# Homework Help: Give the Formal Proof

1. May 22, 2009

### tgt

1. The problem statement, all variables and given/known data
GIve a formal proof or derivation of:

(a iff (b iff c)) iff ((a iff b) iff c))

3. The attempt at a solution
I've tried it for a long time. Idea is to assume (a iff (b iff c)) then get ((a iff b) iff c)) and then vice versa. BUt no matter how I try, I always get some if conditions istead of the iff.

Last edited: May 22, 2009
2. May 22, 2009

### Hurkyl

Staff Emeritus
Well, the most brain-dead way is to write down a truth table, and invoke the fact that a proposition is a tautology if and only if all of the entries in its truth table are "true".

3. May 22, 2009

### tgt

The aim was to use deduction rules to prove it. In other words to deduce it all. That is the method I'm trying to use to prove it.

4. May 22, 2009

### tgt

Just thinking about it, deductions are always tautologies. i.e if one is asked to deduce B from A, A and B are the same statement but in different form.

5. May 22, 2009

### Hurkyl

Staff Emeritus
Not so. The simplest counterexample is probably to let A be "contradiction" and to let B be "tautology". You can deduce B from A, but I doubt you could argue that they are "the same statement but in a different form".

(I chose contradiction and tautology because they appear in the two trivial proofs: you can deduce anything from a contradiction, and from anything you can deduce a tautology)

6. May 23, 2009

### tgt

How about the solution to the OP? It seems impossible but it is so obvious.

7. May 24, 2009

### tgt

What do you mean by being able to deduce anything from a contradiction?

Is that this: Suppose we have a contradiction then we add the negation of the thing we are trying to deduce at the beginning of the contradiction. Then the end result is that we have deduced the thing we are trying to prove.

i.e Assume

Tim does not have a bread

But we could just as easily prove John is dead by reversing things.

8. May 24, 2009

### Hurkyl

Staff Emeritus
That is exactly correct. Such is the nature of contradictions.

9. May 24, 2009

### matt grime

Just use the definition:

A=>B is not(A) OR B

Plug things in and reduce both sides to some canonical form.

10. May 24, 2009

### tgt

Phil is gone,
Tim does not have a bread

Now we could choose between concluding
John is not dead and Phil is not gone?

Any choice would be okay?

Suppose there are finitely many primes,
Conclude a contradiction that the product of those primes plus 1 cannot be a prime nor not a prime.

Now we can either conclude John is not dead or There are infinitely many primes. EIther choice being correct?

pi is an irrational number
Suppose there are finitely many primes,
Conclude a contradiction that the product of those primes plus 1 cannot be a prime nor not a prime.

Now we can either conclude pi is rational or There are infinitely many primes. EIther choice being correct?

11. May 24, 2009

### tgt

I've worked out the problem using the technique of deduction.

12. May 24, 2009

### Staff: Mentor

This seems to me to be nonsense, not a contradiction. Presumably, the way this is laid out, the first three statements are the hypothesis, and the fourth is the conclusion. Because Phil being gone and John being dead have nothing to do with Tim having bread, it seems to me that this can be reduced to
This is not a contradiction. If Tim has bread, then it is not true that he doesn't have bread, so the first statement does not imply the second. If Tim indeed doesn't have bread, then the first statement isn't true, so any statement at all could be appear as the conclusion, and the implication would be true, but meaningless.

BTW, in English, we don't say someone has "a bread": we say someone has bread or has a loaf of bread.

13. May 24, 2009

### tgt

You are misunderstood. All 4 statements are intended to be the hypothesis which we arrive at a contradiction. The conclusion is that one of the hypothesis was wrong.

Last edited: May 25, 2009
14. May 25, 2009

### Staff: Mentor

Maybe I misunderstood, but I wasn't misunderstood, since you apparently understood what I was saying.

It wasn't clear to me which statements were the hypothesis, and which were the conclusion. Going back to your syllogism, let's throw out the statements about Phil and John, since they don't have anything to do with anything. That leaves us with a hypothesis of (Tim has bread) AND (Tim does not have bread). This hypothesis can never be true, so you could have any conclusion, and overall the implication would be true.

