Theorem nmzbi 19374

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 19373	. . 3 ⊢ (𝐴 ∈ 𝑁 ↔ (𝐴 ∈ 𝑋 ∧ ∀𝑧 ∈ 𝑋 ((𝐴 + 𝑧) ∈ 𝑆 ↔ (𝑧 + 𝐴) ∈ 𝑆)))
3	2	simprbi 503	. 2 ⊢ (𝐴 ∈ 𝑁 → ∀𝑧 ∈ 𝑋 ((𝐴 + 𝑧) ∈ 𝑆 ↔ (𝑧 + 𝐴) ∈ 𝑆))
4		oveq2 7428	. . . . 5 ⊢ (𝑧 = 𝐵 → (𝐴 + 𝑧) = (𝐴 + 𝐵))
5	4	eleq1d 2845	. . . 4 ⊢ (𝑧 = 𝐵 → ((𝐴 + 𝑧) ∈ 𝑆 ↔ (𝐴 + 𝐵) ∈ 𝑆))
6		oveq1 7427	. . . . 5 ⊢ (𝑧 = 𝐵 → (𝑧 + 𝐴) = (𝐵 + 𝐴))
7	6	eleq1d 2845	. . . 4 ⊢ (𝑧 = 𝐵 → ((𝑧 + 𝐴) ∈ 𝑆 ↔ (𝐵 + 𝐴) ∈ 𝑆))
8	5, 7	bibi12d 348	. . 3 ⊢ (𝑧 = 𝐵 → (((𝐴 + 𝑧) ∈ 𝑆 ↔ (𝑧 + 𝐴) ∈ 𝑆) ↔ ((𝐴 + 𝐵) ∈ 𝑆 ↔ (𝐵 + 𝐴) ∈ 𝑆)))
9	8	rspccva 3575	. 2 ⊢ ((∀𝑧 ∈ 𝑋 ((𝐴 + 𝑧) ∈ 𝑆 ↔ (𝑧 + 𝐴) ∈ 𝑆) ∧ 𝐵 ∈ 𝑋) → ((𝐴 + 𝐵) ∈ 𝑆 ↔ (𝐵 + 𝐴) ∈ 𝑆))
10	3, 9	sylan 592	1 ⊢ ((𝐴 ∈ 𝑁 ∧ 𝐵 ∈ 𝑋) → ((𝐴 + 𝐵) ∈ 𝑆 ↔ (𝐵 + 𝐴) ∈ 𝑆))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1569   ∈ wcel 2144  ∀wral 3076  {crab 3412  (class class class)co 7420

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-ral 3077  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-ov 7423

This theorem is used by:  nmzsubg  19375  nmznsg  19378  conjnmz  19466