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Proof step 2 from proof div-sum-ls-01 (
previous step
)
⊦ 0 ≤ i ∧ 0 ≤ i1 ∧ 0 < i + i1 ∧ 0 < j1 ∧ 0 < j2 ∧ 0 < j ∧ j1 < j ∧ j2 < j → (i * j1 + i1 * j2) / j < i + i1
Lemma
div-sum-ls