.....................................for (a)

If we put:

you agree with me =p, and you will vote for me=q

after that the form of your argument is:

(p----->q & q)====>p i.e the hypothesis p----->q & q logically implies p ,or by using the given hypothesis we canprove p

It so happens thatpropositional calculus is a decidable theory in which by using the true tables we can decide whether an argument isprovable hence a validargument.

But in our case if we form the true table of the said argument under the column of (p----->q & q)====>p you will not get true valuesonly but true and false values as well

Hence our argument is not atautologythus notprovable thus not valid

Howeverhad your argument had the form:

you agree with me if and only if you will vote for me ,then it would be avalid one

Notethe above is a looseproofthat (a) is not avalid argument

I didnot use the appropriate theorems in predicate calculus because the whole case could become a Little more complicated.

BUT if you want me to i Will do so

.......................................for (c)............................................... .....

this argument isvalidbecause it is a simple application of the hypothetical syllogism :

p----->q & q------>r hence p------>r.

where ;P=I go swimming.......q=i will stay in the sun too long.............r=i will sunburn

Now to prove the hypothetical syllogism:

suppose p................................................. ..........................1

from p---->q & p by using M,Ponens we get : q.............................2

from q----->r & q from line 2 and using again M.Ponens we get: r...............................3

Now by using the rule of conditional proof applied to lines from 1 to 3 we get:

......................................p-------->r............................................

.......................................for(d)..... ...............................................

Here we have two small proofs in predicate calculus:

( 2<x------> 4<x^2) and..................................1

( x=<2------->x^2=<4).......................................... ......2.

The theorem used to prove the above is:

[a>0 ----->(a<b<------->a^2<b^2)]

According to that we can easily conclude that 2 is not valid and 1 is.

For example if in 2 we put x = -3 then x^2=9>=4

I will leave the proofs or disproofs of the above and those of question (b) to the forum.

Finally i am surprised to see that you ask such simple questions in predicate calculus while other more complicated questions about predicate calculus are being attended to.

Perhaps have escaped your attention