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Proof step 44 from proof from-append-add (previous step)

# x = n2 + n1, (n2 + n1) + n = # x0, x0[(n1 + n) + n0] = ((σ to n) + ⌜x⌝ )[(n1 + n) + n0], n0 < n2 ⊦ ((σ to n) + ⌜x⌝ )[(n1 + n) + n0] = x[n1 + n0], x = [], # σ < ⌜n⌝

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

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

Additional information

Used simplifier rules: get-append-ls: ⊦ ⌜m⌝ < # σ → (σ + σ1)[m] = σ[m];%append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2;%size-to: ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = ⌜n⌝ ;%size-to: ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = ⌜n⌝ ;%sel-fin: ⊦ ⌜x⌝ [n] = x[n];% ⊦ ⌜m⌝ .n = m; (s)% ⊦ m + n - m = n; (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)% ⊦ ⌜n1⌝ + ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n1 + n + m < n0; (s)% ⊦ ¬ m + n < m; (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: (n1 + n) + n0, ⌜x⌝ , σ to n