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 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 M ψ
6 fzind2.6 y M ..^ N χ θ
7 elfz2 K M N M N K M K K N
8 anass M N K M K K N M N K M K K N
9 df-3an M N K M N K
10 9 anbi1i M N K M K K N M N K M K K N
11 3anass K M K K N K M K K N
12 11 anbi2i M N K M K K N M N K M K K N
13 8 10 12 3bitr4i M N K M K K N M N K M K K N
14 7 13 bitri K M N M N K M K K N
15 eluz2 N M M N M N
16 15 5 sylbir M N M N ψ
17 3anass y M y y < N y M y y < N
18 elfzo y M N y M ..^ N M y y < N
19 18 6 syl6bir y M N M y y < N χ θ
20 19 3coml M N y M y y < N χ θ
21 20 3expa M N y M y y < N χ θ
22 21 impr M N y M y y < N χ θ
23 17 22 sylan2b M N y M y y < N χ θ
24 1 2 3 4 16 23 fzind M N K M K K N τ
25 14 24 sylbi K M N τ