Metamath Proof Explorer


Theorem reuf1od

Description: There is exactly one element in each of two isomorphic sets. (Contributed by AV, 19-Mar-2023)

Ref Expression
Hypotheses reuf1od.f ( 𝜑𝐹 : 𝐶1-1-onto𝐵 )
reuf1od.x ( ( 𝜑𝑥 = ( 𝐹𝑦 ) ) → ( 𝜓𝜒 ) )
Assertion reuf1od ( 𝜑 → ( ∃! 𝑥𝐵 𝜓 ↔ ∃! 𝑦𝐶 𝜒 ) )

Proof

Step Hyp Ref Expression
1 reuf1od.f ( 𝜑𝐹 : 𝐶1-1-onto𝐵 )
2 reuf1od.x ( ( 𝜑𝑥 = ( 𝐹𝑦 ) ) → ( 𝜓𝜒 ) )
3 f1of ( 𝐹 : 𝐶1-1-onto𝐵𝐹 : 𝐶𝐵 )
4 1 3 syl ( 𝜑𝐹 : 𝐶𝐵 )
5 4 ffvelrnda ( ( 𝜑𝑦𝐶 ) → ( 𝐹𝑦 ) ∈ 𝐵 )
6 f1ofveu ( ( 𝐹 : 𝐶1-1-onto𝐵𝑥𝐵 ) → ∃! 𝑦𝐶 ( 𝐹𝑦 ) = 𝑥 )
7 eqcom ( 𝑥 = ( 𝐹𝑦 ) ↔ ( 𝐹𝑦 ) = 𝑥 )
8 7 reubii ( ∃! 𝑦𝐶 𝑥 = ( 𝐹𝑦 ) ↔ ∃! 𝑦𝐶 ( 𝐹𝑦 ) = 𝑥 )
9 6 8 sylibr ( ( 𝐹 : 𝐶1-1-onto𝐵𝑥𝐵 ) → ∃! 𝑦𝐶 𝑥 = ( 𝐹𝑦 ) )
10 1 9 sylan ( ( 𝜑𝑥𝐵 ) → ∃! 𝑦𝐶 𝑥 = ( 𝐹𝑦 ) )
11 5 10 2 reuxfr1d ( 𝜑 → ( ∃! 𝑥𝐵 𝜓 ↔ ∃! 𝑦𝐶 𝜒 ) )