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Metamath Proof Explorer

Theorem nzrpropd

Description: If two structures have the same components (properties), one is a nonzero ring iff the other one is. (Contributed by SN, 21-Jun-2025)

Ref Expression
Hypotheses nzrpropd.1 ⊢ ( 𝜑 → 𝐵 = ( Base ‘ 𝐾 ) )
nzrpropd.2 ⊢ ( 𝜑 → 𝐵 = ( Base ‘ 𝐿 ) )
nzrpropd.3 ⊢ ( ( 𝜑 ∧ ( 𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐵 ) ) → ( 𝑥 ( +g ‘ 𝐾 ) 𝑦 ) = ( 𝑥 ( +g ‘ 𝐿 ) 𝑦 ) )
nzrpropd.4 ⊢ ( ( 𝜑 ∧ ( 𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐵 ) ) → ( 𝑥 ( .r ‘ 𝐾 ) 𝑦 ) = ( 𝑥 ( .r ‘ 𝐿 ) 𝑦 ) )
Assertion nzrpropd ( 𝜑 → ( 𝐾 ∈ NzRing ↔ 𝐿 ∈ NzRing ) )

Proof

Step Hyp Ref Expression
1 nzrpropd.1 ⊢ ( 𝜑 → 𝐵 = ( Base ‘ 𝐾 ) )
2 nzrpropd.2 ⊢ ( 𝜑 → 𝐵 = ( Base ‘ 𝐿 ) )
3 nzrpropd.3 ⊢ ( ( 𝜑 ∧ ( 𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐵 ) ) → ( 𝑥 ( +g ‘ 𝐾 ) 𝑦 ) = ( 𝑥 ( +g ‘ 𝐿 ) 𝑦 ) )
4 nzrpropd.4 ⊢ ( ( 𝜑 ∧ ( 𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐵 ) ) → ( 𝑥 ( .r ‘ 𝐾 ) 𝑦 ) = ( 𝑥 ( .r ‘ 𝐿 ) 𝑦 ) )
5 1 2 3 4 ringpropd ⊢ ( 𝜑 → ( 𝐾 ∈ Ring ↔ 𝐿 ∈ Ring ) )
6 1 2 4 rngidpropd ⊢ ( 𝜑 → ( 1r ‘ 𝐾 ) = ( 1r ‘ 𝐿 ) )
7 1 2 3 grpidpropd ⊢ ( 𝜑 → ( 0g ‘ 𝐾 ) = ( 0g ‘ 𝐿 ) )
8 6 7 neeq12d ⊢ ( 𝜑 → ( ( 1r ‘ 𝐾 ) ≠ ( 0g ‘ 𝐾 ) ↔ ( 1r ‘ 𝐿 ) ≠ ( 0g ‘ 𝐿 ) ) )
9 5 8 anbi12d ⊢ ( 𝜑 → ( ( 𝐾 ∈ Ring ∧ ( 1r ‘ 𝐾 ) ≠ ( 0g ‘ 𝐾 ) ) ↔ ( 𝐿 ∈ Ring ∧ ( 1r ‘ 𝐿 ) ≠ ( 0g ‘ 𝐿 ) ) ) )
10 eqid ⊢ ( 1r ‘ 𝐾 ) = ( 1r ‘ 𝐾 )
11 eqid ⊢ ( 0g ‘ 𝐾 ) = ( 0g ‘ 𝐾 )
12 10 11 isnzr ⊢ ( 𝐾 ∈ NzRing ↔ ( 𝐾 ∈ Ring ∧ ( 1r ‘ 𝐾 ) ≠ ( 0g ‘ 𝐾 ) ) )
13 eqid ⊢ ( 1r ‘ 𝐿 ) = ( 1r ‘ 𝐿 )
14 eqid ⊢ ( 0g ‘ 𝐿 ) = ( 0g ‘ 𝐿 )
15 13 14 isnzr ⊢ ( 𝐿 ∈ NzRing ↔ ( 𝐿 ∈ Ring ∧ ( 1r ‘ 𝐿 ) ≠ ( 0g ‘ 𝐿 ) ) )
16 9 12 15 3bitr4g ⊢ ( 𝜑 → ( 𝐾 ∈ NzRing ↔ 𝐿 ∈ NzRing ) )