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Proof step 3 from proof mod-sub (previous step)

⊦ 0 < i ∧ j < 0 → (i + ~ j) % i = ~(j % i)

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-div inst: name: mod-add seq: ⊦ 0 < i ∧ 0 ≤ j → (i + j) % i = j % i subst: [j, i] → [~ j, i] flags: rewrite, current

Additional information

Used simplifier rules: mod-neg-01: j ≠ 0 ⊦ ~ i % j = ~(i % j);%neg-12: ⊦ ~ i = ~ j ↔ i = j; from specification int-abs%neg-15: ⊦ i + ~ j = i - j; 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%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%negle-07: ⊦ i ≤ ~ j ↔ i + j ≤ 0; from specification int-abs%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%zero-gels-03: ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs Used terms: ~ j, i