Metamath Proof Explorer


Theorem isrim

Description: An isomorphism of rings is a bijective homomorphism. (Contributed by AV, 22-Oct-2019)

Ref Expression
Hypotheses rhmf1o.b 𝐵 = ( Base ‘ 𝑅 )
rhmf1o.c 𝐶 = ( Base ‘ 𝑆 )
Assertion isrim ( ( 𝑅𝑉𝑆𝑊 ) → ( 𝐹 ∈ ( 𝑅 RingIso 𝑆 ) ↔ ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ 𝐹 : 𝐵1-1-onto𝐶 ) ) )

Proof

Step Hyp Ref Expression
1 rhmf1o.b 𝐵 = ( Base ‘ 𝑅 )
2 rhmf1o.c 𝐶 = ( Base ‘ 𝑆 )
3 isrim0 ( ( 𝑅𝑉𝑆𝑊 ) → ( 𝐹 ∈ ( 𝑅 RingIso 𝑆 ) ↔ ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ 𝐹 ∈ ( 𝑆 RingHom 𝑅 ) ) ) )
4 1 2 rhmf1o ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) → ( 𝐹 : 𝐵1-1-onto𝐶 𝐹 ∈ ( 𝑆 RingHom 𝑅 ) ) )
5 4 bicomd ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) → ( 𝐹 ∈ ( 𝑆 RingHom 𝑅 ) ↔ 𝐹 : 𝐵1-1-onto𝐶 ) )
6 5 a1i ( ( 𝑅𝑉𝑆𝑊 ) → ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) → ( 𝐹 ∈ ( 𝑆 RingHom 𝑅 ) ↔ 𝐹 : 𝐵1-1-onto𝐶 ) ) )
7 6 pm5.32d ( ( 𝑅𝑉𝑆𝑊 ) → ( ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ 𝐹 ∈ ( 𝑆 RingHom 𝑅 ) ) ↔ ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ 𝐹 : 𝐵1-1-onto𝐶 ) ) )
8 3 7 bitrd ( ( 𝑅𝑉𝑆𝑊 ) → ( 𝐹 ∈ ( 𝑅 RingIso 𝑆 ) ↔ ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ 𝐹 : 𝐵1-1-onto𝐶 ) ) )