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Proof step 5 from proof div (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n | m, ∀ m0. m0 | m → m0 = 1 ∨ m0 = m ⊦ m = 1, m = 0, n = 1, n = m

Rule argument

Additional information

Used simplifier rules: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s)