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Proof step 2 from proof one-01 (previous step)

m0 * m = 1 ⊦ m0 = 1 ∧ m = 1

User applied rule structural induction

And closed the goal!

Rule argument

indvar: m0

Additional information

Used simplifier rules: mult-base: ⊦ m * 0 = 0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%dist: ⊦ m0 * (m + n) = m0 * m + m0 * n;%one: ⊦ m0 * 1 = m0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%mult-base: ⊦ m * 0 = 0;%nat freely generated by 0, +1;;% ⊦ m +1 = m + 1; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ 1 = m + 1 ↔ m = 0; (s)% ⊦ m ≠ 1 → (m + n = 1 ↔ m = 0 ∧ n = 1); (ws)% ⊦ n ≠ 0 → (m + n = 1 ↔ n = 1 ∧ m = 0); (ws)