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

kar[m0] = key-index(key1, n1), kar[m1] = key-index(key0, n0), m0 < n, m1 < n, valid-idx(kar, ar, n) ⊦ valid-idx(kar[m0, key-index(key0, n0)][m1, key-index(key1, n1)], ar[n0, ar[n0].idx:= m0][n1, ar[n0, ar0[n0].idx:= m0][n1].idx:= m1], n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
kar[m1] = key-index(key0, n0), kar[m0] = key-index(key1, n1), m1 < n, m0 < n, m1 < # kar, m0 < # kar, n ≤ # kar, ∀ m. m < n → kar[m].idx < # ar ∧ ar[kar[m].idx].idx = m ⊦ valid-idx(kar[m0, key-index(key0, n0)][m1, key-index(key1, n1)], ar[n0, ar[n0].idx:= m0][n1, ar[n0, ar0[n0].idx:= m0][n1].idx:= m1], n)

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