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Proof step 4 from proof add-cancel-num-02 (previous step)

⌜n⌝ = ⌜m0 +1⌝ + n∞2, n < m0 +1, ∀ n, n∞2. ¬(⌜n⌝ = ⌜m0⌝ + n∞2 ∧ n < m0) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⌜n⌝ = (⌜m0⌝ + ⌜1⌝ ) + n∞2, n ≤ m0, ∀ n0, n∞. ⌜n0⌝ ≠ ⌜m0⌝ + n∞ ∨ ¬ n0 < m0 ⊦

Rule argument

Additional information

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