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Proof step 19 from proof fill2append (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
⊦ #(ar0 + ar1) = # ar0 + # ar1
Proof step 21:
⊦ (ar + ar0) + ar1 = ar + ar0 + ar1
Proof step 22:
# ar = n2 + n0 + n1, n0 < # ar, n < # ar, n0 + n1 ≤ # ar, n0 ≤ n ⊦ n2 + n1 = 0, (n < n0 + n1 ⊃ a;ar[n]) = (fill(mkarray(n1), a) + subarray(ar, n0 + n1, n2))[n - n0]

Rule argument

Additional information

Used simplifier rules: append-len: ⊦ #(ar0 + ar1) = # ar0 + # ar1;%append-a: ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1;%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)