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Proof step 22 from proof div-eq-zero-01 (previous step)

i = i0 * j + j0 ∧ j0 ≤ 0 ∧ j < j0 ∧ 0 ≤ i0, j < ~1, i < 0, j < i, j ≤ i, j < 0 ⊦ i0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
i0 * j + j0 < i0, j < j0, j < 0, j ≤ i0 * j + j0, j < i0 * j + j0, j < ~1, 0 < i0, j0 < i0, j0 ≤ 0, j < i0, (i0 * j + j0) + 1 < i0 ⊦ i0 = 0

Rule argument

Additional information

Used simplifier rules: lst-04: ⊦ i + j < i0 ∧ j0 < j → i + j0 < i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%modzero-02: ⊦ i < 0 ∧ j0 < j ∧ 0 < j → i * j + j0 < 0; from specification int-mult%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs