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

#((σ to n0 + m) + σ0 from m) = #((σ to m + n0) from m) + # σ0, ⌜n1⌝ + ⌜n0⌝ < #((σ to m + n0) from m), ⌜n1⌝ + ⌜n0⌝ < #((σ to m + n0) from m) + # σ0, ⌜n1⌝ < # σ0 ⊦ σ0[n1] = ((σ to m + n0) from m)[n1 + n0], # σ < ⌜n0⌝ + ⌜m⌝ , # σ < ⌜m⌝ + ⌜n0⌝

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⌝ ;%sel-fin: ⊦ ⌜x⌝ [n] = x[n];%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;%sel-fin: ⊦ ⌜x⌝ [n] = x[n];%stream freely generated by ⌜ ⌝, ⌜ ⌝;;%l-02: ⊦ ¬ m + n < m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%eq-01: ⊦ m + n - m = n; from specification nat%ls-natinf-to-nat-2-1: ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n + m < n0; from specification nat-inf%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%ls-natinf-to-nat-1-2: ⊦ ⌜n⌝ < ⌜m⌝ + ⌜n0⌝ ↔ n < m + n0; from specification nat-inf%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%ls-le-simp: ⊦ m < n → (n ≤ m ↔ false); from specification nat