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Proof step 4 from proof mult (previous step)

1 < m * n ∧ (∀ m0. m0 | m * n → m0 = 1 ∨ m0 = m * n) ⊦ n = 1, m = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ m0. m0 | m * n → m0 = 1 ∨ m0 = m * n ⊦ n = 1, m = 1, m * n = 1, m = 0, n = 0

Rule argument

Additional information

Used simplifier rules: mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%zero-01: ⊦ m0 * m = 0 ↔ m0 = 0 ∨ m = 0; from specification nat-mult%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1