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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 ⊢ ( 𝜑 → 𝐵 = ( Base ‘ 𝐾 ) )
zrhpropd.2 ⊢ ( 𝜑 → 𝐵 = ( Base ‘ 𝐿 ) )
zrhpropd.3 ⊢ ( ( 𝜑 ∧ ( 𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐵 ) ) → ( 𝑥 ( +g ‘ 𝐾 ) 𝑦 ) = ( 𝑥 ( +g ‘ 𝐿 ) 𝑦 ) )
zrhpropd.4 ⊢ ( ( 𝜑 ∧ ( 𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐵 ) ) → ( 𝑥 ( .r ‘ 𝐾 ) 𝑦 ) = ( 𝑥 ( .r ‘ 𝐿 ) 𝑦 ) )
Assertion zrhpropd ( 𝜑 → ( ℤRHom ‘ 𝐾 ) = ( ℤRHom ‘ 𝐿 ) )

Proof

Step Hyp Ref Expression
1 zrhpropd.1 ⊢ ( 𝜑 → 𝐵 = ( Base ‘ 𝐾 ) )
2 zrhpropd.2 ⊢ ( 𝜑 → 𝐵 = ( Base ‘ 𝐿 ) )
3 zrhpropd.3 ⊢ ( ( 𝜑 ∧ ( 𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐵 ) ) → ( 𝑥 ( +g ‘ 𝐾 ) 𝑦 ) = ( 𝑥 ( +g ‘ 𝐿 ) 𝑦 ) )
4 zrhpropd.4 ⊢ ( ( 𝜑 ∧ ( 𝑥 ∈ 𝐵 ∧ 𝑦 ∈ 𝐵 ) ) → ( 𝑥 ( .r ‘ 𝐾 ) 𝑦 ) = ( 𝑥 ( .r ‘ 𝐿 ) 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 ) )