Theorem nmzbi

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

	Ref	Expression
Hypothesis	elnmz.1	⊢ 𝑁 = { 𝑥 ∈ 𝑋 ∣ ∀ 𝑦 ∈ 𝑋 ( ( 𝑥 + 𝑦 ) ∈ 𝑆 ↔ ( 𝑦 + 𝑥 ) ∈ 𝑆 ) }
Assertion	nmzbi	⊢ ( ( 𝐴 ∈ 𝑁 ∧ 𝐵 ∈ 𝑋 ) → ( ( 𝐴 + 𝐵 ) ∈ 𝑆 ↔ ( 𝐵 + 𝐴 ) ∈ 𝑆 ) )

Proof

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