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Proof step 12 from proof div-zero-02 (previous step)

abs(i) < abs(j) ⊦ i / j = 0, 0 < j ∧ i < 0 ∧ ~ i < j, j < 0 ∧ 0 ≤ i ∧ i + j < 0, 0 < j ∧ 0 ≤ i ∧ i < j, 0 < ~ j ∧ 0 ≤ ~ i ∧ ~ i < ~ j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
0 < abs(j) + i, i < 0 ⊦ i / j = 0, i = 0, 0 < j ∧ 0 < j + i, j < i ∧ j < 0

Rule argument

Additional information

Used simplifier rules: div-eq-zero-01: ⊦ 0 ≠ j → (i / j = 0 ↔ abs(i) < abs(j));%zero: j ≠ 0 ⊦ 0 / j = 0;%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs% ⊦ i < j → ¬ j < i; (s)% ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)% ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)% ⊦ i ≤ j → (j ≤ i ↔ i = j); (s)% ⊦ i < j → (j ≤ i ↔ false); (as)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ 0 < abs(i) ↔ i ≠ 0; (s)%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1% ⊦ 0 < j → abs(j) = j; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ ~ i < j ↔ 0 < j + i; (s)% ⊦ ~ j < ~ i ↔ 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 + 0 = i; (s)% ⊦ i ≤ ~ j ↔ i + j ≤ 0; (s)% ⊦ 0 < ~ i ↔ i < 0; (s)