Theorem nmzbi 19207

Description: Defining property of the normalizer. (Contributed by Mario Carneiro, 18-Jan-2015.)

Hypothesis

Ref	Expression
elnmz.1	⊢ 𝑁 = {𝑥 ∈ 𝑋 ∣ ∀𝑦 ∈ 𝑋 ((𝑥 + 𝑦) ∈ 𝑆 ↔ (𝑦 + 𝑥) ∈ 𝑆)}

Assertion

Ref	Expression
nmzbi	⊢ ((𝐴 ∈ 𝑁 ∧ 𝐵 ∈ 𝑋) → ((𝐴 + 𝐵) ∈ 𝑆 ↔ (𝐵 + 𝐴) ∈ 𝑆))

Distinct variable groups:   𝑥,𝐴   𝑥,𝑦,𝑆   𝑥, + ,𝑦   𝑥,𝑋,𝑦

Allowed substitution hints:   𝐴(𝑦)   𝐵(𝑥,𝑦)   𝑁(𝑥,𝑦)

Proof of Theorem nmzbi

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		elnmz.1	. . . 4 ⊢ 𝑁 = {𝑥 ∈ 𝑋 ∣ ∀𝑦 ∈ 𝑋 ((𝑥 + 𝑦) ∈ 𝑆 ↔ (𝑦 + 𝑥) ∈ 𝑆)}
2	1	elnmz 19206	. . 3 ⊢ (𝐴 ∈ 𝑁 ↔ (𝐴 ∈ 𝑋 ∧ ∀𝑧 ∈ 𝑋 ((𝐴 + 𝑧) ∈ 𝑆 ↔ (𝑧 + 𝐴) ∈ 𝑆)))
3	2	simprbi 501	. 2 ⊢ (𝐴 ∈ 𝑁 → ∀𝑧 ∈ 𝑋 ((𝐴 + 𝑧) ∈ 𝑆 ↔ (𝑧 + 𝐴) ∈ 𝑆))
4		oveq2 7406	. . . . 5 ⊢ (𝑧 = 𝐵 → (𝐴 + 𝑧) = (𝐴 + 𝐵))
5	4	eleq1d 2849	. . . 4 ⊢ (𝑧 = 𝐵 → ((𝐴 + 𝑧) ∈ 𝑆 ↔ (𝐴 + 𝐵) ∈ 𝑆))
6		oveq1 7405	. . . . 5 ⊢ (𝑧 = 𝐵 → (𝑧 + 𝐴) = (𝐵 + 𝐴))
7	6	eleq1d 2849	. . . 4 ⊢ (𝑧 = 𝐵 → ((𝑧 + 𝐴) ∈ 𝑆 ↔ (𝐵 + 𝐴) ∈ 𝑆))
8	5, 7	bibi12d 347	. . 3 ⊢ (𝑧 = 𝐵 → (((𝐴 + 𝑧) ∈ 𝑆 ↔ (𝑧 + 𝐴) ∈ 𝑆) ↔ ((𝐴 + 𝐵) ∈ 𝑆 ↔ (𝐵 + 𝐴) ∈ 𝑆)))
9	8	rspccva 3582	. 2 ⊢ ((∀𝑧 ∈ 𝑋 ((𝐴 + 𝑧) ∈ 𝑆 ↔ (𝑧 + 𝐴) ∈ 𝑆) ∧ 𝐵 ∈ 𝑋) → ((𝐴 + 𝐵) ∈ 𝑆 ↔ (𝐵 + 𝐴) ∈ 𝑆))
10	3, 9	sylan 589	1 ⊢ ((𝐴 ∈ 𝑁 ∧ 𝐵 ∈ 𝑋) → ((𝐴 + 𝐵) ∈ 𝑆 ↔ (𝐵 + 𝐴) ∈ 𝑆))

Syntax hints:   → wi 4   ↔ wb 208   ∧ wa 399   = wceq 1562   ∈ wcel 2144  ∀wral 3078  {crab 3416  (class class class)co 7398

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1817  ax-4 1831  ax-5 1932  ax-6 1989  ax-7 2030  ax-8 2146  ax-9 2154  ax-ext 2736

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-3an 1101  df-tru 1565  df-fal 1575  df-ex 1802  df-sb 2093  df-clab 2743  df-cleq 2756  df-clel 2839  df-ral 3079  df-rab 3417  df-v 3458  df-dif 3909  df-un 3911  df-ss 3923  df-nul 4288  df-if 4483  df-sn 4585  df-pr 4587  df-op 4591  df-uni 4868  df-br 5103  df-iota 6479  df-fv 6531  df-ov 7401

This theorem is referenced by:  nmzsubg  19208  nmznsg  19211  conjnmz  19294