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Proof step 6 from proof fill2append (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
⊦ #(ar0 + ar1) = # ar0 + # ar1
Proof step 8:
⊦ (ar + ar0) + ar1 = ar + ar0 + ar1
Proof step 9:
# ar = n2 + n0 + n1, n < # ar, n0 + n1 ≤ # ar ⊦ fill(ar, a, n0, n1)[n] = (subarray(ar, 0, n0) + fill(mkarray(n1), a) + subarray(ar, n0 + n1, n2))[n]

Rule argument

Additional information

Used simplifier rules: append-len: ⊦ #(ar0 + ar1) = # ar0 + # ar1;%append-a: ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1;% ⊦ m < n → n ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ # mkarray(n) = n; (s)% ⊦ # fill(ar, a) = # ar; (s)% ⊦ # subarray(ar, m, n) = n; (s)% ⊦ # fill(ar, a, m, n) = # ar; (s)