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Proof step 12 from proof suffix-lcp (previous step)

ar[n0] = ar[n2], n2 < # ar, n1 ≤ # ar, n0 < # ar, (((n0 ≥ # ar ∨ n1 ≥ # ar ∨ ar[n0] ≠ ar[n1]) ⊃ 0; lcp(ar, n0 + 1, n1 + 1) + 1)) ≤ lcp(ar, n0 + 1, n2 + 1), suffix(ar, n1) < suffix(ar, n2), suffix(ar, n0) < suffix(ar, n1), ∀ ar0, n, n3, n4. lcp(ar, n0 + 1, n2 + 1) ≠ lcp(ar0, n, n3) ∨ ¬ lcp(ar0, n, n4) < lcp(ar0, n, n3) ∨ ¬ suffix(ar0, n4) < suffix(ar0, n3) ∨ ¬ suffix(ar0, n) < suffix(ar0, n4) ∨ ¬ n3 ≤ # ar0 ∨ ¬ n4 ≤ # ar0 ∨ ¬ n ≤ # ar0 ⊦ n2 = # ar, n0 = # ar

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
ar[n0] = ar[n2], suffix(ar, n1) < suffix(ar, n2), suffix(ar, n0) < suffix(ar, n1), n2 < # ar, n1 ≤ # ar, n0 < # ar, (((n1 = # ar ∨ ar[n2] ≠ ar[n1]) ⊃ 0; lcp(ar, n0 + 1, n1 + 1) + 1)) ≤ lcp(ar, n0 + 1, n2 + 1), ∀ ar0, n, n3, n4. lcp(ar, n0 + 1, n2 + 1) ≠ lcp(ar0, n, n3) ∨ ¬ lcp(ar0, n, n4) < lcp(ar0, n, n3) ∨ ¬ suffix(ar0, n4) < suffix(ar0, n3) ∨ ¬ suffix(ar0, n) < suffix(ar0, n4) ∨ ¬ n3 ≤ # ar0 ∨ ¬ n4 ≤ # ar0 ∨ ¬ n ≤ # ar0 ⊦ n2 = # ar, n0 = # ar

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m ≤ n → (n ≤ m ↔ m = n)], [CSIMPSEQ : ⊦ m < n → ¬ n < m], [CSIMPSEQ : ⊦ m ≠ n → (m ≤ n ↔ m < n)], [CSIMPSEQ : ⊦ n ≥ m ↔ m ≤ n])]]