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Proof step 2 from proof add-cancel-02 (previous step)

⊦ n∞1 = n∞1 + n∞2 ↔ n∞1 = ∞ ∨ n∞2 = ⌜0⌝

Heuristic structural induction applied rule structural induction

And closed the goal!

Rule argument

term: n∞1

Additional information

Used simplifier rules: add-all-inf: ⊦ n∞ + ∞ = ∞;%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⌝ ;%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2;%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1;%add-all-zero: ⊦ n∞ + ⌜0⌝ = n∞;%natinf freely generated by ∞, ⌜ ⌝;;%disj: ⊦ ⌜m⌝ ≠ ∞; from specification nat-inf-basic%inj: ⊦ ⌜m⌝ = ⌜m0⌝ ↔ m = m0; from specification nat-inf-basic