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Proof step 19 from proof from-append-tofrom (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 20:
⊦ ⌜n⌝ + ⌜m⌝ < # σ → (σ from n)[m] = σ[n + m]

Rule argument

spec: <toplevel> name: get-from seq: ⊦ ⌜n⌝ + ⌜m⌝ < # σ → (σ from n)[m] = σ[n + m] subst: [m, n, σ] → [n2, n, σ] flags: <all paths>

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%lele-trans: ⊦ ¬ n∞1 < n∞ ∧ ¬ n∞2 < n∞1 → ¬ n∞2 < n∞; (forward) from specification nat-inf% ⊦ ⌜n⌝ + n∞1 < ⌜n⌝ + n∞2 ↔ n∞1 < n∞2; (s)% ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n → (n ≤ m ↔ false); (s)%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 Used terms: n2, n, σ