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Proof step 26 from proof fill2append (previous step)

# ar = n2 + n0 + n1, n0 < # ar, n3 < n2 + n1, n0 + n1 ≤ # ar ⊦ (n3 < n1 ⊃ a;ar[n3 + n0]) = (fill(mkarray(n1), a) + subarray(ar, n0 + n1, n2))[n3], n3 < # fill(mkarray(n1), a)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 27:
# ar = n2 + n0 + n1, n0 < # ar, n3 < n2 + n1, n1 ≤ n3, n0 + n1 ≤ # ar ⊦ ar[n3 + n0] = (fill(mkarray(n1), a) + subarray(ar, n0 + n1, n2))[n3], n2 = 0

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)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ # mkarray(n) = n; (s)% ⊦ # fill(ar, a) = # ar; (s)