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Proof step 14 from proof div-ls-same (previous step)

j0 < j, i0 < j, abs(j0) < j, 1 < j, i0 < 0, 0 < j, j0 ≤ 0, i0 ≤ i0 * j + j0 ⊦ j ≤ 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
0 < j, i0 < 0, 1 < j, abs(j0) < j, i0 < j, j0 < j, i0 ≤ i0 * j + j0, j0 ≤ 0 ⊦ j = 1

Rule argument

Additional information

Used simplifier rules: ⊦ 0 < i → (i ≤ 1 ↔ i = 1); (s)