Theorem nmzbi 19097

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 19096	. . 3 ⊢ (𝐴 ∈ 𝑁 ↔ (𝐴 ∈ 𝑋 ∧ ∀𝑧 ∈ 𝑋 ((𝐴 + 𝑧) ∈ 𝑆 ↔ (𝑧 + 𝐴) ∈ 𝑆)))
3	2	simprbi 496	. 2 ⊢ (𝐴 ∈ 𝑁 → ∀𝑧 ∈ 𝑋 ((𝐴 + 𝑧) ∈ 𝑆 ↔ (𝑧 + 𝐴) ∈ 𝑆))
4		oveq2 7368	. . . . 5 ⊢ (𝑧 = 𝐵 → (𝐴 + 𝑧) = (𝐴 + 𝐵))
5	4	eleq1d 2822	. . . 4 ⊢ (𝑧 = 𝐵 → ((𝐴 + 𝑧) ∈ 𝑆 ↔ (𝐴 + 𝐵) ∈ 𝑆))
6		oveq1 7367	. . . . 5 ⊢ (𝑧 = 𝐵 → (𝑧 + 𝐴) = (𝐵 + 𝐴))
7	6	eleq1d 2822	. . . 4 ⊢ (𝑧 = 𝐵 → ((𝑧 + 𝐴) ∈ 𝑆 ↔ (𝐵 + 𝐴) ∈ 𝑆))
8	5, 7	bibi12d 345	. . 3 ⊢ (𝑧 = 𝐵 → (((𝐴 + 𝑧) ∈ 𝑆 ↔ (𝑧 + 𝐴) ∈ 𝑆) ↔ ((𝐴 + 𝐵) ∈ 𝑆 ↔ (𝐵 + 𝐴) ∈ 𝑆)))
9	8	rspccva 3576	. 2 ⊢ ((∀𝑧 ∈ 𝑋 ((𝐴 + 𝑧) ∈ 𝑆 ↔ (𝑧 + 𝐴) ∈ 𝑆) ∧ 𝐵 ∈ 𝑋) → ((𝐴 + 𝐵) ∈ 𝑆 ↔ (𝐵 + 𝐴) ∈ 𝑆))
10	3, 9	sylan 581	1 ⊢ ((𝐴 ∈ 𝑁 ∧ 𝐵 ∈ 𝑋) → ((𝐴 + 𝐵) ∈ 𝑆 ↔ (𝐵 + 𝐴) ∈ 𝑆))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   = wceq 1542   ∈ wcel 2114  ∀wral 3052  {crab 3400  (class class class)co 7360

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1545  df-fal 1555  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-ral 3053  df-rab 3401  df-v 3443  df-dif 3905  df-un 3907  df-ss 3919  df-nul 4287  df-if 4481  df-sn 4582  df-pr 4584  df-op 4588  df-uni 4865  df-br 5100  df-iota 6449  df-fv 6501  df-ov 7363

This theorem is referenced by:  nmzsubg  19098  nmznsg  19101  conjnmz  19185