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Proof step 2 from proof ls-04 (previous step)

⊦ m0 = 0, m0 < m0 * m ↔ m ≠ 0 ∧ m ≠ 1

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 2

Additional information

Used simplifier rules: zero-06: ⊦ m ≠ 0 → (m * n = 0 ↔ n = 0);%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%one: ⊦ m0 * 1 = m0;%leone: ⊦ n ≠ 0 → (m * n ≤ n ↔ m ≤ 1);%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%one-02: ⊦ m ≠ 0 → (m ≤ 1 ↔ m = 1); from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%ref-1: ⊦ ¬ n < n; from specification nat-basic1