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Proof step 4 from proof less-less-leftsucc (previous step)

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

User applied rule constructor cut

And closed the goal!

Rule argument

var: n∞

Additional information

Used simplifier rules: eq-natinf-to-nat-2-1: ⊦ ⌜n⌝ + ⌜m⌝ = ⌜n0⌝ ↔ n + m = n0;%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2;%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1;%less-fin: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n;%ls-natinf-to-nat-2-1: ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n + m < n0;%natinf freely generated by ∞, ⌜ ⌝;;%disj: ⊦ ⌜m⌝ ≠ ∞; from specification nat-inf-basic%neq-01: ⊦ m ≠ n → (¬ m < n ↔ n < m); from specification nat-basic1%s: ⊦ m < n + 1 ↔ ¬ 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