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Proof step 16 from proof from-append-tofrom (previous step)

n ≤ n0, n0 < n1 + n ⊦ (σ from n)[n0 - n] = σ[n0], n1 + n = 0, n1 = 0, #(σ from n) < ⌜n1⌝ , # σ < ⌜n⌝ + ⌜n1⌝

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 17:
n0 = n2 + n, n ≤ n0, n0 < n1 + n ⊦ (σ from n)[n2] = σ[n0], n1 + n = 0, n1 = 0, #(σ from n) < ⌜n1⌝ , # σ < ⌜n⌝ + ⌜n1⌝

Rule argument

spec: nat inst: name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [n0, n, n2] flags:

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n0, n, n2