Metamath Proof Explorer


Theorem frr3

Description: Law of general well-founded recursion, part three. Finally, we show that F is unique. We do this by showing that any function H with the same properties we proved of F in frr1 and frr2 is identical to F . (Contributed by Scott Fenton, 11-Sep-2023)

Ref Expression
Hypothesis frr.1 𝐹 = frecs ( 𝑅 , 𝐴 , 𝐺 )
Assertion frr3 ( ( ( 𝑅 Fr 𝐴𝑅 Se 𝐴 ) ∧ ( 𝐻 Fn 𝐴 ∧ ∀ 𝑧𝐴 ( 𝐻𝑧 ) = ( 𝑧 𝐺 ( 𝐻 ↾ Pred ( 𝑅 , 𝐴 , 𝑧 ) ) ) ) ) → 𝐹 = 𝐻 )

Proof

Step Hyp Ref Expression
1 frr.1 𝐹 = frecs ( 𝑅 , 𝐴 , 𝐺 )
2 simpl ( ( ( 𝑅 Fr 𝐴𝑅 Se 𝐴 ) ∧ ( 𝐻 Fn 𝐴 ∧ ∀ 𝑧𝐴 ( 𝐻𝑧 ) = ( 𝑧 𝐺 ( 𝐻 ↾ Pred ( 𝑅 , 𝐴 , 𝑧 ) ) ) ) ) → ( 𝑅 Fr 𝐴𝑅 Se 𝐴 ) )
3 1 frr1 ( ( 𝑅 Fr 𝐴𝑅 Se 𝐴 ) → 𝐹 Fn 𝐴 )
4 1 frr2 ( ( ( 𝑅 Fr 𝐴𝑅 Se 𝐴 ) ∧ 𝑧𝐴 ) → ( 𝐹𝑧 ) = ( 𝑧 𝐺 ( 𝐹 ↾ Pred ( 𝑅 , 𝐴 , 𝑧 ) ) ) )
5 4 ralrimiva ( ( 𝑅 Fr 𝐴𝑅 Se 𝐴 ) → ∀ 𝑧𝐴 ( 𝐹𝑧 ) = ( 𝑧 𝐺 ( 𝐹 ↾ Pred ( 𝑅 , 𝐴 , 𝑧 ) ) ) )
6 3 5 jca ( ( 𝑅 Fr 𝐴𝑅 Se 𝐴 ) → ( 𝐹 Fn 𝐴 ∧ ∀ 𝑧𝐴 ( 𝐹𝑧 ) = ( 𝑧 𝐺 ( 𝐹 ↾ Pred ( 𝑅 , 𝐴 , 𝑧 ) ) ) ) )
7 6 adantr ( ( ( 𝑅 Fr 𝐴𝑅 Se 𝐴 ) ∧ ( 𝐻 Fn 𝐴 ∧ ∀ 𝑧𝐴 ( 𝐻𝑧 ) = ( 𝑧 𝐺 ( 𝐻 ↾ Pred ( 𝑅 , 𝐴 , 𝑧 ) ) ) ) ) → ( 𝐹 Fn 𝐴 ∧ ∀ 𝑧𝐴 ( 𝐹𝑧 ) = ( 𝑧 𝐺 ( 𝐹 ↾ Pred ( 𝑅 , 𝐴 , 𝑧 ) ) ) ) )
8 simpr ( ( ( 𝑅 Fr 𝐴𝑅 Se 𝐴 ) ∧ ( 𝐻 Fn 𝐴 ∧ ∀ 𝑧𝐴 ( 𝐻𝑧 ) = ( 𝑧 𝐺 ( 𝐻 ↾ Pred ( 𝑅 , 𝐴 , 𝑧 ) ) ) ) ) → ( 𝐻 Fn 𝐴 ∧ ∀ 𝑧𝐴 ( 𝐻𝑧 ) = ( 𝑧 𝐺 ( 𝐻 ↾ Pred ( 𝑅 , 𝐴 , 𝑧 ) ) ) ) )
9 frr3g ( ( ( 𝑅 Fr 𝐴𝑅 Se 𝐴 ) ∧ ( 𝐹 Fn 𝐴 ∧ ∀ 𝑧𝐴 ( 𝐹𝑧 ) = ( 𝑧 𝐺 ( 𝐹 ↾ Pred ( 𝑅 , 𝐴 , 𝑧 ) ) ) ) ∧ ( 𝐻 Fn 𝐴 ∧ ∀ 𝑧𝐴 ( 𝐻𝑧 ) = ( 𝑧 𝐺 ( 𝐻 ↾ Pred ( 𝑅 , 𝐴 , 𝑧 ) ) ) ) ) → 𝐹 = 𝐻 )
10 2 7 8 9 syl3anc ( ( ( 𝑅 Fr 𝐴𝑅 Se 𝐴 ) ∧ ( 𝐻 Fn 𝐴 ∧ ∀ 𝑧𝐴 ( 𝐻𝑧 ) = ( 𝑧 𝐺 ( 𝐻 ↾ Pred ( 𝑅 , 𝐴 , 𝑧 ) ) ) ) ) → 𝐹 = 𝐻 )