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Proof step 2 from proof rax (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: m

Additional information

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