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Proof step 2 from proof from-append-toinffrom (previous step)

⊦ (σ to n) + ((σ from n) to∞ n∞1) = σ to∞ ⌜n⌝ + n∞1, # σ < ⌜n⌝ + n∞1

User applied rule constructor cut

And obtained the following new goals

Proof step 3:
n∞1 = ⌜n0⌝ ⊦ (σ to n) + ((σ from n) to∞ n∞1) = σ to∞ ⌜n⌝ + n∞1, # σ < ⌜n⌝ + n∞1

Rule argument

var: n∞1

Additional information

Used simplifier rules: toinf-inf: ⊦ σ to∞ ∞ = σ;%to-plus-from-simp: ¬ # σ < ⌜n⌝ ⊦ (σ to n) + (σ from n) ≠ σ ↔ false;%append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2;%natinf freely generated by ∞, ⌜ ⌝;;%add-inf2: ⊦ ⌜m⌝ + ∞ = ∞; 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%inf-less: ⊦ n∞ < ∞ ↔ n∞ ≠ ∞; from specification nat-inf%less-inf1: ⊦ ¬ ∞ < n∞; from specification nat-inf