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Proof step 17 from proof div-minus-onediff-01 (previous step)

i0 < 0, i0 < ~1, 0 < j ⊦ 1 % j = 0, (i0 * j + 1) / j = i0 + 1, j = 0, 0 ≤ i0 * j ∧ 0 ≤ 1 ∧ 0 < j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
0 < j, i0 < ~1, i0 < 0 ⊦ 1 % j = 0, (i0 * j + 1) / j = i0 + 1, j = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); (s)