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Metamath Proof Explorer

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 ) )