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Proof step 3 from proof zero-03 (previous step)

0 < j, 0 < i, i / j < 0 ⊦

User applied rule insert lemma

And obtained the following new goals

Proof step 4:
⊦ 0 < j ∧ 0 ≤ i → 0 ≤ i / j

Rule argument

spec: <toplevel> name: pos seq: ⊦ 0 < j ∧ 0 ≤ i → 0 ≤ i / j subst: [j, i] → [j, i]

Additional information

Used simplifier rules: ⊦ i < j → (j ≤ i ↔ false); (as)% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i < j → ¬ j < i; (s) Used terms: j, i