Metamath Proof Explorer


Theorem isoddgcd1

Description: The predicate "is an odd number". An odd number and 2 have 1 as greatest common divisor. (Contributed by AV, 1-Jul-2020) (Revised by AV, 8-Aug-2021)

Ref Expression
Assertion isoddgcd1 ( 𝑍 ∈ ℤ → ( ¬ 2 ∥ 𝑍 ↔ ( 2 gcd 𝑍 ) = 1 ) )

Proof

Step Hyp Ref Expression
1 2prm 2 ∈ ℙ
2 coprm ( ( 2 ∈ ℙ ∧ 𝑍 ∈ ℤ ) → ( ¬ 2 ∥ 𝑍 ↔ ( 2 gcd 𝑍 ) = 1 ) )
3 1 2 mpan ( 𝑍 ∈ ℤ → ( ¬ 2 ∥ 𝑍 ↔ ( 2 gcd 𝑍 ) = 1 ) )