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Proof step 11 from proof challenge3a (previous step)

∀ m2, n1. ar[m2] ≠ ar[n1] ∨ m2 = n1 ∨ ¬ m2 < # ar ∨ ¬ n1 < # ar, ar[m] = ar[n0], m0 < # ar, m < # ar, n0 < # ar, ∀ n0. m0 < n0 ∧ n0 < # ar → ar[n0] ≠ ar[m0], ∀ m1. m1 < m0 → (∀ n. m1 < n ∧ n < # ar → ar[m1] ≠ ar[n]) ⊦ m = n0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 1, [QUANTTERMLIST : (m, n0)]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (m, n0)]]