Step |
Hyp |
Ref |
Expression |
1 |
|
elnmz.1 |
⊢ 𝑁 = { 𝑥 ∈ 𝑋 ∣ ∀ 𝑦 ∈ 𝑋 ( ( 𝑥 + 𝑦 ) ∈ 𝑆 ↔ ( 𝑦 + 𝑥 ) ∈ 𝑆 ) } |
2 |
|
oveq2 |
⊢ ( 𝑦 = 𝑧 → ( 𝑥 + 𝑦 ) = ( 𝑥 + 𝑧 ) ) |
3 |
2
|
eleq1d |
⊢ ( 𝑦 = 𝑧 → ( ( 𝑥 + 𝑦 ) ∈ 𝑆 ↔ ( 𝑥 + 𝑧 ) ∈ 𝑆 ) ) |
4 |
|
oveq1 |
⊢ ( 𝑦 = 𝑧 → ( 𝑦 + 𝑥 ) = ( 𝑧 + 𝑥 ) ) |
5 |
4
|
eleq1d |
⊢ ( 𝑦 = 𝑧 → ( ( 𝑦 + 𝑥 ) ∈ 𝑆 ↔ ( 𝑧 + 𝑥 ) ∈ 𝑆 ) ) |
6 |
3 5
|
bibi12d |
⊢ ( 𝑦 = 𝑧 → ( ( ( 𝑥 + 𝑦 ) ∈ 𝑆 ↔ ( 𝑦 + 𝑥 ) ∈ 𝑆 ) ↔ ( ( 𝑥 + 𝑧 ) ∈ 𝑆 ↔ ( 𝑧 + 𝑥 ) ∈ 𝑆 ) ) ) |
7 |
6
|
cbvralvw |
⊢ ( ∀ 𝑦 ∈ 𝑋 ( ( 𝑥 + 𝑦 ) ∈ 𝑆 ↔ ( 𝑦 + 𝑥 ) ∈ 𝑆 ) ↔ ∀ 𝑧 ∈ 𝑋 ( ( 𝑥 + 𝑧 ) ∈ 𝑆 ↔ ( 𝑧 + 𝑥 ) ∈ 𝑆 ) ) |
8 |
|
oveq1 |
⊢ ( 𝑥 = 𝐴 → ( 𝑥 + 𝑧 ) = ( 𝐴 + 𝑧 ) ) |
9 |
8
|
eleq1d |
⊢ ( 𝑥 = 𝐴 → ( ( 𝑥 + 𝑧 ) ∈ 𝑆 ↔ ( 𝐴 + 𝑧 ) ∈ 𝑆 ) ) |
10 |
|
oveq2 |
⊢ ( 𝑥 = 𝐴 → ( 𝑧 + 𝑥 ) = ( 𝑧 + 𝐴 ) ) |
11 |
10
|
eleq1d |
⊢ ( 𝑥 = 𝐴 → ( ( 𝑧 + 𝑥 ) ∈ 𝑆 ↔ ( 𝑧 + 𝐴 ) ∈ 𝑆 ) ) |
12 |
9 11
|
bibi12d |
⊢ ( 𝑥 = 𝐴 → ( ( ( 𝑥 + 𝑧 ) ∈ 𝑆 ↔ ( 𝑧 + 𝑥 ) ∈ 𝑆 ) ↔ ( ( 𝐴 + 𝑧 ) ∈ 𝑆 ↔ ( 𝑧 + 𝐴 ) ∈ 𝑆 ) ) ) |
13 |
12
|
ralbidv |
⊢ ( 𝑥 = 𝐴 → ( ∀ 𝑧 ∈ 𝑋 ( ( 𝑥 + 𝑧 ) ∈ 𝑆 ↔ ( 𝑧 + 𝑥 ) ∈ 𝑆 ) ↔ ∀ 𝑧 ∈ 𝑋 ( ( 𝐴 + 𝑧 ) ∈ 𝑆 ↔ ( 𝑧 + 𝐴 ) ∈ 𝑆 ) ) ) |
14 |
7 13
|
syl5bb |
⊢ ( 𝑥 = 𝐴 → ( ∀ 𝑦 ∈ 𝑋 ( ( 𝑥 + 𝑦 ) ∈ 𝑆 ↔ ( 𝑦 + 𝑥 ) ∈ 𝑆 ) ↔ ∀ 𝑧 ∈ 𝑋 ( ( 𝐴 + 𝑧 ) ∈ 𝑆 ↔ ( 𝑧 + 𝐴 ) ∈ 𝑆 ) ) ) |
15 |
14 1
|
elrab2 |
⊢ ( 𝐴 ∈ 𝑁 ↔ ( 𝐴 ∈ 𝑋 ∧ ∀ 𝑧 ∈ 𝑋 ( ( 𝐴 + 𝑧 ) ∈ 𝑆 ↔ ( 𝑧 + 𝐴 ) ∈ 𝑆 ) ) ) |