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Proof step 30 from proof ex-modeq-half (previous step)

odd(j), 0 < j, j / 2 < i0 % j, 0 ≤ i0 % j, 0 ≤ (i0 % j - j) % j ⊦ ∃ i. ≡mod(i, i0, j) ∧ abs(i) ≤ j / 2, (i0 % j - j) |%| j = i0 % j ∧ j ≤ i0 % j + j / 2

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 31:
⊦ 0 < j ∧ i < 0 → (0 ≤ i % j ↔ i % j = 0)
Proof step 32:
0 < j ∧ i0 % j - j < 0, odd(j), 0 < j, j / 2 < i0 % j, 0 ≤ i0 % j, (i0 % j - j) % j = 0 ⊦ ∃ i. ≡mod(i, i0, j) ∧ abs(i) ≤ j / 2, (i0 % j - j) |%| j = i0 % j ∧ j ≤ i0 % j + j / 2

Rule argument

spec: int-div inst: name: mod-ngez seq: ⊦ 0 < j ∧ i < 0 → (0 ≤ i % j ↔ i % j = 0) subst: [j, i] → [j, i0 % j - j] flags: rewrite

Additional information

Used simplifier rules: sub-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%ax-04: ⊦ i + 0 = i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%mod-less-01: ⊦ 0 < j → i % j < j; from specification int-div Used terms: j, i0 % j - j