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Proof step 18 from proof abs-shift (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 19:
⊦ j < 0 → abs(j) = 0 - j

Rule argument

spec: int-abs name: Abs-neg seq: ⊦ j < 0 → abs(j) = 0 - j subst: [j] → [j] flags: <all paths>

Additional information

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