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Proof step 1 from proof add-cancel-03 (conclusion)

⊦ ⌜n⌝ + n∞ = ⌜n⌝ ↔ n∞ = ⌜0⌝

User applied rule structural induction

And closed the goal!

Rule argument

indvar: n∞

Additional information

Used simplifier rules: add-inf2: ⊦ ⌜m⌝ + ∞ = ∞;%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2;%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1;%eq-natinf-to-nat-2-1: ⊦ ⌜n⌝ + ⌜m⌝ = ⌜n0⌝ ↔ n + m = n0;%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 ∞, ⌜ ⌝;;%eq: ⊦ m = m + n ↔ n = 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%inj: ⊦ ⌜m⌝ = ⌜m0⌝ ↔ m = m0; from specification nat-inf-basic%disj: ⊦ ⌜m⌝ ≠ ∞; from specification nat-inf-basic