|
Chris Pollett >
Old Classses > |
HW4 Solutions PageFor this homework, I had the grader recommend some of the better assignments and I chose from them one to use as the homework solution. The students that had their homework chosen received 1 bonus point after curving for having their homework selected. If you were chosen and would rather your homework not be used as the solution let me know and I will choose someone else's homework and they will receive the bonus point instead. I only took a quick glance at people's solutions so they may still have issues. I wasn't satisfied with anyone's solution to Problem 3 so I give my solution below: We use the following abbreviations...
The initial Nixon diamond consists of the facts: `R(Nixon) ^^ Q(Nixon)` together with the rules: `frac{R(x): neg P(x)}{neg P(x)}` and `frac{Q(x): P(x)}{P(x)}`. One can verify there are exactly two ways one can guess a ground model so that one can only derive back that model. One model is `R(Nixon) ^^ Q(Nixon) ^^ neg P(Nixon)` and the other is `R(Nixon) ^^ Q(Nixon) ^^ P(Nixon)`.
To add a rule so that there are exactly three extensions our strategy is to make it so that `R(Nixon) ^^ Q(Nixon) ^^ neg P(Nixon)` is a subset of two models in the new system, but `R(Nixon) ^^ Q(Nixon) ^^ P(Nixon)` is a subset of only one. One way to do this is to add a rule like: `frac{C(x): neg P(x)}{C(x)}` We can only derive `C(Nixon)` if `neg P(Nixon)` and then only if we had initially guessed this as part of the extension. So the three extension of the new system will be: `R(Nixon) ^^ Q(Nixon) ^^ neg P(Nixon) ^^ C(Nixon)`, `R(Nixon) ^^ Q(Nixon) ^^ neg P(Nixon) ^^ neg C(Nixon)`, and `R(Nixon) ^^ Q(Nixon) ^^ P(Nixon) ^^ neg C(Nixon)`. |