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Proof step 4 from proof subarray-append-shift-right (previous step)

m = n0 + # ar0, m ≤ # ar0 + # ar1, # ar0 ≤ m ⊦ subarray(ar0 + ar1, 0, m) = ar0 + subarray(ar1, 0, n0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n0 ≤ # ar1 ⊦ subarray(ar0 + ar1, 0, n0 + # ar0) = ar0 + subarray(ar1, 0, n0)

Rule argument

Additional information

Used simplifier rules: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m ≤ m + n; (s)