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Proof step 6 from proof modlsp-03 (previous step)

0 ≤ i1 + j1, abs(j1) < abs(i1), i1 + j1 < abs(i1), j * i + i1 < j * i0, 0 < j, j ≤ abs(i1), i0 < i, abs(j1) < j, i1 + j1 < j ⊦ i1 = 0

User applied rule insert lemma

And obtained the following new goals

Proof step 7:
j * i + j0 < j * i0, abs(j0) < j ⊦ i ≤ i0
Proof step 8:
0 ≤ i1 + j1, abs(j1) < abs(i1), i1 + j1 < abs(i1), j * i + i1 < j * i0, 0 < j, j ≤ abs(i1), i0 < i, abs(j1) < j, i1 + j1 < j ⊦ j * (i + 1) + (i1 - j) < j * i0 ∧ abs(i1 - j) < j, i1 = 0

Rule argument

spec: int-mult inst: name: modlsp-02 seq: j * i + j0 < j * i0, abs(j0) < j ⊦ i ≤ i0 subst: [j, i, j0, i0] → [j, i + 1, i1 - j, i0] flags: current, []

Additional information

Used simplifier rules: succ-01: ⊦ i + 1 ≤ j ↔ i < j; from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: j, i + 1, i1 - j, i0