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Proof step 21 from proof from-append-add (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
⊦ # σ < ⌜n⌝ , #((σ to n) + ⌜nf⌝ from n1 + n) = ∞ ∧ ((σ to n) + ⌜nf⌝ from n1 + n)[n0] = nf(n0 + n1)

Rule argument

Additional information

Used simplifier rules: size-inf: ⊦ # ⌜nf⌝ = ∞;%from-inf: ⊦ ⌜nf⌝ from n = ⌜nf from n⌝ ;%size-append: ⊦ #(σ + σ1) = # σ + # σ1;%size-to: ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = ⌜n⌝ ;%append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2;%less-size-from: ⌜n⌝ < # σ ⊦ ⌜n0⌝ < #(σ from n) ↔ ⌜n0 + n⌝ < # σ;%sel-inf: ⊦ ⌜nf⌝ [n] = nf(n);%get-from-inf: ⊦ ((nf from n))(m) = nf(m + n);% ⊦ ⌜m⌝ ≠ ∞; (s)% ⊦ ⌜n⌝ + n∞1 = ∞ ↔ n∞1 = ∞; (s)% ⊦ n∞ < ∞ ↔ n∞ ≠ ∞; (s)%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% ⊦ ∞ + ⌜m⌝ = ∞; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)% ⊦ ¬ finp(⌜nf⌝ ); (s)% ⊦ ⌜m⌝ < ∞; (s)% ⊦ ¬ ∞ < n∞; (s)