Metamath Proof Explorer


Theorem ltmod

Description: A sufficient condition for a "less than" relationship for the mod operator. (Contributed by Glauco Siliprandi, 11-Dec-2019)

Ref Expression
Hypotheses ltmod.a ⊢ φ → A ∈ ℝ
ltmod.b ⊢ φ → B ∈ ℝ +
ltmod.c ⊢ φ → C ∈ A − A mod B A
Assertion ltmod ⊢ φ → C mod B < A mod B

Proof

Step Hyp Ref Expression
1 ltmod.a ⊢ φ → A ∈ ℝ
2 ltmod.b ⊢ φ → B ∈ ℝ +
3 ltmod.c ⊢ φ → C ∈ A − A mod B A
4 1 2 modcld ⊢ φ → A mod B ∈ ℝ
5 1 4 resubcld ⊢ φ → A − A mod B ∈ ℝ
6 1 rexrd ⊢ φ → A ∈ ℝ *
7 icossre ⊢ A − A mod B ∈ ℝ ∧ A ∈ ℝ * → A − A mod B A ⊆ ℝ
8 5 6 7 syl2anc ⊢ φ → A − A mod B A ⊆ ℝ
9 8 3 sseldd ⊢ φ → C ∈ ℝ
10 2 rpred ⊢ φ → B ∈ ℝ
11 9 2 rerpdivcld ⊢ φ → C B ∈ ℝ
12 11 flcld ⊢ φ → C B ∈ ℤ
13 12 zred ⊢ φ → C B ∈ ℝ
14 10 13 remulcld ⊢ φ → B ⁢ C B ∈ ℝ
15 5 rexrd ⊢ φ → A − A mod B ∈ ℝ *
16 icoltub ⊢ A − A mod B ∈ ℝ * ∧ A ∈ ℝ * ∧ C ∈ A − A mod B A → C < A
17 15 6 3 16 syl3anc ⊢ φ → C < A
18 9 1 14 17 ltsub1dd ⊢ φ → C − B ⁢ C B < A − B ⁢ C B
19 icossicc ⊢ A − A mod B A ⊆ A − A mod B A
20 19 3 sselid ⊢ φ → C ∈ A − A mod B A
21 1 2 20 lefldiveq ⊢ φ → A B = C B
22 21 eqcomd ⊢ φ → C B = A B
23 22 oveq2d ⊢ φ → B ⁢ C B = B ⁢ A B
24 23 oveq2d ⊢ φ → A − B ⁢ C B = A − B ⁢ A B
25 18 24 breqtrd ⊢ φ → C − B ⁢ C B < A − B ⁢ A B
26 modval ⊢ C ∈ ℝ ∧ B ∈ ℝ + → C mod B = C − B ⁢ C B
27 9 2 26 syl2anc ⊢ φ → C mod B = C − B ⁢ C B
28 modval ⊢ A ∈ ℝ ∧ B ∈ ℝ + → A mod B = A − B ⁢ A B
29 1 2 28 syl2anc ⊢ φ → A mod B = A − B ⁢ A B
30 25 27 29 3brtr4d ⊢ φ → C mod B < A mod B