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Proof step 2 from proof eqls-succ (previous step)

n∞0 + ⌜1⌝ = n∞ ⊦ ⌜n⌝ + ⌜1⌝ < n∞ ↔ ⌜n⌝ < n∞0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: add-ls-cancel-01: ⊦ ⌜n⌝ + n∞1 < ⌜n⌝ + n∞2 ↔ n∞1 < n∞2;%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2;%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1;