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Question: Given the following signatures: $S_1=\{c,f^1,,=^2\}$, $S_2=\{P^1, R^2,=^2\}$, prove/disprove that there exists and algorithm that for a formula $\phi$ in $S_1$, builds a formula $\phi'$ that is satisfiable iff $\phi$ is satisfiable.

My proof: I believe this is true so I defined the algorithm :

  1. Convert any instance of $c$ to and instance of $P(x)$ ($x$ is always a new variable) and to always define $P=\{v(c)\}$ (a singleton relation).
  2. Convert any instance of $f(x)$ to $R(x,f(x))$, and to define $R$ as the sources and images of the function $f$.

I was wondering if this proof is true, or am I missing anything. (Seems good to me, but in logic you can never know some counterexample might pop up)...

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  • $\begingroup$ Please look at my edits to the question. One shouldn't keep alternating in and out of MathJax in the course of a single expression in mathematical notation; rather one should put the whole thing within MathJax. $\qquad$ $\endgroup$ – Michael Hardy Feb 18 '16 at 16:49
  • $\begingroup$ Not sure how detailed you need to be in describing the algorithm, but I think you're currently missing some steps about how to actually transform formulas. For example, if $\phi(x) : x = c$, simply turning $c$ into $P(y)$ gives you $\phi(x,y): x = P(y)$ which is not valid syntax. $\endgroup$ – Dan Simon Feb 18 '16 at 18:24
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Your idea is mostly correct, but you need to be careful that the syntax works. I'm going to assume these signatures are on the same universe, say $A$. (I'm going to ignore the $=$ since it can be interpreted the same in either).

1) Then with $S_1$ we have some interpretation of $c$ in $A$ (say $c_1\in{}A$). We want to define the relation $P$ to be $P=\{{}c_1\}{}$. Then for any formula $\phi$ in signature $S_1$ we can keep all logical symbols the same, but we must change instances of $c$. Let's say $\psi(c)$ is a subformula of $\phi$ that contains $c$, then we replace it with $P(x)\wedge\psi(x)$, which checks that $x=c$ and then replaces all instances of $c$ with $x$.

2) Now we deal with instances of $f$. First we define $R=\{{}(x,f(x))\}{}$, for all $x$ in the domain of $f$. Next, whenever we have a formula $\phi$ in signature $S_1$ with subformula $\psi(f(x))$ that contains $f(x)$, we replace it with $R(x,y)\wedge{}\psi(y)$. If different terms $f(x_1),f(x_2),\dots{},f(x_n)$ appear, then we'll need to replace those accordingly.

Under these changes, if we have a formula $\phi$ in $S_1$, we see that we now get a formula $\phi'$ in $S_2$. Now all that remains is to show $\phi$ is satisfiable iff $\phi'$ is satisfiable. And that will be fairly straightforward given how we defined $P$ and $R$.

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