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?

A shorter and simpler way than the one that I sent some days ago is as follows:

Premise (i):    NOT ( (king -> ace) AND (NOT king -> ace) )
Premise (ii):    king

gives:    (iii):    (king & -(king -> ace)) OR (king & -(-king -> ace))


(iii):    (K & -(K -> A)) v (K & -(-K -> A))
<->    ((K & -(-K v A)) v (K & -(K v A)))
<->    (K & (K & -A)) v (K & (-K & -A))
<->    ((K & K) & (K & -A)) v ((K & -K) & (K & -A))    

The contradiction '(K & -K)' in the 2nd disjunct makes the whole &-condition of that disjunct FALSE.

<->    (K & (K & ~A)) v FALSE
<->    ((K & K) & (K & ~A)) v FALSE)
<->    ((K & ~A) v FALSE)
<->    ((FALSE v K) & (FALSE v ~A))
<->    K&-A

I liked this problem. It is contra-intuitive and sets a good trap.

Posted by ollie on 2016-11-17 13:32:58