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 ⊢ ( 𝜑 → 𝐹 : 𝐶 –1-1-onto→ 𝐵 )
reuf1od.x ⊢ ( ( 𝜑 ∧ 𝑥 = ( 𝐹 ‘ 𝑦 ) ) → ( 𝜓 ↔ 𝜒 ) )
Assertion reuf1od ( 𝜑 → ( ∃! 𝑥 ∈ 𝐵 𝜓 ↔ ∃! 𝑦 ∈ 𝐶 𝜒 ) )

Proof

Step Hyp Ref Expression
1 reuf1od.f ⊢ ( 𝜑 → 𝐹 : 𝐶 –1-1-onto→ 𝐵 )
2 reuf1od.x ⊢ ( ( 𝜑 ∧ 𝑥 = ( 𝐹 ‘ 𝑦 ) ) → ( 𝜓 ↔ 𝜒 ) )
3 f1of ⊢ ( 𝐹 : 𝐶 –1-1-onto→ 𝐵 → 𝐹 : 𝐶 ⟶ 𝐵 )
4 1 3 syl ⊢ ( 𝜑 → 𝐹 : 𝐶 ⟶ 𝐵 )
5 4 ffvelcdmda ⊢ ( ( 𝜑 ∧ 𝑦 ∈ 𝐶 ) → ( 𝐹 ‘ 𝑦 ) ∈ 𝐵 )
6 f1ofveu ⊢ ( ( 𝐹 : 𝐶 –1-1-onto→ 𝐵 ∧ 𝑥 ∈ 𝐵 ) → ∃! 𝑦 ∈ 𝐶 ( 𝐹 ‘ 𝑦 ) = 𝑥 )
7 eqcom ⊢ ( 𝑥 = ( 𝐹 ‘ 𝑦 ) ↔ ( 𝐹 ‘ 𝑦 ) = 𝑥 )
8 7 reubii ⊢ ( ∃! 𝑦 ∈ 𝐶 𝑥 = ( 𝐹 ‘ 𝑦 ) ↔ ∃! 𝑦 ∈ 𝐶 ( 𝐹 ‘ 𝑦 ) = 𝑥 )
9 6 8 sylibr ⊢ ( ( 𝐹 : 𝐶 –1-1-onto→ 𝐵 ∧ 𝑥 ∈ 𝐵 ) → ∃! 𝑦 ∈ 𝐶 𝑥 = ( 𝐹 ‘ 𝑦 ) )
10 1 9 sylan ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐵 ) → ∃! 𝑦 ∈ 𝐶 𝑥 = ( 𝐹 ‘ 𝑦 ) )
11 5 10 2 reuxfr1d ⊢ ( 𝜑 → ( ∃! 𝑥 ∈ 𝐵 𝜓 ↔ ∃! 𝑦 ∈ 𝐶 𝜒 ) )