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Proof step 5 from proof mult-01 (previous step)

1 < m * n, n0 + 1 < n, ∀ m0. m0 | m * n → m0 = 1 ∨ m0 = m * n ⊦ m = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n0 + 1 < n, ∀ m0. m0 | m * n → m0 = 1 ∨ m0 = m * n ⊦ m * n = 1, m = 1, m = 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% ⊦ m < n → n ≠ 0; (s)% ⊦ m0 * m = 0 ↔ m0 = 0 ∨ m = 0; (ws)% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s)