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Metamath Proof Explorer

Theorem nsgbi

Description: Defining property of a normal subgroup. (Contributed by Mario Carneiro, 18-Jan-2015)

Ref Expression
Hypotheses isnsg.1 ⊢ 𝑋 = ( Base ‘ 𝐺 )
isnsg.2 ⊢ + = ( +g ‘ 𝐺 )
Assertion nsgbi ( ( 𝑆 ∈ ( NrmSGrp ‘ 𝐺 ) ∧ 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ) → ( ( 𝐴 + 𝐵 ) ∈ 𝑆 ↔ ( 𝐵 + 𝐴 ) ∈ 𝑆 ) )

Proof

Step Hyp Ref Expression
1 isnsg.1 ⊢ 𝑋 = ( Base ‘ 𝐺 )
2 isnsg.2 ⊢ + = ( +g ‘ 𝐺 )
3 1 2 isnsg ⊢ ( 𝑆 ∈ ( NrmSGrp ‘ 𝐺 ) ↔ ( 𝑆 ∈ ( SubGrp ‘ 𝐺 ) ∧ ∀ 𝑥 ∈ 𝑋 ∀ 𝑦 ∈ 𝑋 ( ( 𝑥 + 𝑦 ) ∈ 𝑆 ↔ ( 𝑦 + 𝑥 ) ∈ 𝑆 ) ) )
4 3 simprbi ⊢ ( 𝑆 ∈ ( NrmSGrp ‘ 𝐺 ) → ∀ 𝑥 ∈ 𝑋 ∀ 𝑦 ∈ 𝑋 ( ( 𝑥 + 𝑦 ) ∈ 𝑆 ↔ ( 𝑦 + 𝑥 ) ∈ 𝑆 ) )
5 oveq1 ⊢ ( 𝑥 = 𝐴 → ( 𝑥 + 𝑦 ) = ( 𝐴 + 𝑦 ) )
6 5 eleq1d ⊢ ( 𝑥 = 𝐴 → ( ( 𝑥 + 𝑦 ) ∈ 𝑆 ↔ ( 𝐴 + 𝑦 ) ∈ 𝑆 ) )
7 oveq2 ⊢ ( 𝑥 = 𝐴 → ( 𝑦 + 𝑥 ) = ( 𝑦 + 𝐴 ) )
8 7 eleq1d ⊢ ( 𝑥 = 𝐴 → ( ( 𝑦 + 𝑥 ) ∈ 𝑆 ↔ ( 𝑦 + 𝐴 ) ∈ 𝑆 ) )
9 6 8 bibi12d ⊢ ( 𝑥 = 𝐴 → ( ( ( 𝑥 + 𝑦 ) ∈ 𝑆 ↔ ( 𝑦 + 𝑥 ) ∈ 𝑆 ) ↔ ( ( 𝐴 + 𝑦 ) ∈ 𝑆 ↔ ( 𝑦 + 𝐴 ) ∈ 𝑆 ) ) )
10 oveq2 ⊢ ( 𝑦 = 𝐵 → ( 𝐴 + 𝑦 ) = ( 𝐴 + 𝐵 ) )
11 10 eleq1d ⊢ ( 𝑦 = 𝐵 → ( ( 𝐴 + 𝑦 ) ∈ 𝑆 ↔ ( 𝐴 + 𝐵 ) ∈ 𝑆 ) )
12 oveq1 ⊢ ( 𝑦 = 𝐵 → ( 𝑦 + 𝐴 ) = ( 𝐵 + 𝐴 ) )
13 12 eleq1d ⊢ ( 𝑦 = 𝐵 → ( ( 𝑦 + 𝐴 ) ∈ 𝑆 ↔ ( 𝐵 + 𝐴 ) ∈ 𝑆 ) )
14 11 13 bibi12d ⊢ ( 𝑦 = 𝐵 → ( ( ( 𝐴 + 𝑦 ) ∈ 𝑆 ↔ ( 𝑦 + 𝐴 ) ∈ 𝑆 ) ↔ ( ( 𝐴 + 𝐵 ) ∈ 𝑆 ↔ ( 𝐵 + 𝐴 ) ∈ 𝑆 ) ) )
15 9 14 rspc2v ⊢ ( ( 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ) → ( ∀ 𝑥 ∈ 𝑋 ∀ 𝑦 ∈ 𝑋 ( ( 𝑥 + 𝑦 ) ∈ 𝑆 ↔ ( 𝑦 + 𝑥 ) ∈ 𝑆 ) → ( ( 𝐴 + 𝐵 ) ∈ 𝑆 ↔ ( 𝐵 + 𝐴 ) ∈ 𝑆 ) ) )
16 4 15 syl5com ⊢ ( 𝑆 ∈ ( NrmSGrp ‘ 𝐺 ) → ( ( 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ) → ( ( 𝐴 + 𝐵 ) ∈ 𝑆 ↔ ( 𝐵 + 𝐴 ) ∈ 𝑆 ) ) )
17 16 3impib ⊢ ( ( 𝑆 ∈ ( NrmSGrp ‘ 𝐺 ) ∧ 𝐴 ∈ 𝑋 ∧ 𝐵 ∈ 𝑋 ) → ( ( 𝐴 + 𝐵 ) ∈ 𝑆 ↔ ( 𝐵 + 𝐴 ) ∈ 𝑆 ) )