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Proof step 2 from proof ls-notlssucc (previous step)

⌜m⌝ < n∞ ⊦ ⌜m⌝ + ⌜1⌝ < n∞ ↔ ⌜m⌝ + ⌜1⌝ ≠ n∞

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ls-natinf-to-nat-2-2: ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ + ⌜n1⌝ ↔ n + m < n0 + n1;%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2;%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1;%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ¬ n < n; (s)% ⊦ m + n < m + n0 ↔ n < n0; (s)