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Proof step 10 from proof nummult-le-03 (previous step)

i1 ≤ i0, i1 * i + j0 < i, i1 * i + j0 < 0, j0 < i, j0 ≤ 0, (i1 * i + j0) + 1 < i, 0 < i, abs(j0) < i ⊦ i1 * i + j0 ≤ i * i0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
i0 + 1 < i, i * i0 < i1 * i + j0, abs(j0) < i, 0 < i, (i1 * i + j0) + 1 < i, j0 ≤ 0, j0 < i, i1 * i + j0 < i, i1 ≤ i0, i * i0 < ~1, i1 < 0, i0 < 0, i0 < i, i1 < i, i1 + 1 < i ⊦

Rule argument

Additional information

Used simplifier rules: modzero-02: ⊦ i < 0 ∧ j0 < j ∧ 0 < j → i * j + j0 < 0; from specification int-mult%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%zero-gels-03: ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%lsone: ⊦ i < 1 ↔ i ≤ 0; from specification int-abs%ls-same: ⊦ 0 < i → (i * j < i ↔ j < 1); from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%zero-ls: ⊦ 0 < i → (i * j < 0 ↔ j < 0); from specification int-mult%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%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; from specification int-abs%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%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%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs