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Proof step 9 from proof ex-modeq-half (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ i < 0 → abs(i) = ~ i
Proof step 11:
i0 |%| j - j < 0, 0 < j, odd(j), 0 < j, j / 2 < abs(i0 |%| j) ⊦ (i0 |%| j - j) |%| j = i0 |%| j ∧ ~(i0 |%| j - j) ≤ j / 2, ∃ i. ≡mod(i, i0, j) ∧ abs(i) ≤ j / 2

Rule argument

spec: int-abs inst: name: abs-08 seq: ⊦ i < 0 → abs(i) = ~ i subst: [i] → [i0 |%| j - j] flags: rewrite

Additional information

Used simplifier rules: mod-le-same: ⊦ 0 < j → ¬ j ≤ i |%| j;%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%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs Used terms: i0 |%| j - j