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Proof step 1 from proof mult-le (conclusion)

⊦ m1 * m0 ≤ m1 ↔ m1 = 0 ∨ m0 ≤ 1

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: leone: ⊦ n ≠ 0 → (m * n ≤ n ↔ m ≤ 1);%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%zero-01: ⊦ m0 * m = 0 ↔ m0 = 0 ∨ m = 0;%ls-04: ⊦ m0 ≠ 0 → (m0 < m0 * m ↔ 1 < m);%one-02: ⊦ m ≠ 0 → (m ≤ 1 ↔ m = 1); from specification nat%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat