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Proof step 3 from proof modmod-add-law (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Rule argument

spec: <toplevel> 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]

Additional information

Used simplifier rules: mod-neg: j ≠ 0 ⊦ i % ~ j = i % j;% ⊦ i - j = i0 ↔ i = i0 + j; (s)% ⊦ i + ~ j = i - j; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i - ~ j = i + j; (s)% ⊦ 0 < ~ i ↔ i < 0; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)% ⊦ i < j → ¬ j < i; (s) Used terms: ~ j, i, i0