For any implication, p ==> q, the implication will be true or valid under three circumstances, and false or invalid under one circumstance. The three true circumstances are:
1. p is true and q is true.
2. p is false and q is true.
3. p is false and q is false.

The invalid circumstance is:
4. p is true and q is false.

Depending on what you have for the conclusion, your syllogism of Tim having and not having bread falls into the 2nd or 3rd category.

15. May 25, 2009

### Hurkyl

Staff Emeritus
tgt: you seem to be confusing this example of a vacuous proof with proof by contradiction.

Here are three valid arguments:

Tim does not have a bread.

Tim does not have a bread.
Therefore, John is alive.

Phil is gone implies Tim has a bread.
Phil is gone imples Tim does not have a bread.
Therefore, Phil is not gone.

The first two are an example of "contradiction implies anything". The third is an example of proof by contradiction.

16. May 25, 2009

### tgt

Well, I understood that you misunderstood which seems valid.

I do acknowledge that my presentation of arguments was not clear.

17. May 25, 2009

### tgt

Wouldn't your proof by contradiction need the extra assumption to get the contradiction. So it would be better if it was:

1. Phil is gone
2. Phil is gone implies Tim has a bread.
3. Phil is gone imples Tim does not have a bread.
Therefore, Phil is not gone.

Where 1,2 and 3 are assumptions.

18. May 25, 2009

### Hurkyl

Staff Emeritus
Heavens no! Those three statements are a contradiction!

19. May 25, 2009

### tgt

We are trying to deduce a contradiction which is what we get after combining those three statements.

To give more detail it's

If:
Phil is gone
Phil is gone implies Tim has a bread.
Phil is gone imples Tim does not have a bread.

Then:
Tim does not have a bread

Conclusion:
One or more of our assumptions was wrong (or is it only one assumption was wrong?)
We say that Phil is not gone.

But couldn't we conclude that the statement 'Phil is gone implies Tim has a bread' is incorrect hence the reverse of that statement is true instead?

20. May 25, 2009

### tgt

With regards to the OP. I realised that I've made a mistake. The technique is to use natural deduction to prove the statement.

http://www.ags.uni-sb.de/~chris/papers/2002-pisa.pdf [Broken]

Last edited by a moderator: May 4, 2017
21. May 25, 2009

### Hurkyl

Staff Emeritus
But look at your argument form:

{ phil is gone and (phil is gone implies tim has a bread) and (phil is gone implies tim does not have a bread) } implies contradiction

Therefore, not { phil is gone and (phil is gone implies tim has a bread) and (phil is gone implies tim does not have a bread) }

That's exactly the form I'm describing:

Therefore, not P

22. May 25, 2009

### tgt

What would not { phil is gone and (phil is gone implies tim has a bread) and (phil is gone implies tim does not have a bread) } be?

Would it be one of { phil is not gone or not(phil is gone implies tim has a bread) or not(phil is gone implies tim does not have a bread) }?

So we have 3 choices to choose from?

Your way of stating things "P implies contradiction Therefore, not P" is a bit too broad. Often we like to conclude an assumption in P which is wrong which is also the thing we are trying to prove.

23. May 25, 2009

### Hurkyl

Staff Emeritus
Right; DeMorgan's laws: not(P and Q and R) = not P or not Q or not R.

Sure, but that's the basic building block. e.g. we could proceed as:

P
Therefore not (P and Q)
Therefore not P or not Q
Therefore P implies not Q
Therefore not Q

(Of course, the actual chain of logic would depend both on which set of rules you are using, and the particular route you decided to use)

24. May 25, 2009

### Hurkyl

Staff Emeritus
I suppose if you wanted to keep explicit track of when you're just using the logical connective and when you're talking about the existence of a proof, you could argue something like:

($\vdash$ is the symbol that means "there is a proof that starts with the left hand side and concludes with the right hand side")

($\bot$ is a symbol denoting a contradiction)

We have a metalogical theorem that
$P, Q \vdash R$ if and only if $P \vdash Q \to R$​

First, prove $P,Q \vdash \bot$
Then invoke the theorem to conclude $P \vdash Q \to \bot$
And apply the law of noncontradiction to get $P \vdash \neg Q$