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?

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)
which leads to a contradiction.
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