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Proof step 16 from proof suffix-lcp (previous step)

m0 + 1 = (((n0 ≥ # ar ∨ n2 ≥ # ar ∨ ar[n0] ≠ ar[n2]) ⊃ 0; lcp(ar, n0 + 1, n2 + 1) + 1)), n0 ≤ # ar, n1 ≤ # ar, n2 ≤ # ar, suffix(ar, n0) < suffix(ar, n1), suffix(ar, n1) < suffix(ar, n2), lcp(ar, n0, n1) < (((n0 ≥ # ar ∨ n2 ≥ # ar ∨ ar[n0] ≠ ar[n2]) ⊃ 0; lcp(ar, n0 + 1, n2 + 1) + 1)), lcp(ar, n0, n1) ≤ m0, ∀ ar0, n, n3, n4. m0 ≠ 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 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
m0 + 1 = (((n0 = # ar ∨ n2 = # ar ∨ ar[n0] ≠ ar[n2]) ⊃ 0; lcp(ar, n0 + 1, n2 + 1) + 1)), lcp(ar, n0, n1) ≤ m0, n0 ≤ # ar, n1 ≤ # ar, n2 ≤ # ar, suffix(ar, n0) < suffix(ar, n1), suffix(ar, n1) < suffix(ar, n2), ∀ ar0, n, n3, n4. m0 ≠ 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 ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m ≤ n → (n < m ↔ false)], [CSIMPSEQ : ⊦ m < n + 1 ↔ ¬ n < m], [CSIMPSEQ : ⊦ m + 1 ≠ 0], [CSIMPFORWARD : ⊦ m < n → n ≠ 0], [CSIMPFORWARD : ⊦ m < n ∧ n = n0 → m < n0], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m ≤ n → (n ≤ m ↔ m = n)], [CSIMPSEQ : ⊦ n ≥ m ↔ m ≤ n])]]