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Proof step 20 from proof suffix-lcp (conclusion)

⊦ n0 ≤ # ar ∧ n1 ≤ # ar ∧ n2 ≤ # ar ∧ suffix(ar, n0) < suffix(ar, n1) ∧ suffix(ar, n1) < suffix(ar, n2) → (lcp(ar, n0, n1) < lcp(ar, n0, n2) ↔ false)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
lcp(ar, n0, n1) < lcp(ar, n0, n2), suffix(ar, n1) < suffix(ar, n2), suffix(ar, n0) < suffix(ar, n1), n2 ≤ # ar, n1 ≤ # ar, n0 ≤ # ar ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m < n → n ≠ 0], [CSIMPFORWARD : ⊦ m < n → n ≠ 0])]]