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 ) ) |