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Proof step 12 from proof from-append-tofrom (previous step)

#(σ from n) < ⌜n1⌝ ⊦ # σ < ⌜n⌝ + ⌜n1⌝ , ⌜n⌝ < # σ ∧ finp(σ), #((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])

User applied rule constructor cut

And closed the goal!

Rule argument

var: σ

Additional information

Used simplifier rules: from-fin: ⊦ ⌜x⌝ from n = ⌜x from n⌝ ;%size-fin: ⊦ # ⌜x⌝ = ⌜# x⌝ ;%length-from: ¬ # x < n ⊦ #(x from n) = # x - n;%from-inf: ⊦ ⌜nf⌝ from n = ⌜nf from n⌝ ;%size-inf: ⊦ # ⌜nf⌝ = ∞;%stream freely generated by ⌜ ⌝, ⌜ ⌝;;% ⊦ ¬ ∞ < n∞; (s)% ⊦ m < n → ⌜m⌝ < ⌜n⌝ + n∞0; (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% ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)%n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; (s)% ⊦ ⌜n⌝ < ⌜m⌝ + ⌜n0⌝ ↔ n < m + n0; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n → (n ≤ m ↔ false); (s)