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Proof step 2 from proof modmod-add-02 (previous step)

i ≤ 0, i0 ≤ 0 ⊦ (i % j + i0) % j = (i + i0) % j, 0 = j

User applied rule insert lemma

And obtained the following new goals

Rule argument

spec: int-div inst: name: modmod-add-01 seq: 0 ≠ j ⊦ 0 ≤ i ∧ 0 ≤ i0 → (i % j + i0) % j = (i + i0) % j subst: [j, i, i0] → [j, ~ i, ~ i0] flags: current, []

Additional information

Used simplifier rules: mod-neg-01: j ≠ 0 ⊦ ~ i % j = ~(i % j);%negle-07: ⊦ i ≤ ~ j ↔ i + j ≤ 0; from specification int-abs%add-zero-rev: ⊦ 0 + i = 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%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%neg-11: ⊦ ~ i + ~ j = ~(i + j); from specification int-abs%neg-12: ⊦ ~ i = ~ j ↔ i = j; from specification int-abs Used terms: j, ~ i, ~ i0