Truth tables are related to Euler circles. Arguments in the form of Euler circles can be translated into statements using the basic
connectives and the negation as follows;
Let p be “The object belongs to set A.” Let q be “the object belongs to B.”
All A is B is equivalent to p > q.
No A is B is equivalent to P > ~q.
Some A is B is equivalent to P ^ q.
Some A is not B is equivalent to P ^ ~ q.
Determine the validity of the next arguments by using Euler circles, then translate the statements into logical statements using the
basic connectives and using truth tables, determine the validity of the argument. Compare your answers
(a) No A is B
Some C is A
__________
.
. . Some C is not B
(b) All B is A
All C is A
___________
.
. . All C is B
Solution:
Truth table problem
Let p be “The object belongs to set A.” Let q be “the object belongs to B.”
Let r be “The object belongs to C.”
All A is B is equivalent to p  q .
No A is B is equivalent to p   q.
Some A is B is equivalent to p  q .
Some A is not B is equivalent to p  q
(A) When we use Euler’s circles for the statements No A is B, Some C is A as in the diagram shown below,we see that
the portion colored in green is not B, specifically the region colored black is not B. Hence the argument is valid.
Now the given statements can be translated as logical statements as
p  q, r  p  r  q
We prove the validity by truth table
p q r
q
p  q
r p
r  q
( p  q ) 
( p  q )  ( r  p )  ( r  q )
(r  p)
T
T
T
F
F
T
F
F
T
T
T
F
F
F
F
F
F
T
T
F
T
T
T
T
T
T
T
T
F
F
T
T
F
F
F
T
F
T
T
F
T
F
F
F
T
F
T
F
F
T
F
F
F
T
F
F
T
T
T
F
T
F
T
F
F
F
T
T
F
F
F
T
This truth table shows that the argument is valid.
Note that the argument is valid in both methods.
(B)When we use Euler’s circles for the statements all B is A and all C is A, we get the diagram as shown below. The
colored region shows “all C is B” is invalid.
A
B
C
Now to prove by logical statements, we have prove
q  p, r  p does not imply r  q
That is we have to prove (q  p)  (r  p)  (r  q) is not a tautology.
For that the truth table is
p q r
q p
rp
(q  p)  (r  p)
rq
(q  p)  (r  p)  (r  q)
T
T
T
T
T
T
T
T
T
T
F
T
T
T
T
T
T
F
T
T
T
T
F
F
T
F
F
T
T
T
T
T
F
T
T
F
F
F
T
T
F
T
F
F
T
F
T
T
F
F
T
T
F
F
F
T
F
F
F
T
T
T
T
T
This shows that it is not a tautology and hence the argument is invalid.
In both methods we get the argument is invalid.