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 = ℤ ≥ M
algrf.2 ⊢ R = seq M F ∘ 1 st Z × A
Assertion seq1st ⊢ M ∈ ℤ ∧ A ∈ V → R = seq M F ∘ 1 st M A

Proof

Step Hyp Ref Expression
1 algrf.1 ⊢ Z = ℤ ≥ M
2 algrf.2 ⊢ R = seq M F ∘ 1 st Z × A
3 seqfn ⊢ M ∈ ℤ → seq M F ∘ 1 st Z × A Fn ℤ ≥ M
4 3 adantr ⊢ M ∈ ℤ ∧ A ∈ V → seq M F ∘ 1 st Z × A Fn ℤ ≥ M
5 seqfn ⊢ M ∈ ℤ → seq M F ∘ 1 st M A Fn ℤ ≥ M
6 5 adantr ⊢ M ∈ ℤ ∧ A ∈ V → seq M F ∘ 1 st M A Fn ℤ ≥ 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 ∈ ℤ → seq M F ∘ 1 st Z × A ⁡ M = Z × A ⁡ M
20 19 adantr ⊢ M ∈ ℤ ∧ A ∈ V → seq M F ∘ 1 st Z × A ⁡ M = Z × A ⁡ M
21 seq1 ⊢ M ∈ ℤ → seq M F ∘ 1 st M A ⁡ M = M A ⁡ M
22 21 adantr ⊢ M ∈ ℤ ∧ A ∈ V → seq M F ∘ 1 st M A ⁡ M = M A ⁡ M
23 id ⊢ A ∈ V → A ∈ V
24 uzid ⊢ M ∈ ℤ → M ∈ ℤ ≥ M
25 24 1 eleqtrrdi ⊢ M ∈ ℤ → M ∈ Z
26 fvconst2g ⊢ A ∈ V ∧ M ∈ Z → Z × A ⁡ M = A
27 23 25 26 syl2anr ⊢ M ∈ ℤ ∧ A ∈ V → Z × A ⁡ M = A
28 fvsng ⊢ M ∈ ℤ ∧ A ∈ V → M A ⁡ M = A
29 27 28 eqtr4d ⊢ M ∈ ℤ ∧ A ∈ V → Z × A ⁡ M = M A ⁡ M
30 22 29 eqtr4d ⊢ M ∈ ℤ ∧ A ∈ V → seq M F ∘ 1 st M A ⁡ M = Z × A ⁡ M
31 20 30 eqtr4d ⊢ M ∈ ℤ ∧ A ∈ V → seq M F ∘ 1 st Z × A ⁡ M = seq M F ∘ 1 st M A ⁡ M
32 31 ex ⊢ M ∈ ℤ → 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 ∈ ℤ ≥ 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 opco1i ⊢ 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 ∈ ℤ ≥ M → seq M F ∘ 1 st Z × A ⁡ x + 1 = F ⁡ seq M F ∘ 1 st Z × A ⁡ x
39 seqp1 ⊢ x ∈ ℤ ≥ 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 opco1i ⊢ 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 ∈ ℤ ≥ M → seq M F ∘ 1 st M A ⁡ x + 1 = F ⁡ seq M F ∘ 1 st M A ⁡ x
44 38 43 eqeq12d ⊢ x ∈ ℤ ≥ 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 ∈ ℤ ≥ 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 imbitrrid ⊢ A ∈ V ∧ x ∈ ℤ ≥ 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 ∈ ℤ ≥ 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 ∈ ℤ ≥ 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 ∈ ℤ ≥ 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 ∈ ℤ ≥ M → seq M F ∘ 1 st Z × A ⁡ x = seq M F ∘ 1 st M A ⁡ x
51 50 adantll ⊢ M ∈ ℤ ∧ A ∈ V ∧ x ∈ ℤ ≥ M → seq M F ∘ 1 st Z × A ⁡ x = seq M F ∘ 1 st M A ⁡ x
52 4 6 51 eqfnfvd ⊢ M ∈ ℤ ∧ A ∈ V → seq M F ∘ 1 st Z × A = seq M F ∘ 1 st M A
53 2 52 eqtrid ⊢ M ∈ ℤ ∧ A ∈ V → R = seq M F ∘ 1 st M A