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Proof step 5 from proof nummult-le-01 (previous step)

j + 1 < i, j < 0, 0 < i, j < i ⊦ i * i0 ≤ j ↔ i0 ≤ (j + 1 - i) / i, 0 < i ∧ 0 ≤ j + 1 - i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
j < i, 0 < i, j < 0, j + 1 < i ⊦ i * i0 ≤ j ↔ i0 ≤ (j + 1 - i) / i, i = j + 1

Rule argument

Additional information

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