# Metamath Proof Explorer

## Theorem 0ringdif

Description: A zero ring is a ring which is not a nonzero ring. (Contributed by AV, 17-Apr-2020)

Ref Expression
Hypotheses 0ringdif.b 𝐵 = ( Base ‘ 𝑅 )
0ringdif.0 0 = ( 0g𝑅 )
Assertion 0ringdif ( 𝑅 ∈ ( Ring ∖ NzRing ) ↔ ( 𝑅 ∈ Ring ∧ 𝐵 = { 0 } ) )

### Proof

Step Hyp Ref Expression
1 0ringdif.b 𝐵 = ( Base ‘ 𝑅 )
2 0ringdif.0 0 = ( 0g𝑅 )
3 eldif ( 𝑅 ∈ ( Ring ∖ NzRing ) ↔ ( 𝑅 ∈ Ring ∧ ¬ 𝑅 ∈ NzRing ) )
4 1 a1i ( 𝑅 ∈ Ring → 𝐵 = ( Base ‘ 𝑅 ) )
5 4 fveqeq2d ( 𝑅 ∈ Ring → ( ( ♯ ‘ 𝐵 ) = 1 ↔ ( ♯ ‘ ( Base ‘ 𝑅 ) ) = 1 ) )
6 1 2 0ring ( ( 𝑅 ∈ Ring ∧ ( ♯ ‘ 𝐵 ) = 1 ) → 𝐵 = { 0 } )
7 6 ex ( 𝑅 ∈ Ring → ( ( ♯ ‘ 𝐵 ) = 1 → 𝐵 = { 0 } ) )
8 fveq2 ( 𝐵 = { 0 } → ( ♯ ‘ 𝐵 ) = ( ♯ ‘ { 0 } ) )
9 2 fvexi 0 ∈ V
10 hashsng ( 0 ∈ V → ( ♯ ‘ { 0 } ) = 1 )
11 9 10 ax-mp ( ♯ ‘ { 0 } ) = 1
12 8 11 eqtrdi ( 𝐵 = { 0 } → ( ♯ ‘ 𝐵 ) = 1 )
13 7 12 impbid1 ( 𝑅 ∈ Ring → ( ( ♯ ‘ 𝐵 ) = 1 ↔ 𝐵 = { 0 } ) )
14 0ringnnzr ( 𝑅 ∈ Ring → ( ( ♯ ‘ ( Base ‘ 𝑅 ) ) = 1 ↔ ¬ 𝑅 ∈ NzRing ) )
15 5 13 14 3bitr3rd ( 𝑅 ∈ Ring → ( ¬ 𝑅 ∈ NzRing ↔ 𝐵 = { 0 } ) )
16 15 pm5.32i ( ( 𝑅 ∈ Ring ∧ ¬ 𝑅 ∈ NzRing ) ↔ ( 𝑅 ∈ Ring ∧ 𝐵 = { 0 } ) )
17 3 16 bitri ( 𝑅 ∈ ( Ring ∖ NzRing ) ↔ ( 𝑅 ∈ Ring ∧ 𝐵 = { 0 } ) )