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Metamath Proof Explorer

Theorem isdomn2

Description: A ring is a domain iff all nonzero elements are regular elements. (Contributed by Mario Carneiro, 28-Mar-2015) (Proof shortened by SN, 21-Jun-2025)

Ref Expression
Hypotheses isdomn2.b ⊢ 𝐵 = ( Base ‘ 𝑅 )
isdomn2.t ⊢ 𝐸 = ( RLReg ‘ 𝑅 )
isdomn2.z ⊢ 0 = ( 0g ‘ 𝑅 )
Assertion isdomn2 ( 𝑅 ∈ Domn ↔ ( 𝑅 ∈ NzRing ∧ ( 𝐵 ∖ { 0 } ) ⊆ 𝐸 ) )

Proof

Step Hyp Ref Expression
1 isdomn2.b ⊢ 𝐵 = ( Base ‘ 𝑅 )
2 isdomn2.t ⊢ 𝐸 = ( RLReg ‘ 𝑅 )
3 isdomn2.z ⊢ 0 = ( 0g ‘ 𝑅 )
4 eqid ⊢ ( .r ‘ 𝑅 ) = ( .r ‘ 𝑅 )
5 1 4 3 isdomn ⊢ ( 𝑅 ∈ Domn ↔ ( 𝑅 ∈ NzRing ∧ ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ 𝐵 ( ( 𝑥 ( .r ‘ 𝑅 ) 𝑦 ) = 0 → ( 𝑥 = 0 ∨ 𝑦 = 0 ) ) ) )
6 eldifi ⊢ ( 𝑥 ∈ ( 𝐵 ∖ { 0 } ) → 𝑥 ∈ 𝐵 )
7 2 1 4 3 isrrg ⊢ ( 𝑥 ∈ 𝐸 ↔ ( 𝑥 ∈ 𝐵 ∧ ∀ 𝑦 ∈ 𝐵 ( ( 𝑥 ( .r ‘ 𝑅 ) 𝑦 ) = 0 → 𝑦 = 0 ) ) )
8 7 baib ⊢ ( 𝑥 ∈ 𝐵 → ( 𝑥 ∈ 𝐸 ↔ ∀ 𝑦 ∈ 𝐵 ( ( 𝑥 ( .r ‘ 𝑅 ) 𝑦 ) = 0 → 𝑦 = 0 ) ) )
9 6 8 syl ⊢ ( 𝑥 ∈ ( 𝐵 ∖ { 0 } ) → ( 𝑥 ∈ 𝐸 ↔ ∀ 𝑦 ∈ 𝐵 ( ( 𝑥 ( .r ‘ 𝑅 ) 𝑦 ) = 0 → 𝑦 = 0 ) ) )
10 9 ralbiia ⊢ ( ∀ 𝑥 ∈ ( 𝐵 ∖ { 0 } ) 𝑥 ∈ 𝐸 ↔ ∀ 𝑥 ∈ ( 𝐵 ∖ { 0 } ) ∀ 𝑦 ∈ 𝐵 ( ( 𝑥 ( .r ‘ 𝑅 ) 𝑦 ) = 0 → 𝑦 = 0 ) )
11 dfss3 ⊢ ( ( 𝐵 ∖ { 0 } ) ⊆ 𝐸 ↔ ∀ 𝑥 ∈ ( 𝐵 ∖ { 0 } ) 𝑥 ∈ 𝐸 )
12 isdomn5 ⊢ ( ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ 𝐵 ( ( 𝑥 ( .r ‘ 𝑅 ) 𝑦 ) = 0 → ( 𝑥 = 0 ∨ 𝑦 = 0 ) ) ↔ ∀ 𝑥 ∈ ( 𝐵 ∖ { 0 } ) ∀ 𝑦 ∈ 𝐵 ( ( 𝑥 ( .r ‘ 𝑅 ) 𝑦 ) = 0 → 𝑦 = 0 ) )
13 10 11 12 3bitr4ri ⊢ ( ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ 𝐵 ( ( 𝑥 ( .r ‘ 𝑅 ) 𝑦 ) = 0 → ( 𝑥 = 0 ∨ 𝑦 = 0 ) ) ↔ ( 𝐵 ∖ { 0 } ) ⊆ 𝐸 )
14 13 anbi2i ⊢ ( ( 𝑅 ∈ NzRing ∧ ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ 𝐵 ( ( 𝑥 ( .r ‘ 𝑅 ) 𝑦 ) = 0 → ( 𝑥 = 0 ∨ 𝑦 = 0 ) ) ) ↔ ( 𝑅 ∈ NzRing ∧ ( 𝐵 ∖ { 0 } ) ⊆ 𝐸 ) )
15 5 14 bitri ⊢ ( 𝑅 ∈ Domn ↔ ( 𝑅 ∈ NzRing ∧ ( 𝐵 ∖ { 0 } ) ⊆ 𝐸 ) )