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Proof step 9 from proof subarray-split-at (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
⊦ #(ar0 + ar1) = # ar0 + # ar1
Proof step 11:
⊦ (ar + ar0) + ar1 = ar + ar0 + ar1
Proof step 12:
# ar = n0 + n, n1 < n0 + n ⊦ ar[n1] = (subarray(ar, 0, n) + subarray(ar, n, n0))[n1], n0 = 0

Rule argument

Additional information

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