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Proof step 11 from proof div-minus-onediff-01 (previous step)

¬(((j0 - 1) * j) % j = 0 ∧ 1 % j = 0 ∧ 0 ≠ j), 0 < j, j0 < 0 ⊦ j = 0, ((j0 - 1) * j + 1) / j = j0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
0 < j, j0 < 0 ⊦ j = 0, ((j0 - 1) * j + 1) / j = j0, 1 % j = 0

Rule argument

Additional information

Used simplifier rules: mod-mult-04: j ≠ 0 ⊦ (i * j) % j = 0;%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult