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Proof step 5 from proof challenge3b (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
ar[m0] = ar[n], ar[m] = ar[n0], m0 < n, n < # ar, m0 < # ar, m < # ar, n0 < # ar, m0 + 1 < # ar, ∀ n0. m0 < n0 ∧ n0 < # ar → ar[n0] ≠ ar[n] ∨ 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

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m +1 = m + 1], [CSIMPFORWARD : ⊦ m < n ∧ n < n0 → m +1 < n0], [CSIMPSEQ : ⊦ ¬ n < n], [CSIMPSEQ : ⊦ m ≤ m])]]