Metamath Proof Explorer


Theorem isrprm

Description: Property for P to be a prime element in the ring R . (Contributed by Thierry Arnoux, 1-Jul-2024)

Ref Expression
Hypotheses isrprm.1 ⊢ 𝐵 = ( Base ‘ 𝑅 )
isrprm.2 ⊢ 𝑈 = ( Unit ‘ 𝑅 )
isrprm.3 ⊢ 0 = ( 0g ‘ 𝑅 )
isrprm.4 ⊢ ∥ = ( ∥r ‘ 𝑅 )
isrprm.5 ⊢ · = ( .r ‘ 𝑅 )
Assertion isrprm ( 𝑅 ∈ 𝑉 → ( 𝑃 ∈ ( RPrime ‘ 𝑅 ) ↔ ( 𝑃 ∈ ( 𝐵 ∖ ( 𝑈 ∪ { 0 } ) ) ∧ ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ 𝐵 ( 𝑃 ∥ ( 𝑥 · 𝑦 ) → ( 𝑃 ∥ 𝑥 ∨ 𝑃 ∥ 𝑦 ) ) ) ) )

Proof

Step Hyp Ref Expression
1 isrprm.1 ⊢ 𝐵 = ( Base ‘ 𝑅 )
2 isrprm.2 ⊢ 𝑈 = ( Unit ‘ 𝑅 )
3 isrprm.3 ⊢ 0 = ( 0g ‘ 𝑅 )
4 isrprm.4 ⊢ ∥ = ( ∥r ‘ 𝑅 )
5 isrprm.5 ⊢ · = ( .r ‘ 𝑅 )
6 1 2 3 5 4 rprmval ⊢ ( 𝑅 ∈ 𝑉 → ( RPrime ‘ 𝑅 ) = { 𝑝 ∈ ( 𝐵 ∖ ( 𝑈 ∪ { 0 } ) ) ∣ ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ 𝐵 ( 𝑝 ∥ ( 𝑥 · 𝑦 ) → ( 𝑝 ∥ 𝑥 ∨ 𝑝 ∥ 𝑦 ) ) } )
7 6 eleq2d ⊢ ( 𝑅 ∈ 𝑉 → ( 𝑃 ∈ ( RPrime ‘ 𝑅 ) ↔ 𝑃 ∈ { 𝑝 ∈ ( 𝐵 ∖ ( 𝑈 ∪ { 0 } ) ) ∣ ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ 𝐵 ( 𝑝 ∥ ( 𝑥 · 𝑦 ) → ( 𝑝 ∥ 𝑥 ∨ 𝑝 ∥ 𝑦 ) ) } ) )
8 breq1 ⊢ ( 𝑝 = 𝑃 → ( 𝑝 ∥ ( 𝑥 · 𝑦 ) ↔ 𝑃 ∥ ( 𝑥 · 𝑦 ) ) )
9 breq1 ⊢ ( 𝑝 = 𝑃 → ( 𝑝 ∥ 𝑥 ↔ 𝑃 ∥ 𝑥 ) )
10 breq1 ⊢ ( 𝑝 = 𝑃 → ( 𝑝 ∥ 𝑦 ↔ 𝑃 ∥ 𝑦 ) )
11 9 10 orbi12d ⊢ ( 𝑝 = 𝑃 → ( ( 𝑝 ∥ 𝑥 ∨ 𝑝 ∥ 𝑦 ) ↔ ( 𝑃 ∥ 𝑥 ∨ 𝑃 ∥ 𝑦 ) ) )
12 8 11 imbi12d ⊢ ( 𝑝 = 𝑃 → ( ( 𝑝 ∥ ( 𝑥 · 𝑦 ) → ( 𝑝 ∥ 𝑥 ∨ 𝑝 ∥ 𝑦 ) ) ↔ ( 𝑃 ∥ ( 𝑥 · 𝑦 ) → ( 𝑃 ∥ 𝑥 ∨ 𝑃 ∥ 𝑦 ) ) ) )
13 12 2ralbidv ⊢ ( 𝑝 = 𝑃 → ( ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ 𝐵 ( 𝑝 ∥ ( 𝑥 · 𝑦 ) → ( 𝑝 ∥ 𝑥 ∨ 𝑝 ∥ 𝑦 ) ) ↔ ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ 𝐵 ( 𝑃 ∥ ( 𝑥 · 𝑦 ) → ( 𝑃 ∥ 𝑥 ∨ 𝑃 ∥ 𝑦 ) ) ) )
14 13 elrab ⊢ ( 𝑃 ∈ { 𝑝 ∈ ( 𝐵 ∖ ( 𝑈 ∪ { 0 } ) ) ∣ ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ 𝐵 ( 𝑝 ∥ ( 𝑥 · 𝑦 ) → ( 𝑝 ∥ 𝑥 ∨ 𝑝 ∥ 𝑦 ) ) } ↔ ( 𝑃 ∈ ( 𝐵 ∖ ( 𝑈 ∪ { 0 } ) ) ∧ ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ 𝐵 ( 𝑃 ∥ ( 𝑥 · 𝑦 ) → ( 𝑃 ∥ 𝑥 ∨ 𝑃 ∥ 𝑦 ) ) ) )
15 7 14 bitrdi ⊢ ( 𝑅 ∈ 𝑉 → ( 𝑃 ∈ ( RPrime ‘ 𝑅 ) ↔ ( 𝑃 ∈ ( 𝐵 ∖ ( 𝑈 ∪ { 0 } ) ) ∧ ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ 𝐵 ( 𝑃 ∥ ( 𝑥 · 𝑦 ) → ( 𝑃 ∥ 𝑥 ∨ 𝑃 ∥ 𝑦 ) ) ) ) )