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Proof step 17 from proof challenge3a (previous step)

∀ n1, m5, ar2, m6, n3, m7, n4. m4 = # ar2 - n1 ∧ ar2[m6] = ar2[n3] ∧ m6 ≠ n3 ∧ n3 < # ar2 ∧ m6 < # ar2 ∧ m5 < # ar2 ∧ (∀ m1. m1 < m5 → (∀ n. m1 < n ∧ n < # ar2 → ar2[m1] ≠ ar2[n])) ∧ m5 < n1 ∧ n1 ≤ # ar2 ∧ (∀ n0. m5 < n0 ∧ n0 < n1 → ar2[n0] ≠ ar2[m5]) → ⦉m0 := m5, bv := false, ar0 := ar2 ; while n1 < # ar0 ∧ ¬ bv do if ar0[m0] = ar0[n1] then m7 := m0, n4 := n1, bv := true else n1 := n1 + 1⦊ ⦉m0 := m0 + 1⦊ ( (bv ⊃ (ar2[m7] = ar2[n4] ∧ m7 ≠ n4 ∧ m7 < # ar2 ∧ n4 < # ar2); (∀ m1. m1 < m0 → (∀ n. m1 < n ∧ n < # ar2 → ar2[m1] ≠ ar2[n]))) ∧ m0 ≤ # ar2 ∧ (∃ m, n. ar0[m] = ar0[n] ∧ m ≠ n ∧ m < # ar0 ∧ n < # ar0) ∧ ¬ m0 ≤ m5), # ar = (m4 + 1) + n, ar[m] = ar[n0], ar[m0] = ar[n], m0 ≠ n, m ≠ n0, m0 < m4 + n, n < # ar, n0 < # ar, m < # ar, m0 < n, m0 < # ar, n0 ≤ m4 + n, m ≤ m4 + n, ∀ m1, n1. ar[m1] ≠ ar[n1] ∨ m1 = n1 ∨ ¬ m1 < # ar ∨ ¬ n1 < # ar, ∀ m1. m1 < m0 → (∀ n. m1 < n ∧ n < # ar → ar[m1] ≠ ar[n]), ∀ n0. m0 < n0 ∧ n0 < n → ar[n0] ≠ ar[n] ⊦

Proof step 17 from proof challenge3a (previous step)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

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