Theorem elgz 16261

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

Assertion

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

Proof of Theorem elgz

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		fveq2 6664	. . . . 5 ⊢ (𝑥 = 𝐴 → (ℜ‘𝑥) = (ℜ‘𝐴))
2	1	eleq1d 2897	. . . 4 ⊢ (𝑥 = 𝐴 → ((ℜ‘𝑥) ∈ ℤ ↔ (ℜ‘𝐴) ∈ ℤ))
3		fveq2 6664	. . . . 5 ⊢ (𝑥 = 𝐴 → (ℑ‘𝑥) = (ℑ‘𝐴))
4	3	eleq1d 2897	. . . 4 ⊢ (𝑥 = 𝐴 → ((ℑ‘𝑥) ∈ ℤ ↔ (ℑ‘𝐴) ∈ ℤ))
5	2, 4	anbi12d 632	. . 3 ⊢ (𝑥 = 𝐴 → (((ℜ‘𝑥) ∈ ℤ ∧ (ℑ‘𝑥) ∈ ℤ) ↔ ((ℜ‘𝐴) ∈ ℤ ∧ (ℑ‘𝐴) ∈ ℤ)))
6		df-gz 16260	. . 3 ⊢ ℤ[i] = {𝑥 ∈ ℂ ∣ ((ℜ‘𝑥) ∈ ℤ ∧ (ℑ‘𝑥) ∈ ℤ)}
7	5, 6	elrab2 3682	. 2 ⊢ (𝐴 ∈ ℤ[i] ↔ (𝐴 ∈ ℂ ∧ ((ℜ‘𝐴) ∈ ℤ ∧ (ℑ‘𝐴) ∈ ℤ)))
8		3anass 1091	. 2 ⊢ ((𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ∈ ℤ ∧ (ℑ‘𝐴) ∈ ℤ) ↔ (𝐴 ∈ ℂ ∧ ((ℜ‘𝐴) ∈ ℤ ∧ (ℑ‘𝐴) ∈ ℤ)))
9	7, 8	bitr4i 280	1 ⊢ (𝐴 ∈ ℤ[i] ↔ (𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ∈ ℤ ∧ (ℑ‘𝐴) ∈ ℤ))

Syntax hints:   ↔ wb 208   ∧ wa 398   ∧ w3a 1083   = wceq 1533   ∈ wcel 2110  ‘cfv 6349  ℂcc 10529  ℤcz 11975  ℜcre 14450  ℑcim 14451  ℤ[i]cgz 16259

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1907  ax-6 1966  ax-7 2011  ax-8 2112  ax-9 2120  ax-10 2141  ax-11 2157  ax-12 2173  ax-ext 2793

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-3an 1085  df-tru 1536  df-ex 1777  df-nf 1781  df-sb 2066  df-clab 2800  df-cleq 2814  df-clel 2893  df-nfc 2963  df-rex 3144  df-rab 3147  df-v 3496  df-dif 3938  df-un 3940  df-in 3942  df-ss 3951  df-nul 4291  df-if 4467  df-sn 4561  df-pr 4563  df-op 4567  df-uni 4832  df-br 5059  df-iota 6308  df-fv 6357  df-gz 16260

This theorem is referenced by:  gzcn  16262  zgz  16263  igz  16264  gznegcl  16265  gzcjcl  16266  gzaddcl  16267  gzmulcl  16268  gzabssqcl  16271  4sqlem4a  16281  2sqlem2  25988  2sqlem3  25990  cntotbnd  35068