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Proof step 23 from proof challenge3b (previous step)

n1 = n + 1, ¬ bv0, ar[m] = ar[n0], # ar = (n5 + 1) + n, a1 = a, ar1 = ar, ar[n3] = ar[m0], m2 = m0, m ≠ n0, ar[n0] ≠ a, ar[m0] ≠ a, ar[m0] ≠ ar[n], m ≤ n5 + n, n0 ≤ n5 + n, m0 < # ar, m < # ar, n0 < # ar, n < # ar, n3 ≤ n, m0 < n3, m0 < n, n3 < # ar, n3 ≤ n5 + n, m0 < n5 + n, ∀ n0. m0 < n0 ∧ n0 < n → ar[n0] ≠ ar[m0] ∨ ar[n0] = a, ∀ m1. m1 < m0 → (∀ n. m1 < n ∧ n < # ar → ar[m1] ≠ ar[n] ∨ ar[m1] = a) ⊦ ⦉while n1 < # ar1 ∧ ¬ bv0 do if ar1[m2] = ar1[n1] ∧ ar1[m2] ≠ a1 then m1 := m2, n2 := n1, bv0 := true else n1 := n1 + 1⦊ ⦉m2 := m2 + 1⦊ ( (bv0 ⊃ (ar[m1] = ar[n2] ∧ ar[m1] ≠ a ∧ m1 ≠ n2 ∧ m1 < # ar ∧ n2 < # ar) ; (∀ m1. m1 < m2 → (∀ n. m1 < n ∧ n < # ar → ar[m1] ≠ ar[n] ∨ ar[m1] = a)) ) ∧ m2 ≤ # ar ∧ (∃ m, n. ar1[m] = ar1[n] ∧ ar[m] ≠ a1 ∧ m ≠ n ∧ m < # ar1 ∧ n < # ar1) ∧ ¬ m2 ≤ m0)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 25, [QUANTTERMLIST : (n3)]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ m < n ↔ n ≤ m], [CSIMPSEQ : ⊦ m ≤ n → (n ≤ m ↔ m = n)], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPFORWARD : ⊦ m < n ∧ m = n0 → n0 < n], [CSIMPFORWARD : ⊦ m < n ∧ n = n0 → m < n0], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1])], [EXTRATERMS : (n3)]]