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Proof step 12 from proof from-append-tofrom (previous step)

¬ #(σ from n) < ⌜n1⌝ ⊦ # σ < ⌜n⌝ + ⌜n1⌝ , ⌜n1⌝ = ⌜n1⌝ ∧ (⌜n0⌝ < ⌜n⌝ + ⌜n1⌝ → ((σ to n) + ((σ from n) to n1))[n0] = (σ to n + n1)[n0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
n0 < n1 + n ⊦ ((σ to n) + ((σ from n) to n1))[n0] = σ[n0], n1 + n = 0, # σ < ⌜n⌝ + ⌜n1⌝ , #(σ from n) < ⌜n1⌝

Rule argument

Additional information

Used simplifier rules: get-to: m < n, ¬ # σ < ⌜n⌝ ⊦ (σ to n)[m] = σ[m];%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf%c: ⊦ m + n = n + m; from specification nat-basic2%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf%ls-natinf-to-nat-1-2: ⊦ ⌜n⌝ < ⌜m⌝ + ⌜n0⌝ ↔ n < m + n0; from specification nat-inf