Metamath Proof Explorer


Theorem fzind2

Description: Induction on the integers from M to N inclusive. The first four hypotheses give us the substitution instances we need; the last two are the basis and the induction step. Version of fzind using integer range definitions. (Contributed by Mario Carneiro, 6-Feb-2016)

Ref Expression
Hypotheses fzind2.1 ⊢ x = M → φ ↔ ψ
fzind2.2 ⊢ x = y → φ ↔ χ
fzind2.3 ⊢ x = y + 1 → φ ↔ θ
fzind2.4 ⊢ x = K → φ ↔ τ
fzind2.5 ⊢ N ∈ ℤ ≥ M → ψ
fzind2.6 ⊢ y ∈ M ..^ N → χ → θ
Assertion fzind2 ⊢ K ∈ M … N → τ

Proof

Step Hyp Ref Expression
1 fzind2.1 ⊢ x = M → φ ↔ ψ
2 fzind2.2 ⊢ x = y → φ ↔ χ
3 fzind2.3 ⊢ x = y + 1 → φ ↔ θ
4 fzind2.4 ⊢ x = K → φ ↔ τ
5 fzind2.5 ⊢ N ∈ ℤ ≥ M → ψ
6 fzind2.6 ⊢ y ∈ M ..^ N → χ → θ
7 elfz2 ⊢ K ∈ M … N ↔ M ∈ ℤ ∧ N ∈ ℤ ∧ K ∈ ℤ ∧ M ≤ K ∧ K ≤ N
8 anass ⊢ M ∈ ℤ ∧ N ∈ ℤ ∧ K ∈ ℤ ∧ M ≤ K ∧ K ≤ N ↔ M ∈ ℤ ∧ N ∈ ℤ ∧ K ∈ ℤ ∧ M ≤ K ∧ K ≤ N
9 df-3an ⊢ M ∈ ℤ ∧ N ∈ ℤ ∧ K ∈ ℤ ↔ M ∈ ℤ ∧ N ∈ ℤ ∧ K ∈ ℤ
10 9 anbi1i ⊢ M ∈ ℤ ∧ N ∈ ℤ ∧ K ∈ ℤ ∧ M ≤ K ∧ K ≤ N ↔ M ∈ ℤ ∧ N ∈ ℤ ∧ K ∈ ℤ ∧ M ≤ K ∧ K ≤ N
11 3anass ⊢ K ∈ ℤ ∧ M ≤ K ∧ K ≤ N ↔ K ∈ ℤ ∧ M ≤ K ∧ K ≤ N
12 11 anbi2i ⊢ M ∈ ℤ ∧ N ∈ ℤ ∧ K ∈ ℤ ∧ M ≤ K ∧ K ≤ N ↔ M ∈ ℤ ∧ N ∈ ℤ ∧ K ∈ ℤ ∧ M ≤ K ∧ K ≤ N
13 8 10 12 3bitr4i ⊢ M ∈ ℤ ∧ N ∈ ℤ ∧ K ∈ ℤ ∧ M ≤ K ∧ K ≤ N ↔ M ∈ ℤ ∧ N ∈ ℤ ∧ K ∈ ℤ ∧ M ≤ K ∧ K ≤ N
14 7 13 bitri ⊢ K ∈ M … N ↔ M ∈ ℤ ∧ N ∈ ℤ ∧ K ∈ ℤ ∧ M ≤ K ∧ K ≤ N
15 eluz2 ⊢ N ∈ ℤ ≥ M ↔ M ∈ ℤ ∧ N ∈ ℤ ∧ M ≤ N
16 15 5 sylbir ⊢ M ∈ ℤ ∧ N ∈ ℤ ∧ M ≤ N → ψ
17 3anass ⊢ y ∈ ℤ ∧ M ≤ y ∧ y < N ↔ y ∈ ℤ ∧ M ≤ y ∧ y < N
18 elfzo ⊢ y ∈ ℤ ∧ M ∈ ℤ ∧ N ∈ ℤ → y ∈ M ..^ N ↔ M ≤ y ∧ y < N
19 18 6 biimtrrdi ⊢ y ∈ ℤ ∧ M ∈ ℤ ∧ N ∈ ℤ → M ≤ y ∧ y < N → χ → θ
20 19 3coml ⊢ M ∈ ℤ ∧ N ∈ ℤ ∧ y ∈ ℤ → M ≤ y ∧ y < N → χ → θ
21 20 3expa ⊢ M ∈ ℤ ∧ N ∈ ℤ ∧ y ∈ ℤ → M ≤ y ∧ y < N → χ → θ
22 21 impr ⊢ M ∈ ℤ ∧ N ∈ ℤ ∧ y ∈ ℤ ∧ M ≤ y ∧ y < N → χ → θ
23 17 22 sylan2b ⊢ M ∈ ℤ ∧ N ∈ ℤ ∧ y ∈ ℤ ∧ M ≤ y ∧ y < N → χ → θ
24 1 2 3 4 16 23 fzind ⊢ M ∈ ℤ ∧ N ∈ ℤ ∧ K ∈ ℤ ∧ M ≤ K ∧ K ≤ N → τ
25 14 24 sylbi ⊢ K ∈ M … N → τ