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Proof step 8 from proof valid-idx-rec (previous step)

kar[n, key-index(key, m)][m0] = key-index(key0, m2), n < # kar, m < # ar, m0 < # kar, m0 ≤ n, ∀ 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 < # ar ∧ ar[m, ar[m].idx:= n][m2].idx = m0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
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 < # ar ∧ ar[m, ar[m].idx:= n][m2].idx = m0

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)% ⊦ m ≠ n → kar[n, ki][m] = kar[m]; (s)