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Proof step 10 from proof subarray-append-consec (previous step)

n < n0 + n1, m0 + n0 + n1 ≤ # ar ⊦ (subarray(ar, m0, n0) + subarray(ar, n0 + m0, n1))[n] = ar[m0 + n], n < # subarray(ar, m0, n0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
n < n0 + n1, n0 ≤ n, m0 + n0 + n1 ≤ # ar ⊦ (subarray(ar, m0, n0) + subarray(ar, n0 + m0, n1))[n] = ar[m0 + n], n1 = 0

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% ⊦ m < m + n ↔ n ≠ 0; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ # subarray(ar, m, n) = n; (s)