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Proof step 41 from proof subarray-eq-elim (previous step)

# ar = (n1 + m) + n, ar = subarray(ar, 0, n) + subarray(ar, n, n1 + m), subarray(ar, n, n1 + m) = subarray(ar, n + 0, m) + subarray(subarray(ar, n, n1 + m), m, n1), n + n1 + m ≤ # ar, 0 + m ≤ n1 + 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 42:
⊦ n0 + n1 ≤ # ar → subarray(ar, 0, n0) + subarray(ar, n0, n1) = subarray(ar, 0, n0 + n1)
Proof step 43:
⊦ (ar + ar0) + ar1 = ar + ar0 + ar1
Proof step 44:
# ar = (n1 + m) + n, subarray(ar, n, n1 + m) = subarray(ar, n, m) + subarray(subarray(ar, n, n1 + m), m, n1) ⊦ 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-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;% ⊦ m ≤ m; (s)% ⊦ m ≤ m + n; (s)%# ar = n ⊦ subarray(ar, 0, n) = ar; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)