$F$ in $F(mathbf{x},y)=G(y,H_{1}(mathbf{x}),H_{2}(y,mathbf{x}))$ is recursive?
Show that $F$ in $F(mathbf{x},y)=G(y,H_{1}(mathbf{x}),H_{2}(y,mathbf{x}))$ is recursive by giving a description of $F$, where $G,H_{1},H_{2}$ are recursive.
I have to find a description of $F$ to show $F$ is recursive. I was thinking that $H_{1}$ is using redundant variables, where $H_{1}'(mathbf{x}, y) = H_{1}(mathbf{x})$, which is primitive recursive, and $H_{2}$ is using permutation of variables, where $H'_{2}(mathbf{x},y)=H_{2}(y,mathbf{x})$, which is also primitive recursive. But this is as far as I can go. How do I describe $F$?
logic recursion
asked Nov 30 '18 at 3:02
numericalorangenumericalorange
1,726311
|
show 3 more comments
Show that $F$ in $F(mathbf{x},y)=G(y,H_{1}(mathbf{x}),H_{2}(y,mathbf{x}))$ is recursive by giving a description of $F$, where $G,H_{1},H_{2}$ are recursive.
I have to find a description of $F$ to show $F$ is recursive. I was thinking that $H_{1}$ is using redundant variables, where $H_{1}'(mathbf{x}, y) = H_{1}(mathbf{x})$, which is primitive recursive, and $H_{2}$ is using permutation of variables, where $H'_{2}(mathbf{x},y)=H_{2}(y,mathbf{x})$, which is also primitive recursive. But this is as far as I can go. How do I describe $F$?
logic recursion
asked Nov 30 '18 at 3:02
numericalorangenumericalorange
1,726311
1
So $H_1'$ and $H_2'$ are compositions of recursive functions, so recursive. Then show $f(mathbf x,y) = y$ is (primative) recursive, and then $F$ is a composition of recursive functions, and thus recursive.
– spaceisdarkgreen
Nov 30 '18 at 3:16
Oh I see, so I was actually describing F all along...interesting! Thanks for the guidance.
– numericalorange
Nov 30 '18 at 3:20
1
To be fair, I'm not familiar with 'give a description' as a technical term, but I would guess it just means to show at some fairly fine level of detail that $F$ can be formed from the formation rules for recursive functions.
– spaceisdarkgreen
Nov 30 '18 at 3:31
@spaceisdarkgreen No worries! What do I do with the case $f(mathbf{x},y)=y$? Is this a projection?
– numericalorange
Nov 30 '18 at 3:43
1
f is the projection function. The way I learned it, the projection functions are one of the basic pr functions from which the others are built. I'm not sure what this bijectivity requirement is referring to, but I don't think it's for this situation here.
– spaceisdarkgreen
Nov 30 '18 at 5:01
|
show 3 more comments
Show that $F$ in $F(mathbf{x},y)=G(y,H_{1}(mathbf{x}),H_{2}(y,mathbf{x}))$ is recursive by giving a description of $F$, where $G,H_{1},H_{2}$ are recursive.
I have to find a description of $F$ to show $F$ is recursive. I was thinking that $H_{1}$ is using redundant variables, where $H_{1}'(mathbf{x}, y) = H_{1}(mathbf{x})$, which is primitive recursive, and $H_{2}$ is using permutation of variables, where $H'_{2}(mathbf{x},y)=H_{2}(y,mathbf{x})$, which is also primitive recursive. But this is as far as I can go. How do I describe $F$?
logic recursion
asked Nov 30 '18 at 3:02
numericalorangenumericalorange
1,726311
Show that $F$ in $F(mathbf{x},y)=G(y,H_{1}(mathbf{x}),H_{2}(y,mathbf{x}))$ is recursive by giving a description of $F$, where $G,H_{1},H_{2}$ are recursive.
I have to find a description of $F$ to show $F$ is recursive. I was thinking that $H_{1}$ is using redundant variables, where $H_{1}'(mathbf{x}, y) = H_{1}(mathbf{x})$, which is primitive recursive, and $H_{2}$ is using permutation of variables, where $H'_{2}(mathbf{x},y)=H_{2}(y,mathbf{x})$, which is also primitive recursive. But this is as far as I can go. How do I describe $F$?
logic recursion
logic recursion
asked Nov 30 '18 at 3:02
numericalorangenumericalorange
1,726311
asked Nov 30 '18 at 3:02
numericalorangenumericalorange
1,726311
asked Nov 30 '18 at 3:02
numericalorangenumericalorange
1,726311
asked Nov 30 '18 at 3:02
numericalorangenumericalorange
1,726311
asked Nov 30 '18 at 3:02
numericalorangenumericalorange
1,726311
1,726311
1
So $H_1'$ and $H_2'$ are compositions of recursive functions, so recursive. Then show $f(mathbf x,y) = y$ is (primative) recursive, and then $F$ is a composition of recursive functions, and thus recursive.
– spaceisdarkgreen
Nov 30 '18 at 3:16
Oh I see, so I was actually describing F all along...interesting! Thanks for the guidance.
