Recent content by Knark

That's what I wasn't sure of, hence my first attempts included no use of Universal or Existential Quantifiers. I believe it actually wants just a formula and not a sentence since question #3 of our homework is the question that specifically states "Formalize the two SENTENCES". It makes sense...

(∃x)(∀y)[~Fx (∃w1)(Sxw1 & (∃z1)(Pw1z1 & (∃w)(Pz1w & (∃z)(Swz & Pzy))))] Did you only use w1? My answer needs 6 variables so I added z1 to account for that. Good catch on the parentheses, are they properly closed now?

Does this seem similar to your answer? (∃x)(∀y)[~Fx (∃u)(Sxu & (∃v)(Puv & (∃w)(Pvw & (∃z)(Swz & Pzy)] I'm worried because I had to use u, v, w, x, y, z to fill in all the needed people but I seem to remember something about only being allowed to use w, x, y, z. Did you use 6 variables also...

Woops, in c) we have "~Fx" but it should actually be "Fx" since x is an aunt. What did you get for e)?

I like it. So for b) and d) something like: b: (∃x)(∀y)[Fx & (∃w)(Swx & (∃z)(Pwz & Pzy)] d: (∃x)(∃y)[~Fx & (∃w)(Swx & (∃z)(Pwz & Pzy)]

How would you contrast your answer for a) with the answer for c). Would you switch up the quantifiers into something like: (∃y)(∀x)[~Fx & (∃z)(Sxz & Pzy)] How exactly would you show that "x is an aunt" and not "x is an aunt of y".

Well every thing that can be done using SD+ can also be done using SD, it would simply take a little more time. But I believe you are right and that it should be done in only SD so I will rework it. Thanks for catching that.

So I would have to add in (∃y)?

Homework Statement Formalize (in PL) the relations/predicates stated in (a)-(e) using just these relations/predicates: 1) Pxy: x is a parent of y 2) Fx: x is a female 3) Sxy: x is a sibling of y (a) x is an uncle of y (b) x is a great-aunt of y (c) x is an aunt (d) x is a great-uncle (e) x...

Since he already made an attempt to find the solution to the first question I'm going to help out and post my answer for you guys to help me out and him at the same time. Question: Show that [(A -> B) -> A] -> A is a theorem of SD. Relevant Equations: All rules of SD found in the Logic Book...

Since he already made an attempt to find the solution to the first question I'm going to help out and post my answer for you guys to help me out and him at the same time. Question: Show that [(A -> B) -> A] -> A is a theorem of SD. Relevant Equations: All rules of SD found in the Logic...

I get the feeling that I am in this person's class. If so here is the question: Suppose we dropped from SD the rule for vE (Disjunction Elimination), and adopted in its place the rule of Disjunctive Syllogism, thus giving a modified systemd of SD*. Show that you can derive the SD rule for vE...