Why is $F(a,b) = G(a,b, bar F (a,b) )$ unique if $bar F (a,b)$ is a sequence number applied to each element...












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I was reading these notes and on page 88 (of the paper version) it said:



enter image description here



however it wasn't clear to me why the function would be unique (or why it mattered). It seems entirely possible depending on the definition of $G$ that it might not be unique. Also, how does the explanation of making the explicit point by point what $F$ is explain anything? I think that is the key point I don't understand.










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  • I still don't know why uniqueness is obvious.
    – Pinocchio
    Nov 28 at 5:00
















0














I was reading these notes and on page 88 (of the paper version) it said:



enter image description here



however it wasn't clear to me why the function would be unique (or why it mattered). It seems entirely possible depending on the definition of $G$ that it might not be unique. Also, how does the explanation of making the explicit point by point what $F$ is explain anything? I think that is the key point I don't understand.










share|cite|improve this question






















  • I still don't know why uniqueness is obvious.
    – Pinocchio
    Nov 28 at 5:00














0












0








0







I was reading these notes and on page 88 (of the paper version) it said:



enter image description here



however it wasn't clear to me why the function would be unique (or why it mattered). It seems entirely possible depending on the definition of $G$ that it might not be unique. Also, how does the explanation of making the explicit point by point what $F$ is explain anything? I think that is the key point I don't understand.










share|cite|improve this question













I was reading these notes and on page 88 (of the paper version) it said:



enter image description here



however it wasn't clear to me why the function would be unique (or why it mattered). It seems entirely possible depending on the definition of $G$ that it might not be unique. Also, how does the explanation of making the explicit point by point what $F$ is explain anything? I think that is the key point I don't understand.







functions logic computability






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asked Nov 24 at 18:21









Pinocchio

1,88021754




1,88021754












  • I still don't know why uniqueness is obvious.
    – Pinocchio
    Nov 28 at 5:00


















  • I still don't know why uniqueness is obvious.
    – Pinocchio
    Nov 28 at 5:00
















I still don't know why uniqueness is obvious.
– Pinocchio
Nov 28 at 5:00




I still don't know why uniqueness is obvious.
– Pinocchio
Nov 28 at 5:00










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It should be pretty intuitively clear why $F$ is uniquely defined. If you want to know what $F(a,b)$ is for any $a$ and $b$, you simply apply the $F(a,b+1)$ equation repeatedly until $b=0$ and then you apply the $F(a,0)$ equation. This will unfold $F(a,b)$ out into a $b$-deep nested applications of $G$ to itself. Now, we'll still have occurrences of $F$ in this unfolded expression, but since they are always applied to smaller values of $b$ we can use strong induction to show that they too can be rewritten in terms of $G$ only. The end result is that $F(a,b)$ is defined as an expression solely in terms of $G$, $a$, and $b$. If you do the proof, you'll find that strong induction fits extremely nicely. This is no surprise as this is the computational analogue of strong induction (with parameters) itself.






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  • why is uniqueness obvious? Thats usually requires proof.
    – Pinocchio
    Nov 26 at 21:04











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It should be pretty intuitively clear why $F$ is uniquely defined. If you want to know what $F(a,b)$ is for any $a$ and $b$, you simply apply the $F(a,b+1)$ equation repeatedly until $b=0$ and then you apply the $F(a,0)$ equation. This will unfold $F(a,b)$ out into a $b$-deep nested applications of $G$ to itself. Now, we'll still have occurrences of $F$ in this unfolded expression, but since they are always applied to smaller values of $b$ we can use strong induction to show that they too can be rewritten in terms of $G$ only. The end result is that $F(a,b)$ is defined as an expression solely in terms of $G$, $a$, and $b$. If you do the proof, you'll find that strong induction fits extremely nicely. This is no surprise as this is the computational analogue of strong induction (with parameters) itself.






share|cite|improve this answer





















  • why is uniqueness obvious? Thats usually requires proof.
    – Pinocchio
    Nov 26 at 21:04
















0














It should be pretty intuitively clear why $F$ is uniquely defined. If you want to know what $F(a,b)$ is for any $a$ and $b$, you simply apply the $F(a,b+1)$ equation repeatedly until $b=0$ and then you apply the $F(a,0)$ equation. This will unfold $F(a,b)$ out into a $b$-deep nested applications of $G$ to itself. Now, we'll still have occurrences of $F$ in this unfolded expression, but since they are always applied to smaller values of $b$ we can use strong induction to show that they too can be rewritten in terms of $G$ only. The end result is that $F(a,b)$ is defined as an expression solely in terms of $G$, $a$, and $b$. If you do the proof, you'll find that strong induction fits extremely nicely. This is no surprise as this is the computational analogue of strong induction (with parameters) itself.






share|cite|improve this answer





















  • why is uniqueness obvious? Thats usually requires proof.
    – Pinocchio
    Nov 26 at 21:04














0












0








0






It should be pretty intuitively clear why $F$ is uniquely defined. If you want to know what $F(a,b)$ is for any $a$ and $b$, you simply apply the $F(a,b+1)$ equation repeatedly until $b=0$ and then you apply the $F(a,0)$ equation. This will unfold $F(a,b)$ out into a $b$-deep nested applications of $G$ to itself. Now, we'll still have occurrences of $F$ in this unfolded expression, but since they are always applied to smaller values of $b$ we can use strong induction to show that they too can be rewritten in terms of $G$ only. The end result is that $F(a,b)$ is defined as an expression solely in terms of $G$, $a$, and $b$. If you do the proof, you'll find that strong induction fits extremely nicely. This is no surprise as this is the computational analogue of strong induction (with parameters) itself.






share|cite|improve this answer












It should be pretty intuitively clear why $F$ is uniquely defined. If you want to know what $F(a,b)$ is for any $a$ and $b$, you simply apply the $F(a,b+1)$ equation repeatedly until $b=0$ and then you apply the $F(a,0)$ equation. This will unfold $F(a,b)$ out into a $b$-deep nested applications of $G$ to itself. Now, we'll still have occurrences of $F$ in this unfolded expression, but since they are always applied to smaller values of $b$ we can use strong induction to show that they too can be rewritten in terms of $G$ only. The end result is that $F(a,b)$ is defined as an expression solely in terms of $G$, $a$, and $b$. If you do the proof, you'll find that strong induction fits extremely nicely. This is no surprise as this is the computational analogue of strong induction (with parameters) itself.







share|cite|improve this answer












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answered Nov 24 at 19:36









Derek Elkins

16.1k11337




16.1k11337












  • why is uniqueness obvious? Thats usually requires proof.
    – Pinocchio
    Nov 26 at 21:04


















  • why is uniqueness obvious? Thats usually requires proof.
    – Pinocchio
    Nov 26 at 21:04
















why is uniqueness obvious? Thats usually requires proof.
– Pinocchio
Nov 26 at 21:04




why is uniqueness obvious? Thats usually requires proof.
– Pinocchio
Nov 26 at 21:04


















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