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Proof step 2 from proof valid-idx-last-valid (previous step)

kar[m] = key-index(key, n), m < m0, valid-idx(kar, ar, m0), # ar ≤ n ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
kar[m] = key-index(key, n), m < m0, m < # kar, # ar ≤ n, m0 ≤ # kar, ∀ m. m < m0 → kar[m].idx < # ar ∧ ar[kar[m].idx].idx = m ⊦

Rule argument

Additional information

Used simplifier rules: valid-idx: ⊦ valid-idx(kar, ar, n) ↔ n ≤ # kar ∧ (∀ m. m < n → kar[m].idx < # ar ∧ ar[kar[m].idx].idx = m);% ⊦ m < n → n ≠ 0; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic