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Metamath Proof Explorer

Theorem elgz

Description: Elementhood in the gaussian integers. (Contributed by Mario Carneiro, 14-Jul-2014)

Ref Expression
Assertion elgz ( 𝐴 ∈ ℤ[i] ↔ ( 𝐴 ∈ ℂ ∧ ( ℜ ‘ 𝐴 ) ∈ ℤ ∧ ( ℑ ‘ 𝐴 ) ∈ ℤ ) )

Proof

Step Hyp Ref Expression
1 fveq2 ⊢ ( 𝑥 = 𝐴 → ( ℜ ‘ 𝑥 ) = ( ℜ ‘ 𝐴 ) )
2 1 eleq1d ⊢ ( 𝑥 = 𝐴 → ( ( ℜ ‘ 𝑥 ) ∈ ℤ ↔ ( ℜ ‘ 𝐴 ) ∈ ℤ ) )
3 fveq2 ⊢ ( 𝑥 = 𝐴 → ( ℑ ‘ 𝑥 ) = ( ℑ ‘ 𝐴 ) )
4 3 eleq1d ⊢ ( 𝑥 = 𝐴 → ( ( ℑ ‘ 𝑥 ) ∈ ℤ ↔ ( ℑ ‘ 𝐴 ) ∈ ℤ ) )
5 2 4 anbi12d ⊢ ( 𝑥 = 𝐴 → ( ( ( ℜ ‘ 𝑥 ) ∈ ℤ ∧ ( ℑ ‘ 𝑥 ) ∈ ℤ ) ↔ ( ( ℜ ‘ 𝐴 ) ∈ ℤ ∧ ( ℑ ‘ 𝐴 ) ∈ ℤ ) ) )
6 df-gz ⊢ ℤ[i] = { 𝑥 ∈ ℂ ∣ ( ( ℜ ‘ 𝑥 ) ∈ ℤ ∧ ( ℑ ‘ 𝑥 ) ∈ ℤ ) }
7 5 6 elrab2 ⊢ ( 𝐴 ∈ ℤ[i] ↔ ( 𝐴 ∈ ℂ ∧ ( ( ℜ ‘ 𝐴 ) ∈ ℤ ∧ ( ℑ ‘ 𝐴 ) ∈ ℤ ) ) )
8 3anass ⊢ ( ( 𝐴 ∈ ℂ ∧ ( ℜ ‘ 𝐴 ) ∈ ℤ ∧ ( ℑ ‘ 𝐴 ) ∈ ℤ ) ↔ ( 𝐴 ∈ ℂ ∧ ( ( ℜ ‘ 𝐴 ) ∈ ℤ ∧ ( ℑ ‘ 𝐴 ) ∈ ℤ ) ) )
9 7 8 bitr4i ⊢ ( 𝐴 ∈ ℤ[i] ↔ ( 𝐴 ∈ ℂ ∧ ( ℜ ‘ 𝐴 ) ∈ ℤ ∧ ( ℑ ‘ 𝐴 ) ∈ ℤ ) )