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Proof step 10 from proof div-ls (previous step)

j2 < j, i0 * j + j2 < j1 * j, 0 < j1, 0 < j0, 0 < j, 0 < i0, 0 ≤ i0, 0 ≤ j2, j1 ≤ j0, j0 ≤ i0, j1 ≤ i0 ⊦ i0 = j1, j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
0 < j, 0 < j0, 0 < j1, i0 * j + j2 < j1 * j, j2 < j, j1 < i0, 1 < i0, 0 < i0, j0 ≤ i0, j1 ≤ j0, 0 ≤ j2, 0 ≤ i0 ⊦ i0 = j1, j = 0

Rule argument

Additional information

Used simplifier rules: forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)