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Proof step 22 from proof fill2append (previous step)

# ar = n2 + n0 + n1, n0 < # ar, n < # ar, n0 + n1 ≤ # ar, n0 ≤ n ⊦ n2 + n1 = 0, (n < n0 + n1 ⊃ a;ar[n]) = (fill(mkarray(n1), a) + subarray(ar, n0 + n1, n2))[n - n0]

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 23:
# ar = n2 + n0 + n1, n = n3 + n0, n < # ar, n0 < # ar, n0 ≤ n, n0 + n1 ≤ # ar ⊦ n2 + n1 = 0, (n < n0 + n1 ⊃ a;ar[n]) = (fill(mkarray(n1), a) + subarray(ar, n0 + n1, n2))[n3]

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [n, n0, n3]

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n, n0, n3