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Proof step 5 from proof valid-idx-rec (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
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 ⊦ kar[n, key-index(key, m)][m0].idx < # ar ∧ ar[m, ar[m].idx:= n][kar[n, key-index(key, m)][m0].idx].idx = m0

Rule argument

Additional information

Used simplifier rules: fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ # ar[n, ei] = # ar; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ # kar[n, ki] = # kar; (s)