Theorem elgz 17109

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 6885	. . . . 5 ⊢ (𝑥 = 𝐴 → (ℜ‘𝑥) = (ℜ‘𝐴))
2	1	eleq1d 2845	. . . 4 ⊢ (𝑥 = 𝐴 → ((ℜ‘𝑥) ∈ ℤ ↔ (ℜ‘𝐴) ∈ ℤ))
3		fveq2 6885	. . . . 5 ⊢ (𝑥 = 𝐴 → (ℑ‘𝑥) = (ℑ‘𝐴))
4	3	eleq1d 2845	. . . 4 ⊢ (𝑥 = 𝐴 → ((ℑ‘𝑥) ∈ ℤ ↔ (ℑ‘𝐴) ∈ ℤ))
5	2, 4	anbi12d 644	. . 3 ⊢ (𝑥 = 𝐴 → (((ℜ‘𝑥) ∈ ℤ ∧ (ℑ‘𝑥) ∈ ℤ) ↔ ((ℜ‘𝐴) ∈ ℤ ∧ (ℑ‘𝐴) ∈ ℤ)))
6		df-gz 17108	. . 3 ⊢ ℤ[i] = {𝑥 ∈ ℂ ∣ ((ℜ‘𝑥) ∈ ℤ ∧ (ℑ‘𝑥) ∈ ℤ)}
7	5, 6	elrab2 3648	. 2 ⊢ (𝐴 ∈ ℤ[i] ↔ (𝐴 ∈ ℂ ∧ ((ℜ‘𝐴) ∈ ℤ ∧ (ℑ‘𝐴) ∈ ℤ)))
8		3anass 1111	. 2 ⊢ ((𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ∈ ℤ ∧ (ℑ‘𝐴) ∈ ℤ) ↔ (𝐴 ∈ ℂ ∧ ((ℜ‘𝐴) ∈ ℤ ∧ (ℑ‘𝐴) ∈ ℤ)))
9	7, 8	bitr4i 281	1 ⊢ (𝐴 ∈ ℤ[i] ↔ (𝐴 ∈ ℂ ∧ (ℜ‘𝐴) ∈ ℤ ∧ (ℑ‘𝐴) ∈ ℤ))

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401   ∧ w3a 1103   = wceq 1569   ∈ wcel 2144  ‘cfv 6538  ℂcc 11198  ℤcz 12693  ℜcre 15264  ℑcim 15265  ℤ[i]cgz 17107

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-rab 3413  df-v 3452  df-ss 3896  df-dif 4047  df-un 4151  df-nul 4279  df-if 4482  df-sn 4584  df-pr 4586  df-op 4590  df-uni 4867  df-br 5103  df-iota 6494  df-fv 6546  df-gz 17108

This theorem is used by:  gzcn  17110  zgz  17111  igz  17112  gznegcl  17113  gzcjcl  17114  gzaddcl  17115  gzmulcl  17116  gzabssqcl  17119  4sqlem4a  17129  2sqlem2  27745  2sqlem3  27747  cntotbnd  38731