Theorem fpr3

Description: Law of well-founded recursion over a partial order, 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 fpr1 and fpr2 is identical to F . (Contributed by Scott Fenton, 11-Sep-2023)

	Ref	Expression
Hypothesis	fprr.1	⊢ 𝐹 = frecs ( 𝑅 , 𝐴 , 𝐺 )
Assertion	fpr3	⊢ ( ( ( 𝑅 Fr 𝐴 ∧ 𝑅 Po 𝐴 ∧ 𝑅 Se 𝐴 ) ∧ ( 𝐻 Fn 𝐴 ∧ ∀ 𝑧 ∈ 𝐴 ( 𝐻 ‘ 𝑧 ) = ( 𝑧 𝐺 ( 𝐻 ↾ Pred ( 𝑅 , 𝐴 , 𝑧 ) ) ) ) ) → 𝐹 = 𝐻 )

Proof

Step	Hyp	Ref	Expression
1		fprr.1	⊢ 𝐹 = frecs ( 𝑅 , 𝐴 , 𝐺 )
2		simpl	⊢ ( ( ( 𝑅 Fr 𝐴 ∧ 𝑅 Po 𝐴 ∧ 𝑅 Se 𝐴 ) ∧ ( 𝐻 Fn 𝐴 ∧ ∀ 𝑧 ∈ 𝐴 ( 𝐻 ‘ 𝑧 ) = ( 𝑧 𝐺 ( 𝐻 ↾ Pred ( 𝑅 , 𝐴 , 𝑧 ) ) ) ) ) → ( 𝑅 Fr 𝐴 ∧ 𝑅 Po 𝐴 ∧ 𝑅 Se 𝐴 ) )
3	1	fpr1	⊢ ( ( 𝑅 Fr 𝐴 ∧ 𝑅 Po 𝐴 ∧ 𝑅 Se 𝐴 ) → 𝐹 Fn 𝐴 )
4	1	fpr2	⊢ ( ( ( 𝑅 Fr 𝐴 ∧ 𝑅 Po 𝐴 ∧ 𝑅 Se 𝐴 ) ∧ 𝑧 ∈ 𝐴 ) → ( 𝐹 ‘ 𝑧 ) = ( 𝑧 𝐺 ( 𝐹 ↾ Pred ( 𝑅 , 𝐴 , 𝑧 ) ) ) )
5	4	ralrimiva	⊢ ( ( 𝑅 Fr 𝐴 ∧ 𝑅 Po 𝐴 ∧ 𝑅 Se 𝐴 ) → ∀ 𝑧 ∈ 𝐴 ( 𝐹 ‘ 𝑧 ) = ( 𝑧 𝐺 ( 𝐹 ↾ Pred ( 𝑅 , 𝐴 , 𝑧 ) ) ) )
6	3 5	jca	⊢ ( ( 𝑅 Fr 𝐴 ∧ 𝑅 Po 𝐴 ∧ 𝑅 Se 𝐴 ) → ( 𝐹 Fn 𝐴 ∧ ∀ 𝑧 ∈ 𝐴 ( 𝐹 ‘ 𝑧 ) = ( 𝑧 𝐺 ( 𝐹 ↾ Pred ( 𝑅 , 𝐴 , 𝑧 ) ) ) ) )
7	6	adantr	⊢ ( ( ( 𝑅 Fr 𝐴 ∧ 𝑅 Po 𝐴 ∧ 𝑅 Se 𝐴 ) ∧ ( 𝐻 Fn 𝐴 ∧ ∀ 𝑧 ∈ 𝐴 ( 𝐻 ‘ 𝑧 ) = ( 𝑧 𝐺 ( 𝐻 ↾ Pred ( 𝑅 , 𝐴 , 𝑧 ) ) ) ) ) → ( 𝐹 Fn 𝐴 ∧ ∀ 𝑧 ∈ 𝐴 ( 𝐹 ‘ 𝑧 ) = ( 𝑧 𝐺 ( 𝐹 ↾ Pred ( 𝑅 , 𝐴 , 𝑧 ) ) ) ) )
8		simpr	⊢ ( ( ( 𝑅 Fr 𝐴 ∧ 𝑅 Po 𝐴 ∧ 𝑅 Se 𝐴 ) ∧ ( 𝐻 Fn 𝐴 ∧ ∀ 𝑧 ∈ 𝐴 ( 𝐻 ‘ 𝑧 ) = ( 𝑧 𝐺 ( 𝐻 ↾ Pred ( 𝑅 , 𝐴 , 𝑧 ) ) ) ) ) → ( 𝐻 Fn 𝐴 ∧ ∀ 𝑧 ∈ 𝐴 ( 𝐻 ‘ 𝑧 ) = ( 𝑧 𝐺 ( 𝐻 ↾ Pred ( 𝑅 , 𝐴 , 𝑧 ) ) ) ) )
9		fpr3g	⊢ ( ( ( 𝑅 Fr 𝐴 ∧ 𝑅 Po 𝐴 ∧ 𝑅 Se 𝐴 ) ∧ ( 𝐹 Fn 𝐴 ∧ ∀ 𝑧 ∈ 𝐴 ( 𝐹 ‘ 𝑧 ) = ( 𝑧 𝐺 ( 𝐹 ↾ Pred ( 𝑅 , 𝐴 , 𝑧 ) ) ) ) ∧ ( 𝐻 Fn 𝐴 ∧ ∀ 𝑧 ∈ 𝐴 ( 𝐻 ‘ 𝑧 ) = ( 𝑧 𝐺 ( 𝐻 ↾ Pred ( 𝑅 , 𝐴 , 𝑧 ) ) ) ) ) → 𝐹 = 𝐻 )
10	2 7 8 9	syl3anc	⊢ ( ( ( 𝑅 Fr 𝐴 ∧ 𝑅 Po 𝐴 ∧ 𝑅 Se 𝐴 ) ∧ ( 𝐻 Fn 𝐴 ∧ ∀ 𝑧 ∈ 𝐴 ( 𝐻 ‘ 𝑧 ) = ( 𝑧 𝐺 ( 𝐻 ↾ Pred ( 𝑅 , 𝐴 , 𝑧 ) ) ) ) ) → 𝐹 = 𝐻 )