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Proof step 1 from proof ls-notlssucc (conclusion)

⊦ ⌜m⌝ < n∞ → (¬ ⌜m⌝ + ⌜1⌝ < n∞ ↔ ⌜m⌝ + ⌜1⌝ = n∞)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: add-ls-cancel-num-04: ⊦ m ≤ n → (⌜(* m)⌝ + n∞2 < ⌜(* n)⌝ + n∞1 ↔ n∞2 < ⌜n - m⌝ + n∞1);%add-all-zero: ⊦ n∞ + ⌜0⌝ = n∞;%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;%less-less-leftsucc: ⊦ n∞ < n∞0 → (¬ n∞ + ⌜1⌝ < n∞0 ↔ ¬ ¬(n∞ + ⌜1⌝ = n∞0 ∧ n∞ ≠ ∞));%ref-1: ⊦ ¬ n < n; from specification nat-basic1