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

n < # kar, m < # ar, no-dups-idx(kar[n, key-index(key, m)], n + 1), valid-idx(kar, ar, n) ⊦ ¬ valid-idx(kar[n, key-index(key, m)], ar[m, ar[m].idx:= n], n + 1) ↔ false

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n < # kar, m < # ar, ∀ 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 ⊦ valid-idx(kar[n, key-index(key, m)], ar[m, ar[m].idx:= n], n + 1)

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);%no-dups-idx: ⊦ no-dups-idx(kar, n) ↔ (∀ m0, m1. m0 < # kar ∧ m0 < n ∧ m1 < # kar ∧ m1 < n ∧ m0 ≠ m1 → kar[m0].idx ≠ kar[m1].idx);% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ # kar[n, ki] = # kar; (s)% ⊦ m < n → m ≤ n; (s)