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Proof step 2 from proof less-less-leftsucc (previous step)

n∞ < n∞0 ⊦ n∞ = ∞, n∞ + ⌜1⌝ = n∞0, n∞ + ⌜1⌝ < n∞0

User applied rule constructor cut

And obtained the following new goals

Proof step 3:
n∞0 = ⌜m⌝ , n∞ < n∞0 ⊦ n∞ = ∞, n∞ + ⌜1⌝ = n∞0, n∞ + ⌜1⌝ < n∞0

Rule argument

var: n∞0

Additional information

Used simplifier rules: addeqinf-cancel: ⊦ ⌜m⌝ + n∞ = ∞ ↔ n∞ = ∞;%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2;%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1;%inf-less: ⊦ n∞ < ∞ ↔ n∞ ≠ ∞;%natinf freely generated by ∞, ⌜ ⌝;;