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Proof step 4 from proof from-append-tofrom (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

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;%add-cancel-01: ⊦ ⌜n⌝ + n∞1 = ⌜n⌝ + n∞2 ↔ n∞1 = n∞2; from specification nat-inf%add-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%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%l-02: ⊦ ¬ m + n < m; from specification nat-basic2%ls-natinf-to-nat-2-1: ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n + m < n0; from specification nat-inf%lels-trans: ⊦ ¬ n∞1 < n∞ ∧ n∞1 < n∞2 → n∞ < n∞2; (forward) from specification nat-inf