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Proof step 1 from proof suffix-lcp (
previous step
)
⊦ lcp(ar, n0, n1) = (((n0 ≥ # ar ∨ n1 ≥ # ar ∨ ar[n0] ≠ ar[n1]) ⊃ 0; lcp(ar, n0 + 1, n1 + 1) + 1))
A spec-lemma