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Proof step 2 from proof mult-eq-succ (previous step)

m < m1 ⊦ m + 1 = m1 * m0 ↔ m0 = 1 ∧ m1 = m + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
m + 1 = m1 * m0, m < m1, m1 * m0 ≤ m1 ⊦ m0 = 1

Rule argument

Additional information

Used simplifier rules: 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;% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat% ⊦ m + n = m + n0 ↔ n = n0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic