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Proof step 14 from proof exdivmod (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
j < 0, i ≤ 0 → i + j1 = j * i1 ∧ 0 ≤ i1 ∧ 0 ≤ j1 ∧ j1 + j < 0 ⊦ j = 0, ∃ i0, j0. i = i0 * j + j0 ∧ abs(j0) + j < 0

Rule argument

Additional information

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