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Proof step 20 from proof subarray-eq-elim (previous step)

# ar = n0 + n, ar = subarray(ar, 0, n) + subarray(ar, n, n0), m ≤ n0, m < # subarray(ar, n, n0) → subarray(ar, n, n0) = subarray(subarray(ar, n, n0), 0, m) + subarray(subarray(ar, n, n0), m, # subarray(ar, n, n0) - m) ⊦ subarray(ar, n, m) = ar1 ↔ (∃ ar0, ar2. ar = ar0 + ar1 + ar2 ∧ # ar0 = n ∧ # ar2 = # ar1 + # ar2 - m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
# ar = n0 + n, ar = subarray(ar, 0, n) + subarray(ar, n, n0), m ≤ n0, m < n0 → subarray(ar, n, n0) = subarray(subarray(ar, n, n0), 0, m) + subarray(subarray(ar, n, n0), m, n0 - m) ⊦ subarray(ar, n, m) = ar1 ↔ (∃ ar0, ar2. ar = ar0 + ar1 + ar2 ∧ # ar0 = n ∧ # ar2 = # ar1 + # ar2 - m)

Rule argument

Additional information

Used simplifier rules: ⊦ # subarray(ar, m, n) = n; (s)