Theorem nmzbi 17457

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 17456	. . 3 ⊢ (𝐴 ∈ 𝑁 ↔ (𝐴 ∈ 𝑋 ∧ ∀𝑧 ∈ 𝑋 ((𝐴 + 𝑧) ∈ 𝑆 ↔ (𝑧 + 𝐴) ∈ 𝑆)))
3	2	simprbi 479	. 2 ⊢ (𝐴 ∈ 𝑁 → ∀𝑧 ∈ 𝑋 ((𝐴 + 𝑧) ∈ 𝑆 ↔ (𝑧 + 𝐴) ∈ 𝑆))
4		oveq2 6557	. . . . 5 ⊢ (𝑧 = 𝐵 → (𝐴 + 𝑧) = (𝐴 + 𝐵))
5	4	eleq1d 2672	. . . 4 ⊢ (𝑧 = 𝐵 → ((𝐴 + 𝑧) ∈ 𝑆 ↔ (𝐴 + 𝐵) ∈ 𝑆))
6		oveq1 6556	. . . . 5 ⊢ (𝑧 = 𝐵 → (𝑧 + 𝐴) = (𝐵 + 𝐴))
7	6	eleq1d 2672	. . . 4 ⊢ (𝑧 = 𝐵 → ((𝑧 + 𝐴) ∈ 𝑆 ↔ (𝐵 + 𝐴) ∈ 𝑆))
8	5, 7	bibi12d 334	. . 3 ⊢ (𝑧 = 𝐵 → (((𝐴 + 𝑧) ∈ 𝑆 ↔ (𝑧 + 𝐴) ∈ 𝑆) ↔ ((𝐴 + 𝐵) ∈ 𝑆 ↔ (𝐵 + 𝐴) ∈ 𝑆)))
9	8	rspccva 3281	. 2 ⊢ ((∀𝑧 ∈ 𝑋 ((𝐴 + 𝑧) ∈ 𝑆 ↔ (𝑧 + 𝐴) ∈ 𝑆) ∧ 𝐵 ∈ 𝑋) → ((𝐴 + 𝐵) ∈ 𝑆 ↔ (𝐵 + 𝐴) ∈ 𝑆))
10	3, 9	sylan 487	1 ⊢ ((𝐴 ∈ 𝑁 ∧ 𝐵 ∈ 𝑋) → ((𝐴 + 𝐵) ∈ 𝑆 ↔ (𝐵 + 𝐴) ∈ 𝑆))

Syntax hints:   → wi 4   ↔ wb 195   ∧ wa 383   = wceq 1475   ∈ wcel 1977  ∀wral 2896  {crab 2900  (class class class)co 6549

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590

This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-3an 1033  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-nfc 2740  df-ral 2901  df-rex 2902  df-rab 2905  df-v 3175  df-dif 3543  df-un 3545  df-in 3547  df-ss 3554  df-nul 3875  df-if 4037  df-sn 4126  df-pr 4128  df-op 4132  df-uni 4373  df-br 4584  df-iota 5768  df-fv 5812  df-ov 6552

This theorem is referenced by:  nmzsubg  17458  nmznsg  17461  conjnmz  17517