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Proof step 5 from proof suffix-lcp (previous step)

ar[n2] = ar[n1], ar[n0] = ar[n1], n0 < # ar, n2 < # ar, n1 < # ar, lcp(ar, n0 + 1, n1 + 1) < lcp(ar, n0 + 1, n2 + 1), suffix(ar, n0 + 1) < suffix(ar, n1 + 1), suffix(ar, n1 + 1) < suffix(ar, 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 ⊦ n0 = # ar, n2 = # ar, n1 = # ar

User applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 9, [extquanttermlist: (ar, n0 + 1, n2 + 1, n1 + 1) abort:F , discard:T , computed:T ]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ m + 1 ≤ n ↔ m < n], [CSIMPSEQ : ⊦ m ≠ n → (¬ m < n ↔ n < m)], [CSIMPSEQ : ⊦ m < n → ¬ n < m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1])], [EXTRATERMS : (ar, n0 + 1, n2 + 1, n1 + 1)]]