## Logical Paradox

Posted by luv2fap on February 18, 2011 – 2:06 pm

A well{known logical joke is a card saying on one side “The statement

on the other side is false” and on the other, “The statement on the

other side is true”. Clearly, it is impossible for either of the statements

to be true or false without leading to an inconsistency. However, if

both sides said “The statement on the other side is false” or “The

statement on the other side is true”, it is possible for these statements

to be logically consistent.

Now suppose you are given a stack of n cards, each of which says either

“The statement on the next card is false” or “The statement on the

next card is true”. Is it possible to arrange the cards into an order

which makes the whole pack logically consistent?

February 18th, 2011 at 3:13 pm

I would only say no as the last card will not have a following card to validate the statement.

If n however, is infinite, then yes.

February 18th, 2011 at 3:20 pm

I’m going to guess you mean that the final card will refer you to this first (circular referencing).

In this case, as long as when you see a “false” card, you follow it with another “false” you should be ok.

could look something like this:

the statement below is true

the statement below is true

the statement below is true

the statement below is false

the statement below is false

the statement below is true

the statement below is false

the statement below is false

the first statement is true

February 18th, 2011 at 9:47 pm

Using the shorthand t and f denote the literal message printed on the cards, t => “the next is true”, f => “the next is false”, and T and F to represent the cumulative message that is effectively printed on the card.

Then Tf => TF => F, Tt => Tt => T, Ff => FT => T, Ft => FF => F

where the single card is the net effect of the two cards.

It is pretty obvious that T’s have no effect on the state of the effective messages on the subsequent cards, and that each f reverse the effective message on the subsequent cards.

To determine the effective message on a given card, count how many f cards there are preceding it. If #f is an even number, then the effective message = the actual message and if #f is an odd number, the effective message is the reverse of the actual message.

If the deck is not circular, any sequence of cards is valid, as no contradictions arise. If the deck is circular, then it only seems sensible that each card has a unique state i.e. it must be the same every time you go round the loop. So there must be an even number of f cards. That conveniently means that it doesn’t matter where you start in the circle provided you take the first card to be literally telling the truth.

It gets a bit silly if you assume the first card is preceded by an F for the circular case.

February 18th, 2011 at 9:51 pm

Rats. Third para, first line should have been t’s not T’s

February 18th, 2011 at 10:44 pm

Double rats. Second para, I wrote a Tt when I meant a TT.

In the circular case, if you start on a t (and assume it’s a T) then you’ll get one set of effective messages. But if you start on an f and assume it’s an F, all the effective messages will be the reverse of the t starting case. But as long as the number of f’s is even, there are no inconsistencies.

I think I’ve got that right; I’m too tired to think straight.

February 20th, 2011 at 6:29 pm

We assume that the last card in an arrangement refers to the rst

card. We are going to show that regardless of the order, the cards can

be logically consistent if and only if the number of cards which state

that the next card is false is even.

Suppose we have the deck of cards together with some consistent assignment

of truth value to each card. Label each card by T if it says

that the statement on the next card is true, and by F if it says that

the statement on the next card is false. Also label each card by t if it

is true, and by f if it is false. Thus each card has one of the labels Tt,

Tf, Ft, or Ff.

We rst show that any card labelled T can be removed from the deck

without altering the remaining labels. If it is labelled Tt, the preceding

card must be either Tt or Ff and the next card must be either Tt or

Ft. Thus removing the given Tt card does not change anything in the

rest. If it is labelled Tf, then the preceding card must be labelled Tf or

Ft and next card must be either Ft or Ff. Once again, removing the

given Tf card does not change the rest.

Hence we now assume that the remaining cards are all labelled Ft or

Ff. Next we show that removing a pair of adjacent F cards does not

alter the remaining labels. First note that a pair (Ft, Ft) cannot be

consistent, since Ft truthfully claims that the statement on the next

card is false. Similarly, (Ff, Ff) cannot be consistent, since Ff falsely

claims that the statement on the next card is false, so that statement

must be true. Hence any adjacent pair must be (Ft, Ff), in which case

the preceding card is Ff and the next card is Ft; or (Ff, Ft) in which

case the preceding card is Ft and the next card is Ff. In both cases,

the pair can be removed.

February 20th, 2011 at 6:30 pm

*Thus we finally reach a pair (Ft, Ff) or (Ff, Ft) which are consistent, or

a single Ft or Ff which corresponds to `The statement on the other side

is false’ which is inconsistent, a contradiction. Hence the original deck

was consistent if there are an even number of F cards, and inconsistent

if there are an odd number.

February 24th, 2011 at 9:01 am

No, the N+1 card will never exist and therefore a statement about it is implicitly false.