Metamath Proof Explorer


Theorem dfodd7

Description: Alternate definition for odd numbers. (Contributed by AV, 1-Jul-2020)

Ref Expression
Assertion dfodd7 Odd = { 𝑧 ∈ ℤ ∣ ( 2 gcd 𝑧 ) = 1 }

Proof

Step Hyp Ref Expression
1 isodd7 ( 𝑥 ∈ Odd ↔ ( 𝑥 ∈ ℤ ∧ ( 2 gcd 𝑥 ) = 1 ) )
2 oveq2 ( 𝑧 = 𝑥 → ( 2 gcd 𝑧 ) = ( 2 gcd 𝑥 ) )
3 2 eqeq1d ( 𝑧 = 𝑥 → ( ( 2 gcd 𝑧 ) = 1 ↔ ( 2 gcd 𝑥 ) = 1 ) )
4 3 elrab ( 𝑥 ∈ { 𝑧 ∈ ℤ ∣ ( 2 gcd 𝑧 ) = 1 } ↔ ( 𝑥 ∈ ℤ ∧ ( 2 gcd 𝑥 ) = 1 ) )
5 1 4 bitr4i ( 𝑥 ∈ Odd ↔ 𝑥 ∈ { 𝑧 ∈ ℤ ∣ ( 2 gcd 𝑧 ) = 1 } )
6 5 eqriv Odd = { 𝑧 ∈ ℤ ∣ ( 2 gcd 𝑧 ) = 1 }