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Proof step 6 from proof challenge3b (previous step)

ar[m1] = ar[n], ar[m] = ar[n0], m1 < m0, m1 + 1 < # ar, m1 < # ar, m1 < n, n < # ar, m0 < # ar, m < # ar, n0 < # ar, ∀ n. m1 < n ∧ n < # ar → ar[m1] ≠ ar[n] ∨ ar[m1] = a, ∀ n0. m0 < n0 ∧ n0 < # ar → ar[n0] ≠ ar[m0] ∨ ar[n0] = a, ∀ m1. m1 < m0 → (∀ n. m1 < n ∧ n < # ar → ar[m1] ≠ ar[n] ∨ ar[m1] = a) ⊦ ar[n0] = a, m = n0, ar[n] = a, m1 = m0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 11, [QUANTTERMLIST : (n)]]

Additional information

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