logic 3 problems?

Question

1. prove p --> p is a tautology

2. prove [(p-->q)^p]-->q is a tautology

3.prove [(p-->q)^p]-->q is NOT a tautology

done in truth table
note: --> is a conditional and ^ is a conjunction

Answer 1

note: !A is (not A), v is a disjunction, ^ is conjunction
tautology: something that is independently true, or always true

1. A --> B is logically equivalent to (B v !A). Make a truth table for A --> B and for (B v !A) to check this.

So p --> p is logically equivalent to (p v !p), which is always true, regardless of whether p is true.

2. For this problem, if you don't know the rules of symbolic logic, you can use truth tables to prove to yourself that two things are logically equivalent.
Follow these steps to prove [(p-->q)^p]-->q is a tautology

[(p-->q)^p]-->q
q v ![(q v !p) ^ p] conditional to disjunction
q v [!(q v !p) v !p] DeMorgan
q v [(!q ^ p) v !p] DeMorgan
q v [ (!q v !p) ^ (p v !p)] Distribution of conj. and disj.
q v (!q v !p) Eliminate tautology ^(p v !p)
(q v !q) v (q v !p) Distribution of disjunctions
(true) v (q v !p) Identify tautology (q v !q) v
(true) (true) v (anything) is always true

Since all those statements are equivalent, then [(p--q is logically equivalent to true, and so is a tautology.

3. This is the exact same statement as in (2), so you can't prove that it is not a tautology, since it is one. Did you mean to type a different statement here?

Answer 2

A tautology is something that is always true, so if you draw out the truth tables for each, and end up with T in every box, that's a tautology.

The truth table for ^ is
p q p^q
f f f
f t f
t f f
t t t

The truth table for p->q is the same as for (q or ~p).
So do the table for ~p first, then or it with q.

Does this help?

Answer 3

2 and 3 can't both be tautologies. The argument is the same but the conclusions are opposite.

Answer 4

I havn't studied tautology but i'll give it a shot.

#1 is a tautology because whatever p is, the other p will have to be the same as it.

I think #2 can be rewritten as:
[(p-->q)^p]-->q
[pp-->pq]-->q
q-->q ...'p's cancel each other out
same rule as number one then applies

#3
alternatively:
[(p-->q)^p]-->q
[pp-->pq]-->q
p-->q-->q ...only one 'p' will cancel out