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Proof step 15 from proof suffix-lcp (previous step)

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 ⊦

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 14:
¬ (n0 = # ar ∨ n2 = # ar ∨ ar[n0] ≠ ar[n2]), m0 + 1 = 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

[CASEDARG : pos: left 1, ()]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m + 1 ≠ 0], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n = n + m])]]