Theorem elgz 16964

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 6906	. . . . 5 ⊢ (𝑥 = 𝐴 → (ℜ‘𝑥) = (ℜ‘𝐴))
2	1	eleq1d 2823	. . . 4 ⊢ (𝑥 = 𝐴 → ((ℜ‘𝑥) ∈ ℤ ↔ (ℜ‘𝐴) ∈ ℤ))
3		fveq2 6906	. . . . 5 ⊢ (𝑥 = 𝐴 → (ℑ‘𝑥) = (ℑ‘𝐴))
4	3	eleq1d 2823	. . . 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 16963	. . 3 ⊢ ℤ[i] = {𝑥 ∈ ℂ ∣ ((ℜ‘𝑥) ∈ ℤ ∧ (ℑ‘𝑥) ∈ ℤ)}
7	5, 6	elrab2 3697	. 2 ⊢ (𝐴 ∈ ℤ[i] ↔ (𝐴 ∈ ℂ ∧ ((ℜ‘𝐴) ∈ ℤ ∧ (ℑ‘𝐴) ∈ ℤ)))
8		3anass 1094	. 2 ⊢ ((𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ∈ ℤ ∧ (ℑ‘𝐴) ∈ ℤ) ↔ (𝐴 ∈ ℂ ∧ ((ℜ‘𝐴) ∈ ℤ ∧ (ℑ‘𝐴) ∈ ℤ)))
9	7, 8	bitr4i 278	1 ⊢ (𝐴 ∈ ℤ[i] ↔ (𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ∈ ℤ ∧ (ℑ‘𝐴) ∈ ℤ))

Syntax hints:   ↔ wb 206   ∧ wa 395   ∧ w3a 1086   = wceq 1536   ∈ wcel 2105  ‘cfv 6562  ℂcc 11150  ℤcz 12610  ℜcre 15132  ℑcim 15133  ℤ[i]cgz 16962

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1791  ax-4 1805  ax-5 1907  ax-6 1964  ax-7 2004  ax-8 2107  ax-9 2115  ax-ext 2705

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1539  df-fal 1549  df-ex 1776  df-sb 2062  df-clab 2712  df-cleq 2726  df-clel 2813  df-rab 3433  df-v 3479  df-dif 3965  df-un 3967  df-ss 3979  df-nul 4339  df-if 4531  df-sn 4631  df-pr 4633  df-op 4637  df-uni 4912  df-br 5148  df-iota 6515  df-fv 6570  df-gz 16963

This theorem is referenced by:  gzcn  16965  zgz  16966  igz  16967  gznegcl  16968  gzcjcl  16969  gzaddcl  16970  gzmulcl  16971  gzabssqcl  16974  4sqlem4a  16984  2sqlem2  27476  2sqlem3  27478  cntotbnd  37782