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Proof step 2 from proof auxexdivmod-01 (
previous step
)
i0 * j ≤ i, i < (i0 + i1) * j, 0 ≤ i0, 0 < j, 0 ≤ i, 0 ≤ i1 ⊦ ∃ i0. i0 * j ≤ i ∧ i < (i0 + 1) * j ∧ 0 ≤ i0
Lemma
auxexdivmod