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Proof step 4 from proof suffix-lcp (
previous step
)
⊦ n0 < # ar ∧ n1 < # ar → (suffix(ar, n0) < suffix(ar, n1) ↔ ar[n0] < ar[n1] ∨ suffix(ar, n0 + 1) < suffix(ar, n1 + 1) ∧ ar[n0] = ar[n1])
A spec-lemma