Theorem elgz 16997

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 6881	. . . . 5 ⊢ (𝑥 = 𝐴 → (ℜ‘𝑥) = (ℜ‘𝐴))
2	1	eleq1d 2847	. . . 4 ⊢ (𝑥 = 𝐴 → ((ℜ‘𝑥) ∈ ℤ ↔ (ℜ‘𝐴) ∈ ℤ))
3		fveq2 6881	. . . . 5 ⊢ (𝑥 = 𝐴 → (ℑ‘𝑥) = (ℑ‘𝐴))
4	3	eleq1d 2847	. . . 4 ⊢ (𝑥 = 𝐴 → ((ℑ‘𝑥) ∈ ℤ ↔ (ℑ‘𝐴) ∈ ℤ))
5	2, 4	anbi12d 643	. . 3 ⊢ (𝑥 = 𝐴 → (((ℜ‘𝑥) ∈ ℤ ∧ (ℑ‘𝑥) ∈ ℤ) ↔ ((ℜ‘𝐴) ∈ ℤ ∧ (ℑ‘𝐴) ∈ ℤ)))
6		df-gz 16996	. . 3 ⊢ ℤ[i] = {𝑥 ∈ ℂ ∣ ((ℜ‘𝑥) ∈ ℤ ∧ (ℑ‘𝑥) ∈ ℤ)}
7	5, 6	elrab2 3653	. 2 ⊢ (𝐴 ∈ ℤ[i] ↔ (𝐴 ∈ ℂ ∧ ((ℜ‘𝐴) ∈ ℤ ∧ (ℑ‘𝐴) ∈ ℤ)))
8		3anass 1110	. 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 400   ∧ w3a 1102   = wceq 1569   ∈ wcel 2142  ‘cfv 6536  ℂcc 11104  ℤcz 12597  ℜcre 15155  ℑcim 15156  ℤ[i]cgz 16995

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-8 2144  ax-9 2152  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1104  df-tru 1572  df-fal 1582  df-ex 1809  df-sb 2096  df-clab 2741  df-cleq 2754  df-clel 2837  df-rab 3416  df-v 3456  df-dif 3907  df-un 3909  df-ss 3921  df-nul 4286  df-if 4487  df-sn 4589  df-pr 4591  df-op 4595  df-uni 4872  df-br 5109  df-iota 6492  df-fv 6544  df-gz 16996

This theorem is used by:  gzcn  16998  zgz  16999  igz  17000  gznegcl  17001  gzcjcl  17002  gzaddcl  17003  gzmulcl  17004  gzabssqcl  17007  4sqlem4a  17017  2sqlem2  27593  2sqlem3  27595  cntotbnd  38475