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Proof step 12 from proof valid-idx-rec (previous step)

kar[m0] = key-index(key0, m2), m0 + 1 < # kar, m0 < # kar, m < # ar, n < # kar, m0 < n, ∀ m. m < n → kar[m].idx < # ar ∧ ar[kar[m].idx].idx = m, ∀ m0, m1. m0 < # kar ∧ ¬ n < m0 ∧ m1 < # kar ∧ ¬ n < m1 ∧ m0 ≠ m1 → kar[n, key-index(key, m)][m0].idx ≠ kar[n, key-index(key, m)][m1].idx ⊦ m0 = n, m2 = m, m2 < # ar ∧ ar[m, ar[m].idx:= n][m2].idx = m0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
kar[m0] = key-index(key0, m2), m0 < n, n < # kar, m < # ar, m0 < # kar, m0 + 1 < # kar, ∀ m0, m1. m0 < # kar ∧ ¬ n < m0 ∧ m1 < # kar ∧ ¬ n < m1 ∧ m0 ≠ m1 → kar[n, key-index(key, m)][m0].idx ≠ kar[n, key-index(key, m)][m1].idx, ∀ m. m < n → kar[m].idx < # ar ∧ ar[kar[m].idx].idx = m ⊦ m0 = n, m2 = m, m2 < # ar ∧ ar[m2].idx = m0

Rule argument

Additional information

Used simplifier rules: ⊦ m ≠ n → ar[n, ei][m] = ar[m]; (s)