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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 ) )