Theorem fprfung

Description: A "function" defined by well-founded recursion is indeed a function when the relationship is a partial order. Avoids the axiom of replacement. (Contributed by Scott Fenton, 18-Nov-2024)

	Ref	Expression
Hypothesis	fprfung.1	⊢ 𝐹 = frecs ( 𝑅 , 𝐴 , 𝐺 )
Assertion	fprfung	⊢ ( ( 𝑅 Fr 𝐴 ∧ 𝑅 Po 𝐴 ∧ 𝑅 Se 𝐴 ) → Fun 𝐹 )

Proof

Step	Hyp	Ref	Expression
1		fprfung.1	⊢ 𝐹 = frecs ( 𝑅 , 𝐴 , 𝐺 )
2		eqid	⊢ { 𝑓 ∣ ∃ 𝑥 ( 𝑓 Fn 𝑥 ∧ ( 𝑥 ⊆ 𝐴 ∧ ∀ 𝑦 ∈ 𝑥 Pred ( 𝑅 , 𝐴 , 𝑦 ) ⊆ 𝑥 ) ∧ ∀ 𝑦 ∈ 𝑥 ( 𝑓 ‘ 𝑦 ) = ( 𝑦 𝐺 ( 𝑓 ↾ Pred ( 𝑅 , 𝐴 , 𝑦 ) ) ) ) } = { 𝑓 ∣ ∃ 𝑥 ( 𝑓 Fn 𝑥 ∧ ( 𝑥 ⊆ 𝐴 ∧ ∀ 𝑦 ∈ 𝑥 Pred ( 𝑅 , 𝐴 , 𝑦 ) ⊆ 𝑥 ) ∧ ∀ 𝑦 ∈ 𝑥 ( 𝑓 ‘ 𝑦 ) = ( 𝑦 𝐺 ( 𝑓 ↾ Pred ( 𝑅 , 𝐴 , 𝑦 ) ) ) ) }
3	2 1	fprlem1	⊢ ( ( ( 𝑅 Fr 𝐴 ∧ 𝑅 Po 𝐴 ∧ 𝑅 Se 𝐴 ) ∧ ( 𝑔 ∈ { 𝑓 ∣ ∃ 𝑥 ( 𝑓 Fn 𝑥 ∧ ( 𝑥 ⊆ 𝐴 ∧ ∀ 𝑦 ∈ 𝑥 Pred ( 𝑅 , 𝐴 , 𝑦 ) ⊆ 𝑥 ) ∧ ∀ 𝑦 ∈ 𝑥 ( 𝑓 ‘ 𝑦 ) = ( 𝑦 𝐺 ( 𝑓 ↾ Pred ( 𝑅 , 𝐴 , 𝑦 ) ) ) ) } ∧ ℎ ∈ { 𝑓 ∣ ∃ 𝑥 ( 𝑓 Fn 𝑥 ∧ ( 𝑥 ⊆ 𝐴 ∧ ∀ 𝑦 ∈ 𝑥 Pred ( 𝑅 , 𝐴 , 𝑦 ) ⊆ 𝑥 ) ∧ ∀ 𝑦 ∈ 𝑥 ( 𝑓 ‘ 𝑦 ) = ( 𝑦 𝐺 ( 𝑓 ↾ Pred ( 𝑅 , 𝐴 , 𝑦 ) ) ) ) } ) ) → ( ( 𝑥 𝑔 𝑢 ∧ 𝑥 ℎ 𝑣 ) → 𝑢 = 𝑣 ) )
4	2 1 3	frrlem9	⊢ ( ( 𝑅 Fr 𝐴 ∧ 𝑅 Po 𝐴 ∧ 𝑅 Se 𝐴 ) → Fun 𝐹 )