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Proof step 3 from proof zero-07 (previous step)

1 < i / j, i < i / j, 0 < i, 0 < i / j ⊦ j = 0

User applied rule insert lemma

And obtained the following new goals

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

Rule argument

spec: <toplevel> name: zero-05 seq: j ≠ 0 ⊦ i < 0 → i ≤ i / j subst: [j, i] → [j, ~ i]

Additional information

Used simplifier rules: neg-01: j ≠ 0 ⊦ ~ i / j = ~(i / j);% ⊦ ~ i < 0 ↔ 0 < i; (s)% ⊦ ~ i ≤ j ↔ 0 ≤ i + j; (s)% ⊦ i + ~ j = i - j; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; (s)% ⊦ i + 0 = i; (s)% ⊦ i < j → (j ≤ i ↔ false); (as) Used terms: j, ~ i