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Proof step 4 from proof div (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1