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Proof step 9 from proof subarray-elim (previous step)

n + m ≤ # ar ⊦ ar = subarray(ar, 0, n + m) + subarray(ar, n + m, # ar - n + m)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
# ar = n0 + n + m, n + m ≤ # ar ⊦ ar = subarray(ar, 0, n + m) + subarray(ar, n + m, n0)

Rule argument

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

Additional information

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