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Proof step 3 from proof fill-append-shift (previous step)

# ar0 ≤ n ⊦ fill(ar0 + ar1, a, n, m) = ar0 + fill(ar1, a, n - # ar0, m)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
n = n0 + # ar0, # ar0 ≤ n ⊦ fill(ar0 + ar1, a, n, m) = ar0 + fill(ar1, a, n0, m)

Rule argument

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

Additional information

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