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Proof step 3 from proof subarray-append-elim-left (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
n0 = n1 + # ar0, n0 + m ≤ # ar0 + # ar1, # ar0 ≤ n0 ⊦ subarray(ar0 + ar1, n0, m) = subarray(ar1, n1, m)

Rule argument

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

Additional information

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