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Proof step 3 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 elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
m = n0 + # ar0, m ≤ # ar0 + # ar1, # ar0 ≤ m ⊦ subarray(ar0 + ar1, 0, m) = ar0 + subarray(ar1, 0, n0)

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [m, # ar0, n]

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: m, # ar0, n