All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Your conclusion requested (Posted on 2016-11-03)
i. If there is a king in the hand then there is an ace, or if there isn’t a king in the hand then there is an ace, but not both.
ii. There is a king in the hand.

Given the above premises, what can you infer?

Comments: ( Back to comment list | You must be logged in to post comments.)
 At least one thing we can infer | Comment 8 of 9 |
Let's write C1 for the first condition of premise (i), and let's write C2 for the second condition.
Premise (i) could be written as a negation of a biconditional:
–(C1<->C2)
which is true in case either C1 is true and C2 is false, or vice versa.
A better formalization of this exclusive sense of 'or' is:
((C1 & –C2) or (C2 & –C1))

Now let us write K1 for the first condition of this disjunction, and let us write K2 for the second condition.
Then, in principle we have:

i.    K1 or K2
ii.    K

Therefore: K & (K1 or K2) which is

(K & K1) or (K & K2)

In our case:

i.    [(K -> A) & –(–K -> A)] or [(–K -> A) & –(K -> A)]
ii.    K

Therefore:

iii.
[ K & ((K -> A) & –(–K -> A)) ]  or  [ K & ((–K -> A) & –(K -> A)) ]

We can easily see that the first part (K & K1) leads to the contradiction: A & –A

as we have: K & ((K -> A) & (K -> –A))

The true conclusion of this first part (=false) would be: –(K & K1)

At this point, we can see that the whole statement
–(K & K1) or (K & K2)
would be true and would be the same as
(K & K1) -> (K & K2)
As the first part of this implication would be false, the whole implication would be true, no matter of the truth value of (K & K2).

But we are interested in this detail, that is
K & ((–K -> A) & –(K -> A)),
that is the second part of our disjunctive statement in (iii).

We simplify:

K & ((–K -> A) & –(K -> A))

<-> ( (K & (KvA)) & –(–KvA) )

<-> ( K & (KvA) & (K&–A) )

<-> ( (K&–A) & (KvA) )

<-> ( (K&–A) & (–A->K) )

Suppose, we still do not see the conclusion.
Then we could assume –(K&–A)
Therefore, the negation of –(K&–A) is true,
which is obviously 'king and not ace'.

 Posted by ollie on 2016-11-13 15:38:53

 Search: Search body:
Forums (0)