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Proof step 11 from proof abs-shift (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

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

Additional information

Used simplifier rules: mod-zero-06: ⊦ 0 ≤ i ∧ j ≠ 0 → ¬ i % j < 0;%mod-neg-01: j ≠ 0 ⊦ ~ i % j = ~(i % j);%neg-06: ⊦ 0 - i = ~ i; from specification int-abs%neg-03: ⊦ ~ i + i = 0; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%Abs-pos: ⊦ ¬ j < 0 → abs(j) = j; from specification int-abs%add-mult-two: ⊦ i + i = 2 * i; from specification int-mult%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 Used terms: i