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Proof step 29 from proof exdivmod (previous step)

i + i0 * j + j0 = 0, 0 < j, i < 0, j0 < j, i ≤ 0, 0 ≤ i0, 0 ≤ j0 ⊦ ∃ i1, j1. i = i1 * j + j1 ∧ abs(j1) < j, i = (~ i0 - 1) * j + (j - j0) ∧ abs(j - j0) < j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 30:
i + i0 * j + j0 = 0, j0 < j, i < 0, 0 < j, 0 ≤ j0, 0 ≤ i0, i ≤ 0 ⊦ j - j0 < j, ∃ i1, j1. i = i1 * j + j1 ∧ abs(j1) < j

Rule argument

Additional information

Used simplifier rules: minus-left: ⊦ ~ i * j = ~(i * j);%dist-lsplit: ⊦ (i + i0) * j = i * j + i0 * j;%one: ⊦ 1 * i = i;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;% ⊦ j < i → abs(i - j) = i - j; (s)% ⊦ ~ i = j ↔ i + j = 0; (s)% ⊦ i - j + i = ~ j; (s)% ⊦ i - i1 - j = i - i1 + j; (s)% ⊦ i + ~ j = i - j; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ ~ i - j = ~(i + j); (s)