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Proof step 2 from proof div-sum-ls-01 (previous step)

0 < i, 0 < i1, 0 < j1, 0 < j2, 0 < j, j1 < j, j2 < j ⊦ (i * j1 + i1 * j2) / j < i + i1

User applied rule insert lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: div-sum-ls seq: ⊦ 0 ≤ i ∧ 0 ≤ i1 ∧ 0 < i + i1 ∧ 0 < j1 ∧ 0 < j2 ∧ 0 < j ∧ j1 < j ∧ j2 < j → (i * j1 + i1 * j2) / j < i + i1 subst: [i, i1, j1, j2, j] → [i, i1, j1, j2, j]

Additional information

Used simplifier rules: ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ 0 < i ∧ 0 < j → 0 < 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 → ¬ j < i; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i < j → ¬ j < i; (s) Used terms: i, i1, j1, j2, j