seq M F 1 st M A x + 1 = seq M F 1 st M A x F 1 st M A x + 1
40 fvex seq M F 1 st M A x V
41 fvex M A x + 1 V
42 40 41 algrflem seq M F 1 st M A x F 1 st M A x + 1 = F seq M F 1 st M A x
43 39 42 eqtrdi x M seq M F 1 st M A x + 1 = F seq M F 1 st M A x
44 38 43 eqeq12d x M seq M F 1 st Z × A x + 1 = seq M F 1 st M A x + 1 F seq M F 1 st Z × A x = F seq M F 1 st M A x
45 44 adantl A V x M seq M F 1 st Z × A x + 1 = seq M F 1 st M A x + 1 F seq M F 1 st Z × A x = F seq M F 1 st M A x
46 33 45 syl5ibr A V x M seq M F 1 st Z × A x = seq M F 1 st M A x seq M F 1 st Z × A x + 1 = seq M F 1 st M A x + 1
47 46 expcom x M A V seq M F 1 st Z × A x = seq M F 1 st M A x seq M F 1 st Z × A x + 1 = seq M F 1 st M A x + 1
48 47 a2d x M A V seq M F 1 st Z × A x = seq M F 1 st M A x A V seq M F 1 st Z × A x + 1 = seq M F 1 st M A x + 1
49 10 14 18 14 32 48 uzind4 x M A V seq M F 1 st Z × A x = seq M F 1 st M A x
50 49 impcom A V x M seq M F 1 st Z × A x = seq M F 1 st M A x
51 50 adantll M A V x M seq M F 1 st Z × A x = seq M F 1 st M A x
52 4 6 51 eqfnfvd M A V seq M F 1 st Z × A = seq M F 1 st M A
53 2 52 syl5eq M A V R = seq M F 1 st M A