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Proof step 2 from proof challenge3b (previous step)

ar[m] = ar[n], n < # ar, m < # ar, ∀ m0, n0. ar[m0] ≠ ar[n0] ∨ ar[m0] = a ∨ m0 = n0 ∨ ¬ m0 < # ar ∨ ¬ n0 < # ar ⊦ ar[n] = a, m = n

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 4, [QUANTTERMLIST : (n, m)]]

Additional information

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