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

kar[m] = key-index(key, m1), m < # kar, valid-idx(kar, ar, m + 1) ⊦ valid-idx(kar[m0, key-index(key, m1)], ar[m1, ar[m1].idx:= m0], m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
kar[m] = key-index(key, m1), m < # kar, ∀ m2. ¬ m < m2 → kar[m2].idx < # ar ∧ ar[kar[m2].idx].idx = m2 ⊦ valid-idx(kar[m0, key-index(key, m1)], ar[m1, ar[m1].idx:= m0], 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)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ 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)