Metamath Proof Explorer
Description: In an integral domain, if a prime element divides another, they are
associates. (Contributed by Thierry Arnoux, 27-May-2025)
|
|
Ref |
Expression |
|
Hypotheses |
rprmasso.b |
⊢ 𝐵 = ( Base ‘ 𝑅 ) |
|
|
rprmasso.p |
⊢ 𝑃 = ( RPrime ‘ 𝑅 ) |
|
|
rprmasso.d |
⊢ ∥ = ( ∥r ‘ 𝑅 ) |
|
|
rprmasso.r |
⊢ ( 𝜑 → 𝑅 ∈ IDomn ) |
|
|
rprmasso.x |
⊢ ( 𝜑 → 𝑋 ∈ 𝑃 ) |
|
|
rprmasso.1 |
⊢ ( 𝜑 → 𝑋 ∥ 𝑌 ) |
|
|
rprmasso2.y |
⊢ ( 𝜑 → 𝑌 ∈ 𝑃 ) |
|
|
rprmasso3.1 |
⊢ · = ( .r ‘ 𝑅 ) |
|
|
rprmasso3.u |
⊢ 𝑈 = ( Unit ‘ 𝑅 ) |
|
Assertion |
rprmasso3 |
⊢ ( 𝜑 → ∃ 𝑡 ∈ 𝑈 ( 𝑡 · 𝑋 ) = 𝑌 ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
rprmasso.b |
⊢ 𝐵 = ( Base ‘ 𝑅 ) |
2 |
|
rprmasso.p |
⊢ 𝑃 = ( RPrime ‘ 𝑅 ) |
3 |
|
rprmasso.d |
⊢ ∥ = ( ∥r ‘ 𝑅 ) |
4 |
|
rprmasso.r |
⊢ ( 𝜑 → 𝑅 ∈ IDomn ) |
5 |
|
rprmasso.x |
⊢ ( 𝜑 → 𝑋 ∈ 𝑃 ) |
6 |
|
rprmasso.1 |
⊢ ( 𝜑 → 𝑋 ∥ 𝑌 ) |
7 |
|
rprmasso2.y |
⊢ ( 𝜑 → 𝑌 ∈ 𝑃 ) |
8 |
|
rprmasso3.1 |
⊢ · = ( .r ‘ 𝑅 ) |
9 |
|
rprmasso3.u |
⊢ 𝑈 = ( Unit ‘ 𝑅 ) |
10 |
1 2 3 4 5 6 7
|
rprmasso2 |
⊢ ( 𝜑 → 𝑌 ∥ 𝑋 ) |
11 |
|
eqid |
⊢ ( RSpan ‘ 𝑅 ) = ( RSpan ‘ 𝑅 ) |
12 |
1 2 4 5
|
rprmcl |
⊢ ( 𝜑 → 𝑋 ∈ 𝐵 ) |
13 |
1 2 4 7
|
rprmcl |
⊢ ( 𝜑 → 𝑌 ∈ 𝐵 ) |
14 |
1 11 3 12 13 9 8 4
|
dvdsruasso |
⊢ ( 𝜑 → ( ( 𝑋 ∥ 𝑌 ∧ 𝑌 ∥ 𝑋 ) ↔ ∃ 𝑡 ∈ 𝑈 ( 𝑡 · 𝑋 ) = 𝑌 ) ) |
15 |
6 10 14
|
mpbi2and |
⊢ ( 𝜑 → ∃ 𝑡 ∈ 𝑈 ( 𝑡 · 𝑋 ) = 𝑌 ) |