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Proof step 4 from proof to-append-add (previous step)

⊦ #((σ to n) + σ0 to n1 + n) = #((σ to n) + (σ0 to n1)) ∧ (∀ n0. ⌜n0⌝ < #((σ to n) + σ0 to n1 + n) → ((σ to n) + σ0 to n1 + n)[n0] = ((σ to n) + (σ0 to n1))[n0]), # σ0 < ⌜n1⌝ , # σ < ⌜n⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n0 < n + n1 ⊦ ((σ to n) + σ0)[n0] = ((σ to n) + (σ0 to n1))[n0], n + n1 = 0, # σ0 < ⌜n1⌝ , # σ < ⌜n⌝

Rule argument

Additional information

Used simplifier rules: size-append: ⊦ #(σ + σ1) = # σ + # σ1;%size-to: ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = ⌜n⌝ ;%append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2;%get-to: m < n, ¬ # σ < ⌜n⌝ ⊦ (σ to n)[m] = σ[m];%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%ls-natinf-to-nat-1-2: ⊦ ⌜n⌝ < ⌜m⌝ + ⌜n0⌝ ↔ n < m + n0; from specification nat-inf%eq-natinf-to-nat-2-2: ⊦ ⌜n⌝ + ⌜m⌝ = ⌜n0⌝ + ⌜n1⌝ ↔ n + m = n0 + n1; 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%add-ls-cancel-01: ⊦ ⌜n⌝ + n∞1 < ⌜n⌝ + n∞2 ↔ n∞1 < n∞2; from specification nat-inf%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf