Philosophy 2340
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13.4
> Dr. Brown, I did not include 13.4-if ~ax(Cube x) with the conclusion
> ~ax(Cube x and Small x). I know that it must be true because if all x
> are not Cubes then all x's can not be small cubes.
Right. ~Ax(Cube(x)) means that not everything is a cube (some things could be, but not everything). Not certain whether this is what you meant by "all x are not cubes" since that could also be read as meaning that there are no cubes.
Anyway, I think you're on the right track. Here's one way to think of it: if everything *were* a small cube, then everything would be a cube, contrary to our premise. So it must *not* be the case that everything is a small cube.
Pursuing this informal reasoning, you might try assuming the opposite of the conclusion (i.e. assume Ax(Small(x) & Cube(x))) and then derive a contradiction so you can use negation introduction to derive the final conclusion.
> 13.4, I didn't really understand how to utilize the negation
> in front of the
> universal quantifier.
Right -- you can't use the quantifier rules on a formula that's actually a negation. (This is just like all the other rules: you can't use them on subformulas, only on formulas that match the form of the rule. ~Ax Cube(x) is not a universal but a negation, so you can't use universal elimination on it.
(So what to do? Try negation introduction.)
13.6
> In 13.6, I am very close, but as you
> can probably see
> I need a Tet(a) which I had a hard time proving within one of
> the subproofs.
That's because you can't prove it inside that subproof! At least not directly. The idea of doing disjunction elimination is not a bad one. In one of the subproofs you can prove Tet(a) and use it to get BackOf(a,c) which is what you really want. In the other subproof, you can derive a contradiction and then use contradiction elimination to get what you want.
13.8
> 13.8 I'm not sure how I can use the rules to turn a
> conditional statement into a conjunction, so that's a big
> sticking point.
This may not be quite the right way to think about it -- you can't really turn a conditional into a conjunction (P->Q doesn't imply P & Q). But the conjunction you want is actually the antecedent of a conditional. (If this were propositional logic, you'd need to show that from P->(Q->R) it follows that (P & Q)->R.) It's a little trickier with the quantifiers thrown in, but basically it'll be like proving any other conditional -- you just have to show that *if* P & Q, then R.
13.9
> On 13.9 I keep getting close to the answer,
> but I can never get the right variables in the solution. I
> know I'm close on both of these, but I'll just have to talk
> to you to figure out why I'm not getting them.
Don't forget the two hints! (1) keep in mind that the arguments may not be valid, in which case you need a counterexample rather than a proof; (2) a universally quantified conditional Ax(P(x)->Q(x)) is true if nothing is P (since then the antecedent isn't true for any objects, and a conditional with a false antecedent is true).
13.30
13.31
13.40
Note that the premise is not an existential, it's a conjunction. So you can't use existential elimination on it -- you need to do conjunction elimination to get the existential separately, then do ExElim on that.
13.45
This is *very* close, but you're trying to close off two subproofs at the same time -- can't do that, because each subproof works differently. You have an ExElim subproof inside a negation introduction subproof -- need to conclude the existential elimination subproof first (still inside the negation introduction one), then conclude the negation introduction subproof.
Last update: March 27,
2007. |