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Proof step 1 from proof div-nle (conclusion)

⊦ 0 < j ∧ i < i0 ∧ 0 ≤ i → ¬ i0 ≤ i / j

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: div-ls-same: ⊦ 0 < j → (i < i / j ↔ i < 0 ∧ 1 < j);%lenls: ⊦ i ≤ j → ¬ j < i; from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs