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Proof step 3 from proof array-split2 (previous step)

n + m ≤ # ar ⊦ ∃ ar0, ar1, ar2. ar = ar0 + ar1 + ar2 ∧ # ar0 = n ∧ # ar1 = m

User applied rule exists right

And obtained the following new goals

Proof step 4:
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

Rule argument

pos: right 1 terms: subarray(ar, 0, n), subarray(ar, n, m), subarray(ar, n + m, # ar - m + n) abort: F discard: T computed: F

Additional information

Used terms: subarray(ar, 0, n), subarray(ar, n, m), subarray(ar, n + m, # ar - m + n)