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Proof step 2 from proof sub-ls (previous step)

0 < i0, i < j ⊦ (i - j) / i0 ≤ 0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: div-lezero: ⊦ 0 < j → (i / j ≤ 0 ↔ i < j);%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i < j → ¬ j < i; (s)% ⊦ i + j < j ↔ i < 0; (s)%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i - j < i0 ↔ i < j + i0; (s)