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Proof step 2 from proof subarray-append-shift-right (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# ar0 ≤ m, m ≤ # ar0 + # ar1 ⊦ subarray(ar0 + ar1, 0, m) = ar0 + subarray(ar1, 0, m - # ar0)

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; (s)%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat