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Proof step 13 from proof fill2append (previous step)

# ar = n2 + n0 + n1, n < # ar, n0 + n1 ≤ # ar ⊦ (n0 ≤ n ∧ n < n0 + n1 ⊃ a;ar[n]) = (subarray(ar, 0, n0) + fill(mkarray(n1), a) + subarray(ar, n0 + n1, n2))[n], n < # subarray(ar, 0, n0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
# ar = n2 + n0 + n1, n < # ar, n0 < # ar, n0 + n1 ≤ # ar, n0 ≤ n ⊦ n2 + n1 = 0, (n < n0 + n1 ⊃ a;ar[n]) = (subarray(ar, 0, n0) + fill(mkarray(n1), a) + subarray(ar, n0 + n1, n2))[n]

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < m + n ↔ n ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ # subarray(ar, m, n) = n; (s)