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Proof step 39 from proof from-append-add (previous step)

# x = n2 + n1, # ⌜x0⌝ = #((σ to n) + ⌜x⌝ ) ∧ (∀ n0. ⌜n0⌝ < # ⌜x0⌝ → ⌜x0⌝ [n0] = ((σ to n) + ⌜x⌝ )[n0]), n0 < n2 ⊦ (x0 from n1 + n)[n0] = x[n1 + n0], n2 + n1 = 0, x = [], n2 = 0, # σ < ⌜n⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 40:
(n2 + n1) + n = # x0, # x = n2 + n1, n0 < n2, ∀ n3. n3 < # x0 → x0[n3] = ((σ to n) + ⌜x⌝ )[n3] ⊦ n2 = 0, x = [], n2 + n1 = 0, x0[(n1 + n) + n0] = x[n1 + n0], # σ < ⌜n⌝

Rule argument

Additional information

Used simplifier rules: size-fin: ⊦ # ⌜x⌝ = ⌜# x⌝ ;%size-append: ⊦ #(σ + σ1) = # σ + # σ1;%size-to: ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = ⌜n⌝ ;%append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2;%sel-fin: ⊦ ⌜x⌝ [n] = x[n];%get-from-01: m + m0 < # x ⊦ (x from m)[m0] = x[m + m0];%l-06: ⊦ m = m0 + n0 ∧ n < n0 → m0 + n < m; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%less-fin: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; from specification nat-inf%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf%eq-natinf-to-nat-2-1: ⊦ ⌜n⌝ + ⌜m⌝ = ⌜n0⌝ ↔ n + m = n0; from specification nat-inf