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Proof step 1 from proof modeq-pos-01 (conclusion)

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

User applied rule insert lemma

And obtained the following new goals

Rule argument

spec: int-div inst: 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] flags: current, []

Additional information

Used simplifier rules: succ-rmerge: ⊦ i * j + i = i * (j + 1); from specification int-mult%Sub-def: ⊦ 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%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%neg-13: ⊦ ~ i < j ↔ 0 < j + i; from specification int-abs%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1 Used terms: i0 - 1, j, j + j0