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Proof step 1 from proof gcd-mod-one (conclusion)

gcd(m, n) = 1, 1 < n, m ≠ 0 ⊦ ∃ m0. (m * m0) % n = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
gcd(m, n) = 1 ⊦ m = 0, n = 1, n = 0, ∃ m0. (m * m0) % n = 1

Rule argument

Additional information

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