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Proof step 1 from proof div-not-ls (conclusion)

⊦ i < j ∧ 0 < i → ¬ j | i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
j | i, i < j, 1 < j, 0 < j, 0 < i ⊦

Rule argument

Additional information

Used simplifier rules: forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1