Step |
Hyp |
Ref |
Expression |
1 |
|
rrgval.e |
⊢ 𝐸 = ( RLReg ‘ 𝑅 ) |
2 |
|
rrgval.b |
⊢ 𝐵 = ( Base ‘ 𝑅 ) |
3 |
|
rrgval.t |
⊢ · = ( .r ‘ 𝑅 ) |
4 |
|
rrgval.z |
⊢ 0 = ( 0g ‘ 𝑅 ) |
5 |
1 2 3 4
|
rrgeq0i |
⊢ ( ( 𝑋 ∈ 𝐸 ∧ 𝑌 ∈ 𝐵 ) → ( ( 𝑋 · 𝑌 ) = 0 → 𝑌 = 0 ) ) |
6 |
5
|
3adant1 |
⊢ ( ( 𝑅 ∈ Ring ∧ 𝑋 ∈ 𝐸 ∧ 𝑌 ∈ 𝐵 ) → ( ( 𝑋 · 𝑌 ) = 0 → 𝑌 = 0 ) ) |
7 |
|
simp1 |
⊢ ( ( 𝑅 ∈ Ring ∧ 𝑋 ∈ 𝐸 ∧ 𝑌 ∈ 𝐵 ) → 𝑅 ∈ Ring ) |
8 |
1 2 3 4
|
rrgval |
⊢ 𝐸 = { 𝑥 ∈ 𝐵 ∣ ∀ 𝑦 ∈ 𝐵 ( ( 𝑥 · 𝑦 ) = 0 → 𝑦 = 0 ) } |
9 |
8
|
ssrab3 |
⊢ 𝐸 ⊆ 𝐵 |
10 |
|
simp2 |
⊢ ( ( 𝑅 ∈ Ring ∧ 𝑋 ∈ 𝐸 ∧ 𝑌 ∈ 𝐵 ) → 𝑋 ∈ 𝐸 ) |
11 |
9 10
|
sselid |
⊢ ( ( 𝑅 ∈ Ring ∧ 𝑋 ∈ 𝐸 ∧ 𝑌 ∈ 𝐵 ) → 𝑋 ∈ 𝐵 ) |
12 |
2 3 4
|
ringrz |
⊢ ( ( 𝑅 ∈ Ring ∧ 𝑋 ∈ 𝐵 ) → ( 𝑋 · 0 ) = 0 ) |
13 |
7 11 12
|
syl2anc |
⊢ ( ( 𝑅 ∈ Ring ∧ 𝑋 ∈ 𝐸 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑋 · 0 ) = 0 ) |
14 |
|
oveq2 |
⊢ ( 𝑌 = 0 → ( 𝑋 · 𝑌 ) = ( 𝑋 · 0 ) ) |
15 |
14
|
eqeq1d |
⊢ ( 𝑌 = 0 → ( ( 𝑋 · 𝑌 ) = 0 ↔ ( 𝑋 · 0 ) = 0 ) ) |
16 |
13 15
|
syl5ibrcom |
⊢ ( ( 𝑅 ∈ Ring ∧ 𝑋 ∈ 𝐸 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑌 = 0 → ( 𝑋 · 𝑌 ) = 0 ) ) |
17 |
6 16
|
impbid |
⊢ ( ( 𝑅 ∈ Ring ∧ 𝑋 ∈ 𝐸 ∧ 𝑌 ∈ 𝐵 ) → ( ( 𝑋 · 𝑌 ) = 0 ↔ 𝑌 = 0 ) ) |