Let p be "The object belongs to set A." Let q be "the object belongs to set 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 arguments. Compare your answers.
No A is B.
Some C is A.
Some C is not B.
All B is A.
All C is A
All C is B
In the above given figure the red area is a part of both A and C. But, it is not part of B at all, therefore we can say the statement some C is not B holds.
Now, let r ne an object that belongs to set C .
So, lets make a truth table for the given equation, ...
The validity of the arguments by using Euler Circles are examined. The complete step-by-step solution included in the attached file.