Metamath Proof Explorer


Theorem fzo0opth

Description: Equality for a half open integer range starting at zero is the same as equality of its upper bound, analogous to fzopth and fzoopth . (Contributed by Thierry Arnoux, 27-May-2025)

Ref Expression
Hypotheses fzo0opth.1 φ M 0
fzo0opth.2 φ N 0
Assertion fzo0opth φ 0 ..^ M = 0 ..^ N M = N

Proof

Step Hyp Ref Expression
1 fzo0opth.1 φ M 0
2 fzo0opth.2 φ N 0
3 0z 0
4 1 nn0zd φ M
5 simpr φ 0 < M 0 < M
6 fzoopth 0 M 0 < M 0 ..^ M = 0 ..^ N 0 = 0 M = N
7 3 4 5 6 mp3an2ani φ 0 < M 0 ..^ M = 0 ..^ N 0 = 0 M = N
8 eqid 0 = 0
9 8 biantrur M = N 0 = 0 M = N
10 7 9 bitr4di φ 0 < M 0 ..^ M = 0 ..^ N M = N
11 simpr φ 0 = M 0 = M
12 11 oveq2d φ 0 = M 0 ..^ 0 = 0 ..^ M
13 fzo0 0 ..^ 0 =
14 12 13 eqtr3di φ 0 = M 0 ..^ M =
15 14 eqeq1d φ 0 = M 0 ..^ M = 0 ..^ N = 0 ..^ N
16 eqcom = 0 ..^ N 0 ..^ N =
17 15 16 bitrdi φ 0 = M 0 ..^ M = 0 ..^ N 0 ..^ N =
18 0zd φ 0 = M 0
19 2 nn0zd φ N
20 19 adantr φ 0 = M N
21 fzon 0 N N 0 0 ..^ N =
22 18 20 21 syl2anc φ 0 = M N 0 0 ..^ N =
23 nn0le0eq0 N 0 N 0 N = 0
24 23 biimpa N 0 N 0 N = 0
25 2 24 sylan φ N 0 N = 0
26 25 adantlr φ 0 = M N 0 N = 0
27 id N = 0 N = 0
28 0le0 0 0
29 27 28 eqbrtrdi N = 0 N 0
30 29 adantl φ 0 = M N = 0 N 0
31 26 30 impbida φ 0 = M N 0 N = 0
32 eqcom N = 0 0 = N
33 32 a1i φ 0 = M N = 0 0 = N
34 11 eqeq1d φ 0 = M 0 = N M = N
35 31 33 34 3bitrd φ 0 = M N 0 M = N
36 17 22 35 3bitr2d φ 0 = M 0 ..^ M = 0 ..^ N M = N
37 1 nn0ge0d φ 0 M
38 0red φ 0
39 1 nn0red φ M
40 38 39 leloed φ 0 M 0 < M 0 = M
41 37 40 mpbid φ 0 < M 0 = M
42 10 36 41 mpjaodan φ 0 ..^ M = 0 ..^ N M = N