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Proof step 18 from proof div-mult-two-nls-01 (
previous step
)
⊦ 0 ≤ i ∧ 0 ≤ i0 ∧ 0 ≤ j0 ∧ 0 ≤ j1 ∧ j0 < j ∧ j1 < j → (i * j + j0 = i0 * j + j1 ↔ i = i0 ∧ j0 = j1)
A spec-lemma