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

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 19:
⊦ ⌜n⌝ + ⌜m⌝ < # σ → (σ from n)[m] = σ[n + m]

Rule argument

spec: inst: name: get-from seq: ⊦ ⌜n⌝ + ⌜m⌝ < # σ → (σ from n)[m] = σ[n + m] subst: [m, n, σ] → [n2, n, σ] flags: rewrite, current

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%lele-trans: ⊦ ¬ n∞1 < n∞ ∧ ¬ n∞2 < n∞1 → ¬ n∞2 < n∞; (forward) from specification nat-inf%ls-natinf-to-nat-2-2: ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ + ⌜n1⌝ ↔ n + m < n0 + n1; from specification nat-inf%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%ls-le-simp: ⊦ m < n → (n ≤ m ↔ false); from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%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