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Proof step 1 from proof gcd-mod-one-01 (conclusion)
1 < j, 1 < i, gcd(i, j) = 1 ⊦ ∃ i0. (i * i0) % j = 1 ∧ 0 < i0
System applied rule normalize
And obtained the following new goals
Proof step 2
:
gcd(i, j) = 1, 1 < i, 1 < j ⊦ ∃ i0. (i * i0) % j = 1 ∧ 0 < i0
Rule argument
Additional information