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Proof step 3 from proof subarray-elim (previous step)

n + m ≤ # ar ⊦ ar1 = subarray(ar, n, m) ↔ (∃ ar0, ar2. ar = ar0 + ar1 + ar2 ∧ # ar0 = n ∧ # ar1 = m)

User applied rule case distinction

And obtained the following new goals

Proof step 4:
ar1 = subarray(ar, n, m), n + m ≤ # ar ⊦ ∃ ar0, ar2. ar = ar0 + ar1 + ar2 ∧ # ar0 = n ∧ # ar1 = m
Proof step 14:
n + m ≤ # ar, ∃ ar0, ar2. ar = ar0 + ar1 + ar2 ∧ # ar0 = n ∧ # ar1 = m ⊦ ar1 = subarray(ar, n, m)

Rule argument

pos: right 1

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