Metamath Proof Explorer


Theorem sumrb

Description: Rebase the starting point of a sum. (Contributed by Mario Carneiro, 14-Jul-2013) (Revised by Mario Carneiro, 9-Apr-2014)

Ref Expression
Hypotheses summo.1 ⊢ F = k ∈ ℤ ⟼ if k ∈ A B 0
summo.2 ⊢ φ ∧ k ∈ A → B ∈ ℂ
sumrb.4 ⊢ φ → M ∈ ℤ
sumrb.5 ⊢ φ → N ∈ ℤ
sumrb.6 ⊢ φ → A ⊆ ℤ ≥ M
sumrb.7 ⊢ φ → A ⊆ ℤ ≥ N
Assertion sumrb ⊢ φ → seq M + F ⇝ C ↔ seq N + F ⇝ C

Proof

Step Hyp Ref Expression
1 summo.1 ⊢ F = k ∈ ℤ ⟼ if k ∈ A B 0
2 summo.2 ⊢ φ ∧ k ∈ A → B ∈ ℂ
3 sumrb.4 ⊢ φ → M ∈ ℤ
4 sumrb.5 ⊢ φ → N ∈ ℤ
5 sumrb.6 ⊢ φ → A ⊆ ℤ ≥ M
6 sumrb.7 ⊢ φ → A ⊆ ℤ ≥ N
7 4 adantr ⊢ φ ∧ N ∈ ℤ ≥ M → N ∈ ℤ
8 seqex ⊢ seq M + F ∈ V
9 climres ⊢ N ∈ ℤ ∧ seq M + F ∈ V → seq M + F ↾ ℤ ≥ N ⇝ C ↔ seq M + F ⇝ C
10 7 8 9 sylancl ⊢ φ ∧ N ∈ ℤ ≥ M → seq M + F ↾ ℤ ≥ N ⇝ C ↔ seq M + F ⇝ C
11 2 adantlr ⊢ φ ∧ N ∈ ℤ ≥ M ∧ k ∈ A → B ∈ ℂ
12 simpr ⊢ φ ∧ N ∈ ℤ ≥ M → N ∈ ℤ ≥ M
13 1 11 12 sumrblem ⊢ φ ∧ N ∈ ℤ ≥ M ∧ A ⊆ ℤ ≥ N → seq M + F ↾ ℤ ≥ N = seq N + F
14 6 13 mpidan ⊢ φ ∧ N ∈ ℤ ≥ M → seq M + F ↾ ℤ ≥ N = seq N + F
15 14 breq1d ⊢ φ ∧ N ∈ ℤ ≥ M → seq M + F ↾ ℤ ≥ N ⇝ C ↔ seq N + F ⇝ C
16 10 15 bitr3d ⊢ φ ∧ N ∈ ℤ ≥ M → seq M + F ⇝ C ↔ seq N + F ⇝ C
17 2 adantlr ⊢ φ ∧ M ∈ ℤ ≥ N ∧ k ∈ A → B ∈ ℂ
18 simpr ⊢ φ ∧ M ∈ ℤ ≥ N → M ∈ ℤ ≥ N
19 1 17 18 sumrblem ⊢ φ ∧ M ∈ ℤ ≥ N ∧ A ⊆ ℤ ≥ M → seq N + F ↾ ℤ ≥ M = seq M + F
20 5 19 mpidan ⊢ φ ∧ M ∈ ℤ ≥ N → seq N + F ↾ ℤ ≥ M = seq M + F
21 20 breq1d ⊢ φ ∧ M ∈ ℤ ≥ N → seq N + F ↾ ℤ ≥ M ⇝ C ↔ seq M + F ⇝ C
22 3 adantr ⊢ φ ∧ M ∈ ℤ ≥ N → M ∈ ℤ
23 seqex ⊢ seq N + F ∈ V
24 climres ⊢ M ∈ ℤ ∧ seq N + F ∈ V → seq N + F ↾ ℤ ≥ M ⇝ C ↔ seq N + F ⇝ C
25 22 23 24 sylancl ⊢ φ ∧ M ∈ ℤ ≥ N → seq N + F ↾ ℤ ≥ M ⇝ C ↔ seq N + F ⇝ C
26 21 25 bitr3d ⊢ φ ∧ M ∈ ℤ ≥ N → seq M + F ⇝ C ↔ seq N + F ⇝ C
27 uztric ⊢ M ∈ ℤ ∧ N ∈ ℤ → N ∈ ℤ ≥ M ∨ M ∈ ℤ ≥ N
28 3 4 27 syl2anc ⊢ φ → N ∈ ℤ ≥ M ∨ M ∈ ℤ ≥ N
29 16 26 28 mpjaodan ⊢ φ → seq M + F ⇝ C ↔ seq N + F ⇝ C