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Proof step 3 from proof toinf-empty3 (previous step)

σ to∞ n∞ + ⌜1⌝ = [], n∞ < # σ ⊦

User applied rule constructor cut

And closed the goal!

Rule argument

var: n∞

Additional information

Used simplifier rules: toinf-inf: ⊦ σ to∞ ∞ = σ;%toinf-add: ⊦ σ to∞ ⌜n⌝ + ⌜m⌝ = σ to n + m;%to-empty3: ⊦ ⌜n⌝ < # σ → σ to n + 1 ≠ [];%natinf freely generated by ∞, ⌜ ⌝;;%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%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ ∞ + ⌜m⌝ = ∞; (s)% ⊦ ¬ ∞ < n∞; (s)