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Proof step 24 from proof subarray-eq-elim (previous step)

# ar = (n1 + m) + n, ar = subarray(ar, 0, n) + subarray(ar, n, n1 + m), n1 = 0 ⊦ 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 25:
# ar = m + n, ar = subarray(ar, 0, n) + subarray(ar, n, 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: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)