Metamath Proof Explorer


Theorem zrhpropd

Description: The ZZ ring homomorphism depends only on the ring attributes of a structure. (Contributed by Mario Carneiro, 15-Jun-2015)

Ref Expression
Hypotheses zrhpropd.1 ( 𝜑𝐵 = ( Base ‘ 𝐾 ) )
zrhpropd.2 ( 𝜑𝐵 = ( Base ‘ 𝐿 ) )
zrhpropd.3 ( ( 𝜑 ∧ ( 𝑥𝐵𝑦𝐵 ) ) → ( 𝑥 ( +g𝐾 ) 𝑦 ) = ( 𝑥 ( +g𝐿 ) 𝑦 ) )
zrhpropd.4 ( ( 𝜑 ∧ ( 𝑥𝐵𝑦𝐵 ) ) → ( 𝑥 ( .r𝐾 ) 𝑦 ) = ( 𝑥 ( .r𝐿 ) 𝑦 ) )
Assertion zrhpropd ( 𝜑 → ( ℤRHom ‘ 𝐾 ) = ( ℤRHom ‘ 𝐿 ) )

Proof

Step Hyp Ref Expression
1 zrhpropd.1 ( 𝜑𝐵 = ( Base ‘ 𝐾 ) )
2 zrhpropd.2 ( 𝜑𝐵 = ( Base ‘ 𝐿 ) )
3 zrhpropd.3 ( ( 𝜑 ∧ ( 𝑥𝐵𝑦𝐵 ) ) → ( 𝑥 ( +g𝐾 ) 𝑦 ) = ( 𝑥 ( +g𝐿 ) 𝑦 ) )
4 zrhpropd.4 ( ( 𝜑 ∧ ( 𝑥𝐵𝑦𝐵 ) ) → ( 𝑥 ( .r𝐾 ) 𝑦 ) = ( 𝑥 ( .r𝐿 ) 𝑦 ) )
5 eqidd ( 𝜑 → ( Base ‘ ℤring ) = ( Base ‘ ℤring ) )
6 eqidd ( ( 𝜑 ∧ ( 𝑥 ∈ ( Base ‘ ℤring ) ∧ 𝑦 ∈ ( Base ‘ ℤring ) ) ) → ( 𝑥 ( +g ‘ ℤring ) 𝑦 ) = ( 𝑥 ( +g ‘ ℤring ) 𝑦 ) )
7 eqidd ( ( 𝜑 ∧ ( 𝑥 ∈ ( Base ‘ ℤring ) ∧ 𝑦 ∈ ( Base ‘ ℤring ) ) ) → ( 𝑥 ( .r ‘ ℤring ) 𝑦 ) = ( 𝑥 ( .r ‘ ℤring ) 𝑦 ) )
8 5 1 5 2 6 3 7 4 rhmpropd ( 𝜑 → ( ℤring RingHom 𝐾 ) = ( ℤring RingHom 𝐿 ) )
9 8 unieqd ( 𝜑 ( ℤring RingHom 𝐾 ) = ( ℤring RingHom 𝐿 ) )
10 eqid ( ℤRHom ‘ 𝐾 ) = ( ℤRHom ‘ 𝐾 )
11 10 zrhval ( ℤRHom ‘ 𝐾 ) = ( ℤring RingHom 𝐾 )
12 eqid ( ℤRHom ‘ 𝐿 ) = ( ℤRHom ‘ 𝐿 )
13 12 zrhval ( ℤRHom ‘ 𝐿 ) = ( ℤring RingHom 𝐿 )
14 9 11 13 3eqtr4g ( 𝜑 → ( ℤRHom ‘ 𝐾 ) = ( ℤRHom ‘ 𝐿 ) )