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Proof step 23 from proof fill2append (previous step)

# ar = n2 + n0 + n1, n = n3 + n0, n < # ar, n0 < # ar, n0 ≤ n, n0 + n1 ≤ # ar ⊦ n2 + n1 = 0, (n < n0 + n1 ⊃ a;ar[n]) = (fill(mkarray(n1), a) + subarray(ar, n0 + n1, n2))[n3]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
# ar = n2 + n0 + n1, n3 < n2 + n1, n0 < # ar, n0 + n1 ≤ # ar ⊦ (n3 < n1 ⊃ a;ar[n3 + n0]) = (fill(mkarray(n1), a) + subarray(ar, n0 + n1, n2))[n3]

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m = m0 + n0 ∧ n < n0 → m0 + n < m; (s)% ⊦ m < n → n ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m ≤ m + n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s)