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Proof step 10 from proof nummult-le-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ⊦ i < 1 ↔ i ≤ 0; (s)% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)% ⊦ i < j + 1 ↔ i ≤ j; (s)% ⊦ i - j < i0 ↔ i < j + i0; (s)% ⊦ i + 1 ≤ j ↔ i < j; (s)% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; (s)% ⊦ i + j = i + i0 ↔ j = i0; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + (j - i) = j; (s)