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Proof step 20 from proof div-minus-onediff-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
1 % j = 0, j0 < 0, 0 < j ⊦ 0 = j, j = 1

Rule argument

Additional information

Used simplifier rules: mod-mult-04: j ≠ 0 ⊦ (i * j) % j = 0;%div-mult-03: j ≠ 0 ⊦ (i * j) / j = i;%one-02: j ≠ 0 ⊦ 1 / j = 1 ↔ j = 1;% ⊦ i + j = i + i0 ↔ j = i0; (s)% ⊦ i - j = i0 ↔ i = i0 + j; (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 - i0) = i + j - i0; (s)%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult