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Proof step 14 from proof nummult-le-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
i1 < i, i1 < ~1, i0 < i, i0 + 1 < i, i * (i1 + 1) + j0 < i, j0 < i, 0 < i, i * (i1 + 1) + j0 ≤ 0, abs(j0) < i, j0 ≤ 0, i * i0 < i * (i1 + 1) + j0, i1 < i0, i * i0 + 1 < i, i * i0 < i, i0 < 0, i1 < 0 ⊦ i = i * (i1 + 1) + j0

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%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; from specification int-abs%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%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (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