– numericalorange
Nov 30 '18 at 3:20
1
To be fair, I'm not familiar with 'give a description' as a technical term, but I would guess it just means to show at some fairly fine level of detail that $F$ can be formed from the formation rules for recursive functions.
– spaceisdarkgreen
Nov 30 '18 at 3:31
@spaceisdarkgreen No worries! What do I do with the case $f(mathbf{x},y)=y$? Is this a projection?
– numericalorange
Nov 30 '18 at 3:43
1
f is the projection function. The way I learned it, the projection functions are one of the basic pr functions from which the others are built. I'm not sure what this bijectivity requirement is referring to, but I don't think it's for this situation here.
– spaceisdarkgreen
Nov 30 '18 at 5:01
|
show 3 more comments
1
So $H_1'$ and $H_2'$ are compositions of recursive functions, so recursive. Then show $f(mathbf x,y) = y$ is (primative) recursive, and then $F$ is a composition of recursive functions, and thus recursive.
– spaceisdarkgreen
Nov 30 '18 at 3:16
Oh I see, so I was actually describing F all along...interesting! Thanks for the guidance.
– numericalorange
Nov 30 '18 at 3:20
1
To be fair, I'm not familiar with 'give a description' as a technical term, but I would guess it just means to show at some fairly fine level of detail that $F$ can be formed from the formation rules for recursive functions.
– spaceisdarkgreen
Nov 30 '18 at 3:31
@spaceisdarkgreen No worries! What do I do with the case $f(mathbf{x},y)=y$? Is this a projection?
– numericalorange
Nov 30 '18 at 3:43
1
f is the projection function. The way I learned it, the projection functions are one of the basic pr functions from which the others are built. I'm not sure what this bijectivity requirement is referring to, but I don't think it's for this situation here.
– spaceisdarkgreen
Nov 30 '18 at 5:01
1
1
So $H_1'$ and $H_2'$ are compositions of recursive functions, so recursive. Then show $f(mathbf x,y) = y$ is (primative) recursive, and then $F$ is a composition of recursive functions, and thus recursive.
– spaceisdarkgreen
Nov 30 '18 at 3:16
So $H_1'$ and $H_2'$ are compositions of recursive functions, so recursive. Then show $f(mathbf x,y) = y$ is (primative) recursive, and then $F$ is a composition of recursive functions, and thus recursive.
– spaceisdarkgreen
Nov 30 '18 at 3:16
Oh I see, so I was actually describing F all along...interesting! Thanks for the guidance.
– numericalorange
Nov 30 '18 at 3:20
Oh I see, so I was actually describing F all along...interesting! Thanks for the guidance.
– numericalorange
Nov 30 '18 at 3:20
1
1
To be fair, I'm not familiar with 'give a description' as a technical term, but I would guess it just means to show at some fairly fine level of detail that $F$ can be formed from the formation rules for recursive functions.
– spaceisdarkgreen
Nov 30 '18 at 3:31
To be fair, I'm not familiar with 'give a description' as a technical term, but I would guess it just means to show at some fairly fine level of detail that $F$ can be formed from the formation rules for recursive functions.
– spaceisdarkgreen
Nov 30 '18 at 3:31
@spaceisdarkgreen No worries! What do I do with the case $f(mathbf{x},y)=y$? Is this a projection?
– numericalorange
Nov 30 '18 at 3:43
@spaceisdarkgreen No worries! What do I do with the case $f(mathbf{x},y)=y$? Is this a projection?
– numericalorange
Nov 30 '18 at 3:43
1
1
f is the projection function. The way I learned it, the projection functions are one of the basic pr functions from which the others are built. I'm not sure what this bijectivity requirement is referring to, but I don't think it's for this situation here.
– spaceisdarkgreen
Nov 30 '18 at 5:01
f is the projection function. The way I learned it, the projection functions are one of the basic pr functions from which the others are built. I'm not sure what this bijectivity requirement is referring to, but I don't think it's for this situation here.
– spaceisdarkgreen
Nov 30 '18 at 5:01
|
show 3 more comments
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1
So $H_1'$ and $H_2'$ are compositions of recursive functions, so recursive. Then show $f(mathbf x,y) = y$ is (primative) recursive, and then $F$ is a composition of recursive functions, and thus recursive.
– spaceisdarkgreen
Nov 30 '18 at 3:16
Oh I see, so I was actually describing F all along...interesting! Thanks for the guidance.
– numericalorange
Nov 30 '18 at 3:20
1
To be fair, I'm not familiar with 'give a description' as a technical term, but I would guess it just means to show at some fairly fine level of detail that $F$ can be formed from the formation rules for recursive functions.
– spaceisdarkgreen
Nov 30 '18 at 3:31
@spaceisdarkgreen No worries! What do I do with the case $f(mathbf{x},y)=y$? Is this a projection?
– numericalorange
Nov 30 '18 at 3:43
1
f is the projection function. The way I learned it, the projection functions are one of the basic pr functions from which the others are built. I'm not sure what this bijectivity requirement is referring to, but I don't think it's for this situation here.
– spaceisdarkgreen
Nov 30 '18 at 5:01