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Proof step 1 from proof add-inf-01 (conclusion)

⊦ n∞ + n∞1 = ∞ ↔ n∞ = ∞ ∨ n∞1 = ∞

User applied rule structural induction

And closed the goal!

Rule argument

indvar: n∞

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-inf-02: ⊦ ⌜n⌝ + n∞1 = ∞ ↔ n∞1 = ∞;%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2;%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1;%add-inf2: ⊦ ⌜m⌝ + ∞ = ∞;%natinf freely generated by ∞, ⌜ ⌝;;%disj: ⊦ ⌜m⌝ ≠ ∞; from specification nat-inf-basic