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

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

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⌝ ;%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2