Writing a formal description for F to prove recursion.
$begingroup$
I am making an update for finding a formal description using the language of recursion to show that the following is recursive:
$left{begin{matrix}
F(0,y,z)=G(y)\
F(x+1,y,z)=H(z,F(x,y,z))
end{matrix}right.$
One of my classmates helped me come up with this:
Since $H$ depends on $F$, define a new description $bar{mathbf{F}}(y,z,x)$, where $bar{F}(y,z,0)=bar{G}(y,z)=G(y)$ and $bar{F}(y,z,x+1)=bar{H}(y,z,x,bar{F}(y,z,x))=
H(z,bar{F}(y,z,x))$. Then $mathbf{F}=mathbf{C}(bar{mathbf{F}},mathbf{I}_{3,3},mathbf{I}_{3,1},mathbf{I}_{3,2})$, and $mathbf{bar{F}}=mathbf{P}(bar{mathbf{G}},mathbf{bar{H}})$, where $mathbf{bar{G}}=mathbf{C}(mathbf{G},mathbf{I}_{2,1})$ and $mathbf{bar{H}}=mathbf{C}(mathbf{H},mathbf{I}_{4,2},mathbf{I}_{4,4})$, describes F.
I am wondering whether this description is correct, and if so, can someone explain why $bar{F}(y,z,0)=bar{G}(y,z)=G(y)$ and $bar{F}(y,z,x+1)=bar{H}(y,z,x,bar{F}(y,z,x))=
H(z,bar{F}(y,z,x))$? This is the only part I don't understand.
logic recursion computability
asked Dec 4 '18 at 22:56
numericalorangenumericalorange
1,770311
$endgroup$
add a comment |
$begingroup$
I am making an update for finding a formal description using the language of recursion to show that the following is recursive:
$left{begin{matrix}
F(0,y,z)=G(y)\
F(x+1,y,z)=H(z,F(x,y,z))
end{matrix}right.$
One of my classmates helped me come up with this:
Since $H$ depends on $F$, define a new description $bar{mathbf{F}}(y,z,x)$, where $bar{F}(y,z,0)=bar{G}(y,z)=G(y)$ and $bar{F}(y,z,x+1)=bar{H}(y,z,x,bar{F}(y,z,x))=
H(z,bar{F}(y,z,x))$. Then $mathbf{F}=mathbf{C}(bar{mathbf{F}},mathbf{I}_{3,3},mathbf{I}_{3,1},mathbf{I}_{3,2})$, and $mathbf{bar{F}}=mathbf{P}(bar{mathbf{G}},mathbf{bar{H}})$, where $mathbf{bar{G}}=mathbf{C}(mathbf{G},mathbf{I}_{2,1})$ and $mathbf{bar{H}}=mathbf{C}(mathbf{H},mathbf{I}_{4,2},mathbf{I}_{4,4})$, describes F.
I am wondering whether this description is correct, and if so, can someone explain why $bar{F}(y,z,0)=bar{G}(y,z)=G(y)$ and $bar{F}(y,z,x+1)=bar{H}(y,z,x,bar{F}(y,z,x))=
H(z,bar{F}(y,z,x))$? This is the only part I don't understand.
logic recursion computability
asked Dec 4 '18 at 22:56
numericalorangenumericalorange
1,770311
$endgroup$
add a comment |
$begingroup$
I am making an update for finding a formal description using the language of recursion to show that the following is recursive:
$left{begin{matrix}
F(0,y,z)=G(y)\
F(x+1,y,z)=H(z,F(x,y,z))
end{matrix}right.$
One of my classmates helped me come up with this:
Since $H$ depends on $F$, define a new description $bar{mathbf{F}}(y,z,x)$, where $bar{F}(y,z,0)=bar{G}(y,z)=G(y)$ and $bar{F}(y,z,x+1)=bar{H}(y,z,x,bar{F}(y,z,x))=
H(z,bar{F}(y,z,x))$. Then $mathbf{F}=mathbf{C}(bar{mathbf{F}},mathbf{I}_{3,3},mathbf{I}_{3,1},mathbf{I}_{3,2})$, and $mathbf{bar{F}}=mathbf{P}(bar{mathbf{G}},mathbf{bar{H}})$, where $mathbf{bar{G}}=mathbf{C}(mathbf{G},mathbf{I}_{2,1})$ and $mathbf{bar{H}}=mathbf{C}(mathbf{H},mathbf{I}_{4,2},mathbf{I}_{4,4})$, describes F.
I am wondering whether this description is correct, and if so, can someone explain why $bar{F}(y,z,0)=bar{G}(y,z)=G(y)$ and $bar{F}(y,z,x+1)=bar{H}(y,z,x,bar{F}(y,z,x))=
H(z,bar{F}(y,z,x))$? This is the only part I don't understand.
logic recursion computability
asked Dec 4 '18 at 22:56
numericalorangenumericalorange
1,770311
$endgroup$
I am making an update for finding a formal description using the language of recursion to show that the following is recursive:
$left{begin{matrix}
F(0,y,z)=G(y)\
F(x+1,y,z)=H(z,F(x,y,z))
end{matrix}right.$
One of my classmates helped me come up with this:
Since $H$ depends on $F$, define a new description $bar{mathbf{F}}(y,z,x)$, where $bar{F}(y,z,0)=bar{G}(y,z)=G(y)$ and $bar{F}(y,z,x+1)=bar{H}(y,z,x,bar{F}(y,z,x))=
H(z,bar{F}(y,z,x))$. Then $mathbf{F}=mathbf{C}(bar{mathbf{F}},mathbf{I}_{3,3},mathbf{I}_{3,1},mathbf{I}_{3,2})$, and $mathbf{bar{F}}=mathbf{P}(bar{mathbf{G}},mathbf{bar{H}})$, where $mathbf{bar{G}}=mathbf{C}(mathbf{G},mathbf{I}_{2,1})$ and $mathbf{bar{H}}=mathbf{C}(mathbf{H},mathbf{I}_{4,2},mathbf{I}_{4,4})$, describes F.
I am wondering whether this description is correct, and if so, can someone explain why $bar{F}(y,z,0)=bar{G}(y,z)=G(y)$ and $bar{F}(y,z,x+1)=bar{H}(y,z,x,bar{F}(y,z,x))=
H(z,bar{F}(y,z,x))$? This is the only part I don't understand.
logic recursion computability
logic recursion computability
asked Dec 4 '18 at 22:56
numericalorangenumericalorange
1,770311
asked Dec 4 '18 at 22:56
numericalorangenumericalorange
1,770311
asked Dec 4 '18 at 22:56
numericalorangenumericalorange
1,770311
asked Dec 4 '18 at 22:56
numericalorangenumericalorange
1,770311
asked Dec 4 '18 at 22:56
numericalorangenumericalorange
1,770311
1,770311
add a comment |
add a comment |
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