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Proof step 18 from proof swap-from-to (previous step)

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

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: <toplevel> name: get-from seq: ⊦ ⌜n⌝ + ⌜m⌝ < # σ → (σ from n)[m] = σ[n + m] subst: [m, n, σ] → [n0, m, σ to m + 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];% ⊦ m + n < m + n0 ↔ n < n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)% ⊦ ⌜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: n0, m, σ to m + n