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Proof step 15 from proof from-append-tofrom (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
n ≤ n0, n0 < n1 + n ⊦ (σ from n)[n0 - n] = σ[n0], n1 + n = 0, n1 = 0, #(σ from n) < ⌜n1⌝ , # σ < ⌜n⌝ + ⌜n1⌝

Rule argument

Additional information

Used simplifier rules: get-append-notls: ⊦ ¬ ⌜m⌝ < # σ → (σ + σ1)[m] = σ1[m - (# σ).n];%size-to: ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = ⌜n⌝ ;%append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2;%get-to: m < n, ¬ # σ < ⌜n⌝ ⊦ (σ to n)[m] = σ[m];%l-01: ⊦ m < m + n ↔ n ≠ 0; from specification nat-basic2%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%ls-01: n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; from specification nat%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%less-fin: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; from specification nat-inf%sel: ⊦ ⌜m⌝ .n = m; from specification nat-inf-basic%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%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%l-02: ⊦ ¬ m + n < m; from specification nat-basic2%ls-natinf-to-nat-2-1: ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n + m < n0; from specification nat-inf%lels-trans: ⊦ ¬ n∞1 < n∞ ∧ n∞1 < n∞2 → n∞ < n∞2; (forward) from specification nat-inf