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Metamath Proof Explorer

Theorem isrim

Description: An isomorphism of rings is a bijective homomorphism. (Contributed by AV, 22-Oct-2019) Remove sethood antecedent. (Revised by SN, 12-Jan-2025)

Ref Expression
Hypotheses rhmf1o.b ⊢ 𝐵 = ( Base ‘ 𝑅 )
rhmf1o.c ⊢ 𝐶 = ( Base ‘ 𝑆 )
Assertion isrim ( 𝐹 ∈ ( 𝑅 RingIso 𝑆 ) ↔ ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ 𝐹 : 𝐵 –1-1-onto→ 𝐶 ) )

Proof

Step Hyp Ref Expression
1 rhmf1o.b ⊢ 𝐵 = ( Base ‘ 𝑅 )
2 rhmf1o.c ⊢ 𝐶 = ( Base ‘ 𝑆 )
3 isrim0 ⊢ ( 𝐹 ∈ ( 𝑅 RingIso 𝑆 ) ↔ ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ ◡ 𝐹 ∈ ( 𝑆 RingHom 𝑅 ) ) )
4 1 2 rhmf1o ⊢ ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) → ( 𝐹 : 𝐵 –1-1-onto→ 𝐶 ↔ ◡ 𝐹 ∈ ( 𝑆 RingHom 𝑅 ) ) )
5 4 bicomd ⊢ ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) → ( ◡ 𝐹 ∈ ( 𝑆 RingHom 𝑅 ) ↔ 𝐹 : 𝐵 –1-1-onto→ 𝐶 ) )
6 5 pm5.32i ⊢ ( ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ ◡ 𝐹 ∈ ( 𝑆 RingHom 𝑅 ) ) ↔ ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ 𝐹 : 𝐵 –1-1-onto→ 𝐶 ) )
7 3 6 bitri ⊢ ( 𝐹 ∈ ( 𝑅 RingIso 𝑆 ) ↔ ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ 𝐹 : 𝐵 –1-1-onto→ 𝐶 ) )