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Proof step 4 from proof array-split2 (previous step)

n + m ≤ # ar ⊦ ar = subarray(ar, 0, n) + subarray(ar, n, m) + subarray(ar, n + m, # ar - m + n) ∧ # subarray(ar, 0, n) = n ∧ # subarray(ar, n, m) = m

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ n0 + n1 ≤ # ar → subarray(ar, 0, n0) + subarray(ar, n0, n1) = subarray(ar, 0, n0 + n1)
Proof step 6:
⊦ (ar + ar0) + ar1 = ar + ar0 + ar1
Proof step 7:
n + m ≤ # ar ⊦ ar = subarray(ar, 0, n + m) + subarray(ar, n + m, # ar - m + n)

Rule argument

Additional information

Used simplifier rules: subarray-append-consec0: ⊦ n0 + n1 ≤ # ar → subarray(ar, 0, n0) + subarray(ar, n0, n1) = subarray(ar, 0, n0 + n1);%append-a: ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1;% ⊦ # subarray(ar, m, n) = n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic