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Proof step 14 from proof from-append-tofrom (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ ⌜m⌝ < # σ → (σ + σ1)[m] = σ[m]
Proof step 16:
n0 < n1 + n ⊦ ((σ to n) + ((σ from n) to n1))[n0] = σ[n0], #(σ from n) < ⌜n1⌝ , # σ < ⌜n⌝ + ⌜n1⌝ , ⌜n0⌝ < #(σ to n)

Rule argument

spec: <toplevel> name: get-append-ls seq: ⊦ ⌜m⌝ < # σ → (σ + σ1)[m] = σ[m] subst: [m, σ1, σ] → [n0, (σ from n) to n1, σ to n] flags: <all paths>

Additional information

Used simplifier rules: size-to: ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = ⌜n⌝ ;%get-to: m < n, ¬ # σ < ⌜n⌝ ⊦ (σ to n)[m] = σ[m];%lels-trans: ⊦ ¬ n∞1 < n∞ ∧ n∞1 < n∞2 → n∞ < n∞2; (forward) from specification nat-inf% ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n + m < n0; (s)% ⊦ ¬ m + n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf% ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s) Used terms: n0, (σ from n) to n1, σ to n