A property of representable functions?
$begingroup$
Recall that a function $f:mathbb{N}^krightarrowmathbb{N}$ is representable in Peano arithmetic if there exists a formula $varphi(x_1,dots ,x_k,y)$ such that for every $n_1,dots,n_k,minmathbb{N}$,
- If $m=f(n_1,dots,n_k)$ then $mathcal{A}vdash varphi(underline{n_1},dots ,underline{n_k},underline{m})$
- If $mneq f(n_1,dots,n_k)$ then $mathcal{A}vdash negvarphi(underline{n_1},dots ,underline{n_k},underline{m})$
Here, $mathcal{A}$ is the theory (set of axioms) of Peano arithmetic, and $underline{n}$ is the term $S(S(dots S(0)dots))$ (with $n$ operations of $S$).
I have two related questions regarding this definition.
Suppose that $f$ is representable by a formula $varphi(x_1,dots ,x_k,y)$ (as defined above);
- Is it necessarily true that for every $n_1,dots,n_k$, $$mathcal{A}vdashexists !y(varphi(underline{n_1},dots ,underline{n_k},y))$$?
- Is it necessarily true that $$mathcal{A}vdashforall x_1cdotsforall x_kexists !y(varphi(x_1,dots ,x_n,y))$$?
I wanted to prove these by contradiction, i.e assume that $mathcal{A}vdashpsi$ doesn't hold, but the problem is that I can't deduce that $mathcal{A}vdashnegpsi$ since Peano arithmetic isn't complete.
Other than that I'm quite clueless so I'd appriciate your help.
first-order-logic peano-axioms
asked Jan 2 at 12:49
35T4135T41
1,748417
$endgroup$
add a comment |
$begingroup$
Recall that a function $f:mathbb{N}^krightarrowmathbb{N}$ is representable in Peano arithmetic if there exists a formula $varphi(x_1,dots ,x_k,y)$ such that for every $n_1,dots,n_k,minmathbb{N}$,
- If $m=f(n_1,dots,n_k)$ then $mathcal{A}vdash varphi(underline{n_1},dots ,underline{n_k},underline{m})$
- If $mneq f(n_1,dots,n_k)$ then $mathcal{A}vdash negvarphi(underline{n_1},dots ,underline{n_k},underline{m})$
Here, $mathcal{A}$ is the theory (set of axioms) of Peano arithmetic, and $underline{n}$ is the term $S(S(dots S(0)dots))$ (with $n$ operations of $S$).
I have two related questions regarding this definition.
Suppose that $f$ is representable by a formula $varphi(x_1,dots ,x_k,y)$ (as defined above);
- Is it necessarily true that for every $n_1,dots,n_k$, $$mathcal{A}vdashexists !y(varphi(underline{n_1},dots ,underline{n_k},y))$$?
- Is it necessarily true that $$mathcal{A}vdashforall x_1cdotsforall x_kexists !y(varphi(x_1,dots ,x_n,y))$$?
I wanted to prove these by contradiction, i.e assume that $mathcal{A}vdashpsi$ doesn't hold, but the problem is that I can't deduce that $mathcal{A}vdashnegpsi$ since Peano arithmetic isn't complete.
Other than that I'm quite clueless so I'd appriciate your help.
first-order-logic peano-axioms
asked Jan 2 at 12:49
35T4135T41
1,748417
$endgroup$
add a comment |
$begingroup$
Recall that a function $f:mathbb{N}^krightarrowmathbb{N}$ is representable in Peano arithmetic if there exists a formula $varphi(x_1,dots ,x_k,y)$ such that for every $n_1,dots,n_k,minmathbb{N}$,
- If $m=f(n_1,dots,n_k)$ then $mathcal{A}vdash varphi(underline{n_1},dots ,underline{n_k},underline{m})$
- If $mneq f(n_1,dots,n_k)$ then $mathcal{A}vdash negvarphi(underline{n_1},dots ,underline{n_k},underline{m})$
Here, $mathcal{A}$ is the theory (set of axioms) of Peano arithmetic, and $underline{n}$ is the term $S(S(dots S(0)dots))$ (with $n$ operations of $S$).
I have two related questions regarding this definition.
Suppose that $f$ is representable by a formula $varphi(x_1,dots ,x_k,y)$ (as defined above);
- Is it necessarily true that for every $n_1,dots,n_k$, $$mathcal{A}vdashexists !y(varphi(underline{n_1},dots ,underline{n_k},y))$$?
- Is it necessarily true that $$mathcal{A}vdashforall x_1cdotsforall x_kexists !y(varphi(x_1,dots ,x_n,y))$$?
I wanted to prove these by contradiction, i.e assume that $mathcal{A}vdashpsi$ doesn't hold, but the problem is that I can't deduce that $mathcal{A}vdashnegpsi$ since Peano arithmetic isn't complete.
Other than that I'm quite clueless so I'd appriciate your help.
first-order-logic peano-axioms
asked Jan 2 at 12:49
35T4135T41
1,748417
$endgroup$
Recall that a function $f:mathbb{N}^krightarrowmathbb{N}$ is representable in Peano arithmetic if there exists a formula $varphi(x_1,dots ,x_k,y)$ such that for every $n_1,dots,n_k,minmathbb{N}$,
- If $m=f(n_1,dots,n_k)$ then $mathcal{A}vdash varphi(underline{n_1},dots ,underline{n_k},underline{m})$
- If $mneq f(n_1,dots,n_k)$ then $mathcal{A}vdash negvarphi(underline{n_1},dots ,underline{n_k},underline{m})$
Here, $mathcal{A}$ is the theory (set of axioms) of Peano arithmetic, and $underline{n}$ is the term $S(S(dots S(0)dots))$ (with $n$ operations of $S$).
I have two related questions regarding this definition.
Suppose that $f$ is representable by a formula $varphi(x_1,dots ,x_k,y)$ (as defined above);
- Is it necessarily true that for every $n_1,dots,n_k$, $$mathcal{A}vdashexists !y(varphi(underline{n_1},dots ,underline{n_k},y))$$?
- Is it necessarily true that $$mathcal{A}vdashforall x_1cdotsforall x_kexists !y(varphi(x_1,dots ,x_n,y))$$?
I wanted to prove these by contradiction, i.e assume that $mathcal{A}vdashpsi$ doesn't hold, but the problem is that I can't deduce that $mathcal{A}vdashnegpsi$ since Peano arithmetic isn't complete.
Other than that I'm quite clueless so I'd appriciate your help.
first-order-logic peano-axioms
first-order-logic peano-axioms
asked Jan 2 at 12:49
35T4135T41
1,748417
asked Jan 2 at 12:49
35T4135T41
1,748417
asked Jan 2 at 12:49
35T4135T41
1,748417
asked Jan 2 at 12:49
35T4135T41
1,748417
asked Jan 2 at 12:49
35T4135T41
1,748417
1,748417
add a comment |
add a comment |
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