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Proof step 1 from proof less-less-leftsucc (conclusion)

⊦ n∞ < n∞0 → (¬ n∞ + ⌜1⌝ < n∞0 ↔ ¬ ¬(n∞ + ⌜1⌝ = n∞0 ∧ n∞ ≠ ∞))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
n∞ < n∞0 ⊦ n∞ = ∞, n∞ + ⌜1⌝ = n∞0, n∞ + ⌜1⌝ < n∞0

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;%ls-irref: ⊦ ¬ n∞ < n∞;%lsle-trans: ⊦ n∞ < n∞1 ∧ ¬ n∞2 < n∞1 → n∞ < n∞2;%add-less: ⊦ n∞1 < n∞ + n∞1 ↔ ¬ ¬(n∞ ≠ ⌜0⌝ ∧ n∞1 ≠ ∞);%inj: ⊦ ⌜m⌝ = ⌜m0⌝ ↔ m = m0; from specification nat-inf-basic