Metamath Proof Explorer


Theorem seqfveq

Description: Equality of sequences. (Contributed by NM, 17-Mar-2005) (Revised by Mario Carneiro, 27-May-2014)

Ref Expression
Hypotheses seqfveq.1 ⊢ φ → N ∈ ℤ ≥ M
seqfveq.2 ⊢ φ ∧ k ∈ M … N → F ⁡ k = G ⁡ k
Assertion seqfveq ⊢ φ → seq M + ˙ F ⁡ N = seq M + ˙ G ⁡ N

Proof

Step Hyp Ref Expression
1 seqfveq.1 ⊢ φ → N ∈ ℤ ≥ M
2 seqfveq.2 ⊢ φ ∧ k ∈ M … N → F ⁡ k = G ⁡ k
3 eluzel2 ⊢ N ∈ ℤ ≥ M → M ∈ ℤ
4 1 3 syl ⊢ φ → M ∈ ℤ
5 uzid ⊢ M ∈ ℤ → M ∈ ℤ ≥ M
6 4 5 syl ⊢ φ → M ∈ ℤ ≥ M
7 seq1 ⊢ M ∈ ℤ → seq M + ˙ F ⁡ M = F ⁡ M
8 4 7 syl ⊢ φ → seq M + ˙ F ⁡ M = F ⁡ M
9 fveq2 ⊢ k = M → F ⁡ k = F ⁡ M
10 fveq2 ⊢ k = M → G ⁡ k = G ⁡ M
11 9 10 eqeq12d ⊢ k = M → F ⁡ k = G ⁡ k ↔ F ⁡ M = G ⁡ M
12 2 ralrimiva ⊢ φ → ∀ k ∈ M … N F ⁡ k = G ⁡ k
13 eluzfz1 ⊢ N ∈ ℤ ≥ M → M ∈ M … N
14 1 13 syl ⊢ φ → M ∈ M … N
15 11 12 14 rspcdva ⊢ φ → F ⁡ M = G ⁡ M
16 8 15 eqtrd ⊢ φ → seq M + ˙ F ⁡ M = G ⁡ M
17 fzp1ss ⊢ M ∈ ℤ → M + 1 … N ⊆ M … N
18 4 17 syl ⊢ φ → M + 1 … N ⊆ M … N
19 18 sselda ⊢ φ ∧ k ∈ M + 1 … N → k ∈ M … N
20 19 2 syldan ⊢ φ ∧ k ∈ M + 1 … N → F ⁡ k = G ⁡ k
21 6 16 1 20 seqfveq2 ⊢ φ → seq M + ˙ F ⁡ N = seq M + ˙ G ⁡ N