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Proof step 5 from proof mod-zero-06 (previous step)

0 < i, i - i / j * j < 0, 0 ≤ i, abs(j) = 1 ∨ abs(j) ≤ i ⊦ j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
0 < i, i < i / j * j, 0 < i / j * j, 1 < i / j * j, 0 ≤ i, abs(j) = 1 ∨ abs(j) ≤ i ⊦ j = 0

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%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 < j + i0; (s)