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Proof step 8 from proof abs-shift (previous step)

i % j < 0, 0 < j, i % j + 1 < j ⊦ i % j + abs(i) % j = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ ¬ j < 0 → abs(j) = j

Rule argument

spec: int-abs inst: name: Abs-pos seq: ⊦ ¬ j < 0 → abs(j) = j subst: [j] → [i] flags: rewrite

Additional information

Used simplifier rules: mod-neg-01: j ≠ 0 ⊦ ~ i % j = ~(i % j);%mod-zero-01: ⊦ 0 < j ∧ 0 ≤ i → ¬ i % j < 0;%add-mult-two: ⊦ i + i = 2 * i; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%zero-neq-01: ⊦ 0 < i → (i * j = 0 ↔ j = 0); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%neg-03: ⊦ ~ i + i = 0; from specification int-abs Used terms: i