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Proof step 29 from proof swap-from-to (previous step)

⊦ ⌜n⌝ = #((σ to m + n) from m), #(σ from m) < ⌜n⌝ , # σ < ⌜m⌝ + ⌜n⌝

User applied rule constructor cut

And closed the goal!

Rule argument

var: σ

Additional information

Used simplifier rules: to-fin: ⊦ ⌜x⌝ to n = ⌜x to n⌝ ;%from-fin: ⊦ ⌜x⌝ from n = ⌜x from n⌝ ;%size-fin: ⊦ # ⌜x⌝ = ⌜# x⌝ ;%length-to: ¬ # x < n ⊦ #(x to n) = n;%length-from: ¬ # x < n ⊦ #(x from n) = # x - n;%to-inf: ⊦ ⌜nf⌝ to n = ⌜nf to n⌝ ;%from-fin: ⊦ ⌜x⌝ from n = ⌜x from n⌝ ;%size-fin: ⊦ # ⌜x⌝ = ⌜# x⌝ ;%size-inf-to: ⊦ #(nf to n) = n;%length-from: ¬ # x < n ⊦ #(x from n) = # x - n;%stream freely generated by ⌜ ⌝, ⌜ ⌝;;% ⊦ ¬ m + n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + n - m = n; (s)% ⊦ ⌜m⌝ = ⌜m0⌝ ↔ m = m0; (s)% ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ ⌜n⌝ < ⌜m⌝ + ⌜n0⌝ ↔ n < m + n0; (s)% ⊦ m < n → (n ≤ m ↔ false); (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