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Metamath Proof Explorer

Theorem fprfung

Description: A "function" defined by well-founded recursion is indeed a function when the relation 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 )