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Proof step 1 from proof ls-lsadd (conclusion)

⊦ m < n → ⌜m⌝ < ⌜n⌝ + n∞0

User applied rule constructor cut

And closed the goal!

Rule argument

var: n∞0

Additional information

Used simplifier rules: add-inf2: ⊦ ⌜m⌝ + ∞ = ∞;%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2;%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1;%less-inf2: ⊦ ⌜m⌝ < ∞;%ls-natinf-to-nat-1-2: ⊦ ⌜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;%natinf freely generated by ∞, ⌜ ⌝;;%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%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