Metamath Proof Explorer


Theorem wessf1orn

Description: Given a function F on a well-ordered domain A there exists a subset of A such that F restricted to such subset is injective and onto the range of F (without using the axiom of choice). (Contributed by Glauco Siliprandi, 17-Aug-2020)

Ref Expression
Hypotheses wessf1orn.f ⊢ ( 𝜑 → 𝐹 Fn 𝐴 )
wessf1orn.a ⊢ ( 𝜑 → 𝐴 ∈ 𝑉 )
wessf1orn.r ⊢ ( 𝜑 → 𝑅 We 𝐴 )
Assertion wessf1orn ( 𝜑 → ∃ 𝑥 ∈ 𝒫 𝐴 ( 𝐹 ↾ 𝑥 ) : 𝑥 –1-1-onto→ ran 𝐹 )

Proof

Step Hyp Ref Expression
1 wessf1orn.f ⊢ ( 𝜑 → 𝐹 Fn 𝐴 )
2 wessf1orn.a ⊢ ( 𝜑 → 𝐴 ∈ 𝑉 )
3 wessf1orn.r ⊢ ( 𝜑 → 𝑅 We 𝐴 )
4 eqid ⊢ ( 𝑦 ∈ ran 𝐹 ↦ ( ℩ 𝑥 ∈ ( ◡ 𝐹 “ { 𝑦 } ) ∀ 𝑧 ∈ ( ◡ 𝐹 “ { 𝑦 } ) ¬ 𝑧 𝑅 𝑥 ) ) = ( 𝑦 ∈ ran 𝐹 ↦ ( ℩ 𝑥 ∈ ( ◡ 𝐹 “ { 𝑦 } ) ∀ 𝑧 ∈ ( ◡ 𝐹 “ { 𝑦 } ) ¬ 𝑧 𝑅 𝑥 ) )
5 1 2 3 4 wessf1ornlem ⊢ ( 𝜑 → ∃ 𝑥 ∈ 𝒫 𝐴 ( 𝐹 ↾ 𝑥 ) : 𝑥 –1-1-onto→ ran 𝐹 )