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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: zero: ⊦ 0 * n = 0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%rsucc-01: ⊦ (m + 1) * n = m * n + n;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%zero: ⊦ 0 * n = 0;%one: ⊦ m0 * 1 = m0;%nat freely generated by 0, +1;;%succ: ⊦ m +1 = m + 1; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%one-03: ⊦ 1 = m + 1 ↔ m = 0; from specification nat-basic2%one-02: ⊦ m ≠ 1 → (m + n = 1 ↔ m = 0 ∧ n = 1); from specification nat-basic2%one-01: ⊦ n ≠ 0 → (m + n = 1 ↔ n = 1 ∧ m = 0); from specification nat-basic2