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Proof step 5 from proof add-cancel-num-02 (previous step)

⌜n⌝ = (⌜m0⌝ + ⌜1⌝ ) + n∞2, n ≤ m0, ∀ n0, n∞. ⌜n0⌝ ≠ ⌜m0⌝ + n∞ ∨ ¬ n0 < m0 ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: n, n∞2 + ⌜1⌝

Additional information

Used simplifier rules: add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2;%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1;%add-cancel-03: ⊦ ⌜n⌝ + n∞ = ⌜n⌝ ↔ n∞ = ⌜0⌝ ;%sum-zero-fin: ⊦ ⌜n⌝ + n∞ = ⌜0⌝ ↔ n∞ = ⌜0⌝ ∧ n = 0;% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m ≤ n → (n ≤ m ↔ m = n); (s) Used terms: n, n∞2 + ⌜1⌝