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

j < 0, 0 ≠ j ⊦ (i % j + i0 % j) % j = (i + i0) % j ∨ (i % j + i0 % j) % j = (i + i0) % j + j ∨ (i % j + i0 % j) % j = (i + i0) % j - j

User applied rule insert lemma

And obtained the following new goals

Rule argument

spec: int-div inst: name: modmod-add-law-pos seq: 0 < j ⊦ (i % j + i0 % j) % j = (i + i0) % j ∨ (i % j + i0 % j) % j = (i + i0) % j + j ∨ (i % j + i0 % j) % j = (i + i0) % j - j subst: [j, i, i0] → [~ j, i, i0] flags: current, []

Additional information

Used simplifier rules: mod-neg: j ≠ 0 ⊦ i % ~ j = i % j;%sub-eq: ⊦ i - j = i0 ↔ i = i0 + 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%subneg: ⊦ i - ~ j = i + j; from specification int-abs%neg: ⊦ 0 < ~ i ↔ i < 0; from specification int-abs%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs Used terms: ~ j, i, i0