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Proof step 2 from proof modeq-pos-01 (previous step)

abs(j0) < j, j0 < 0, 0 ≤ i0 * j + j0 ⊦ (i0 * j + j0) % j = j + j0

User applied rule insert lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: modeq-pos seq: 0 ≤ i0 * j + j0, 0 ≤ j0, j0 < j ⊦ (i0 * j + j0) % j = j0 subst: [i0, j, j0] → [i0 - 1, j, j + j0]

Additional information

Used simplifier rules: ⊦ i * j + i = i * (j + 1); (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%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ ~ i < j ↔ 0 < j + i; (s)% ⊦ i + j < j ↔ i < 0; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i < j → ¬ j < i; (s) Used terms: i0 - 1, j, j + j0