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Proof step 8 from proof add-cancel-01 (previous step)

n∞1 + ⌜1⌝ = ⌜m⌝ + ⌜1⌝ ⊦ n∞1 = ⌜m⌝

User applied rule structural induction

And closed the goal!

Rule argument

indvar: n∞1

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;%addeqinf-cancel: ⊦ ⌜m⌝ + n∞ = ∞ ↔ n∞ = ∞;%eq-natinf-to-nat-2-2: ⊦ ⌜n⌝ + ⌜m⌝ = ⌜n0⌝ + ⌜n1⌝ ↔ n + m = n0 + n1;%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2;%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1;%natinf freely generated by ∞, ⌜ ⌝;;%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%disj: ⊦ ⌜m⌝ ≠ ∞; from specification nat-inf-basic