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Proof step 1 from proof mult-eq-succ (conclusion)

⊦ m < m1 → (m + 1 = m1 * m0 ↔ m0 = 1 ∧ m1 = m + 1)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: rsucc-01: ⊦ (m + 1) * n = m * n + n;%one: ⊦ m0 * 1 = m0;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%leone: ⊦ n ≠ 0 → (m * n ≤ n ↔ m ≤ 1);%zero: ⊦ 0 * n = 0;%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%lels-02: ⊦ m ≠ n → (m ≤ n ↔ m < n); from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2