Metamath Proof Explorer


Theorem seqfeq2

Description: Equality of sequences. (Contributed by Mario Carneiro, 13-Jul-2013) (Revised by Mario Carneiro, 27-May-2014)

Ref Expression
Hypotheses seqfveq2.1 ⊢ φ → K ∈ ℤ ≥ M
seqfveq2.2 ⊢ φ → seq M + ˙ F ⁡ K = G ⁡ K
seqfeq2.4 ⊢ φ ∧ k ∈ ℤ ≥ K + 1 → F ⁡ k = G ⁡ k
Assertion seqfeq2 ⊢ φ → seq M + ˙ F ↾ ℤ ≥ K = seq K + ˙ G

Proof

Step Hyp Ref Expression
1 seqfveq2.1 ⊢ φ → K ∈ ℤ ≥ M
2 seqfveq2.2 ⊢ φ → seq M + ˙ F ⁡ K = G ⁡ K
3 seqfeq2.4 ⊢ φ ∧ k ∈ ℤ ≥ K + 1 → F ⁡ k = G ⁡ k
4 eluzel2 ⊢ K ∈ ℤ ≥ M → M ∈ ℤ
5 seqfn ⊢ M ∈ ℤ → seq M + ˙ F Fn ℤ ≥ M
6 1 4 5 3syl ⊢ φ → seq M + ˙ F Fn ℤ ≥ M
7 uzss ⊢ K ∈ ℤ ≥ M → ℤ ≥ K ⊆ ℤ ≥ M
8 1 7 syl ⊢ φ → ℤ ≥ K ⊆ ℤ ≥ M
9 fnssres ⊢ seq M + ˙ F Fn ℤ ≥ M ∧ ℤ ≥ K ⊆ ℤ ≥ M → seq M + ˙ F ↾ ℤ ≥ K Fn ℤ ≥ K
10 6 8 9 syl2anc ⊢ φ → seq M + ˙ F ↾ ℤ ≥ K Fn ℤ ≥ K
11 eluzelz ⊢ K ∈ ℤ ≥ M → K ∈ ℤ
12 seqfn ⊢ K ∈ ℤ → seq K + ˙ G Fn ℤ ≥ K
13 1 11 12 3syl ⊢ φ → seq K + ˙ G Fn ℤ ≥ K
14 fvres ⊢ x ∈ ℤ ≥ K → seq M + ˙ F ↾ ℤ ≥ K ⁡ x = seq M + ˙ F ⁡ x
15 14 adantl ⊢ φ ∧ x ∈ ℤ ≥ K → seq M + ˙ F ↾ ℤ ≥ K ⁡ x = seq M + ˙ F ⁡ x
16 1 adantr ⊢ φ ∧ x ∈ ℤ ≥ K → K ∈ ℤ ≥ M
17 2 adantr ⊢ φ ∧ x ∈ ℤ ≥ K → seq M + ˙ F ⁡ K = G ⁡ K
18 simpr ⊢ φ ∧ x ∈ ℤ ≥ K → x ∈ ℤ ≥ K
19 elfzuz ⊢ k ∈ K + 1 … x → k ∈ ℤ ≥ K + 1
20 19 3 sylan2 ⊢ φ ∧ k ∈ K + 1 … x → F ⁡ k = G ⁡ k
21 20 adantlr ⊢ φ ∧ x ∈ ℤ ≥ K ∧ k ∈ K + 1 … x → F ⁡ k = G ⁡ k
22 16 17 18 21 seqfveq2 ⊢ φ ∧ x ∈ ℤ ≥ K → seq M + ˙ F ⁡ x = seq K + ˙ G ⁡ x
23 15 22 eqtrd ⊢ φ ∧ x ∈ ℤ ≥ K → seq M + ˙ F ↾ ℤ ≥ K ⁡ x = seq K + ˙ G ⁡ x
24 10 13 23 eqfnfvd ⊢ φ → seq M + ˙ F ↾ ℤ ≥ K = seq K + ˙ G