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Proof step 18 from proof from-append-tofrom (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
n2 < n1 ⊦ (σ from n)[n2] = σ[n2 + n], # σ < ⌜n⌝ + ⌜n1⌝ , #(σ from n) < ⌜n1⌝

Rule argument

Additional information

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