# Math Help - Proofs

1. ## Proofs

Are the following valid arguments? Explain using the rules of inference.

(a) If you agree with me, then you will vote for me. You vote for me. Therefore, you agree with me.

(b) Everyone who eats fruit every day is healthy. Christine is not healthy. Therefore, Christine does not eat fruit every day.

(c) If I go swimming, then I will stay in the sun too long. If I stay in the sun too long, I will sunburn. Therefore, if I go swimming, then I will sunburn.

(d) If x is a real number with x > 2, then x^2 > 4. Suppose that x <= 2. Then x^2 <= 4.

Thanks

2. .....................................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 can prove p

It so happens that propositional calculus is a decidable theory in which by using the true tables we can decide whether an argument is provable hence a valid argument.

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 values only but true and false values as well

Hence our argument is not a tautology thus not provable thus not valid

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

Note the above is a loose proof that (a) is not a valid argument

I did not 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 is valid because 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:

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

$\forall x$( x=<2------->x^2=<4).......................................... ......2.

The theorem used to prove the above is:

$\forall a$ $\forall b$[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.

a)
Let p represent the statement "you agree with me."

Let q represent the statement "you will vote for me."

If p then q
q
Therefore p.

If, then represents a conditional statement. This statement is false. This type of incorrect reasoning is called the fallace of affirming the conclusion.

This statement would be valid, by the Modus Ponens rule of inference:

p
If p then q
Therefore q.