Metamath Proof Explorer


Theorem pzriprnglem7

Description: Lemma 7 for pzriprng : J is a unital ring. (Contributed by AV, 19-Mar-2025)

Ref Expression
Hypotheses pzriprng.r ⊢ 𝑅 = ( ℤring ×s ℤring )
pzriprng.i ⊢ 𝐼 = ( ℤ × { 0 } )
pzriprng.j ⊢ 𝐽 = ( 𝑅 ↾s 𝐼 )
Assertion pzriprnglem7 𝐽 ∈ Ring

Proof

Step Hyp Ref Expression
1 pzriprng.r ⊢ 𝑅 = ( ℤring ×s ℤring )
2 pzriprng.i ⊢ 𝐼 = ( ℤ × { 0 } )
3 pzriprng.j ⊢ 𝐽 = ( 𝑅 ↾s 𝐼 )
4 1 2 pzriprnglem5 ⊢ 𝐼 ∈ ( SubRng ‘ 𝑅 )
5 3 subrngrng ⊢ ( 𝐼 ∈ ( SubRng ‘ 𝑅 ) → 𝐽 ∈ Rng )
6 4 5 ax-mp ⊢ 𝐽 ∈ Rng
7 1z ⊢ 1 ∈ ℤ
8 c0ex ⊢ 0 ∈ V
9 8 snid ⊢ 0 ∈ { 0 }
10 7 9 opelxpii ⊢ ⟨ 1 , 0 ⟩ ∈ ( ℤ × { 0 } )
11 3 subrngbas ⊢ ( 𝐼 ∈ ( SubRng ‘ 𝑅 ) → 𝐼 = ( Base ‘ 𝐽 ) )
12 4 11 ax-mp ⊢ 𝐼 = ( Base ‘ 𝐽 )
13 12 2 eqtr3i ⊢ ( Base ‘ 𝐽 ) = ( ℤ × { 0 } )
14 10 13 eleqtrri ⊢ ⟨ 1 , 0 ⟩ ∈ ( Base ‘ 𝐽 )
15 oveq1 ⊢ ( 𝑖 = ⟨ 1 , 0 ⟩ → ( 𝑖 ( .r ‘ 𝐽 ) 𝑥 ) = ( ⟨ 1 , 0 ⟩ ( .r ‘ 𝐽 ) 𝑥 ) )
16 15 eqeq1d ⊢ ( 𝑖 = ⟨ 1 , 0 ⟩ → ( ( 𝑖 ( .r ‘ 𝐽 ) 𝑥 ) = 𝑥 ↔ ( ⟨ 1 , 0 ⟩ ( .r ‘ 𝐽 ) 𝑥 ) = 𝑥 ) )
17 16 ovanraleqv ⊢ ( 𝑖 = ⟨ 1 , 0 ⟩ → ( ∀ 𝑥 ∈ ( Base ‘ 𝐽 ) ( ( 𝑖 ( .r ‘ 𝐽 ) 𝑥 ) = 𝑥 ∧ ( 𝑥 ( .r ‘ 𝐽 ) 𝑖 ) = 𝑥 ) ↔ ∀ 𝑥 ∈ ( Base ‘ 𝐽 ) ( ( ⟨ 1 , 0 ⟩ ( .r ‘ 𝐽 ) 𝑥 ) = 𝑥 ∧ ( 𝑥 ( .r ‘ 𝐽 ) ⟨ 1 , 0 ⟩ ) = 𝑥 ) ) )
18 id ⊢ ( ⟨ 1 , 0 ⟩ ∈ ( Base ‘ 𝐽 ) → ⟨ 1 , 0 ⟩ ∈ ( Base ‘ 𝐽 ) )
19 12 eleq2i ⊢ ( 𝑥 ∈ 𝐼 ↔ 𝑥 ∈ ( Base ‘ 𝐽 ) )
20 1 2 3 pzriprnglem6 ⊢ ( 𝑥 ∈ 𝐼 → ( ( ⟨ 1 , 0 ⟩ ( .r ‘ 𝐽 ) 𝑥 ) = 𝑥 ∧ ( 𝑥 ( .r ‘ 𝐽 ) ⟨ 1 , 0 ⟩ ) = 𝑥 ) )
21 19 20 sylbir ⊢ ( 𝑥 ∈ ( Base ‘ 𝐽 ) → ( ( ⟨ 1 , 0 ⟩ ( .r ‘ 𝐽 ) 𝑥 ) = 𝑥 ∧ ( 𝑥 ( .r ‘ 𝐽 ) ⟨ 1 , 0 ⟩ ) = 𝑥 ) )
22 21 a1i ⊢ ( ⟨ 1 , 0 ⟩ ∈ ( Base ‘ 𝐽 ) → ( 𝑥 ∈ ( Base ‘ 𝐽 ) → ( ( ⟨ 1 , 0 ⟩ ( .r ‘ 𝐽 ) 𝑥 ) = 𝑥 ∧ ( 𝑥 ( .r ‘ 𝐽 ) ⟨ 1 , 0 ⟩ ) = 𝑥 ) ) )
23 22 ralrimiv ⊢ ( ⟨ 1 , 0 ⟩ ∈ ( Base ‘ 𝐽 ) → ∀ 𝑥 ∈ ( Base ‘ 𝐽 ) ( ( ⟨ 1 , 0 ⟩ ( .r ‘ 𝐽 ) 𝑥 ) = 𝑥 ∧ ( 𝑥 ( .r ‘ 𝐽 ) ⟨ 1 , 0 ⟩ ) = 𝑥 ) )
24 17 18 23 rspcedvdw ⊢ ( ⟨ 1 , 0 ⟩ ∈ ( Base ‘ 𝐽 ) → ∃ 𝑖 ∈ ( Base ‘ 𝐽 ) ∀ 𝑥 ∈ ( Base ‘ 𝐽 ) ( ( 𝑖 ( .r ‘ 𝐽 ) 𝑥 ) = 𝑥 ∧ ( 𝑥 ( .r ‘ 𝐽 ) 𝑖 ) = 𝑥 ) )
25 14 24 ax-mp ⊢ ∃ 𝑖 ∈ ( Base ‘ 𝐽 ) ∀ 𝑥 ∈ ( Base ‘ 𝐽 ) ( ( 𝑖 ( .r ‘ 𝐽 ) 𝑥 ) = 𝑥 ∧ ( 𝑥 ( .r ‘ 𝐽 ) 𝑖 ) = 𝑥 )
26 eqid ⊢ ( Base ‘ 𝐽 ) = ( Base ‘ 𝐽 )
27 eqid ⊢ ( .r ‘ 𝐽 ) = ( .r ‘ 𝐽 )
28 26 27 isringrng ⊢ ( 𝐽 ∈ Ring ↔ ( 𝐽 ∈ Rng ∧ ∃ 𝑖 ∈ ( Base ‘ 𝐽 ) ∀ 𝑥 ∈ ( Base ‘ 𝐽 ) ( ( 𝑖 ( .r ‘ 𝐽 ) 𝑥 ) = 𝑥 ∧ ( 𝑥 ( .r ‘ 𝐽 ) 𝑖 ) = 𝑥 ) ) )
29 6 25 28 mpbir2an ⊢ 𝐽 ∈ Ring