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Proof step 2 from proof subarray-append-left-01 (previous step)

m ≤ # ar0 ⊦ subarray(ar0 + ar1, 0, m) = subarray(ar0, 0, m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n + m ≤ # ar0 ⊦ subarray(ar0 + ar1, n, m) = subarray(ar0, n, m)
Proof step 4:
⊦ (ar + ar0) + ar1 = ar + ar0 + ar1

Rule argument

Additional information

Used simplifier rules: subarray-elim-right: n + m ≤ # ar0 ⊦ subarray(ar0 + ar1, n, m) = subarray(ar0, n, m);%append-a: ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1;% ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)