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Proof step 17 from proof from-append-tofrom (previous step)

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⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
n2 < n1 ⊦ (σ from n)[n2] = σ[n2 + n], n1 + n = 0, n1 = 0, #(σ from n) < ⌜n1⌝ , # σ < ⌜n⌝ + ⌜n1⌝

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%lep: ⊦ m ≤ m + n; from specification nat