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Proof step 2 from proof div-nls (previous step)

i0 < i / j, 0 < j, 0 ≤ i ⊦ i0 < i

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);% ⊦ i ≤ j → ¬ j < i; (s)%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (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% ⊦ i < j ↔ ¬ j ≤ i; (ss)