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Proof step 21 from proof copy2append (previous step)

# ar1 = (n2 + m) + n1, n1 < # ar1, n < # ar1, n < # subarray(ar1, 0, n1) + #(subarray(ar0, n0, m) + subarray(ar1, n1 + m, n2)), n1 ≤ n, n1 + m ≤ # ar1, n0 + m ≤ # ar0, m + n1 ≤ # ar1, # subarray(ar1, 0, n1) ≤ n ⊦ n2 + m = 0, (n1 + m ≤ n ⊃ ar1[n];ar0[(n - n1) + n0]) = (subarray(ar0, n0, m) + subarray(ar1, n1 + m, n2))[n - # subarray(ar1, 0, n1)]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
⊦ #(ar0 + ar1) = # ar0 + # ar1
Proof step 23:
⊦ (ar + ar0) + ar1 = ar + ar0 + ar1
Proof step 24:
# ar1 = (n2 + m) + n1, n1 < # ar1, n < # ar1, m + n1 ≤ # ar1, n0 + m ≤ # ar0, n1 + m ≤ # ar1, n1 ≤ n ⊦ n2 + m = 0, (n1 + m ≤ n ⊃ ar1[n];ar0[(n - n1) + n0]) = (subarray(ar0, n0, m) + subarray(ar1, n1 + m, n2))[n - n1]

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% ⊦ # subarray(ar, m, n) = n; (s)