MadeEasy Test Series: Mathematical Logic - First Order Logic

Question

Match the following Lists

List-I

A. There are atmost two apples.

B. There are exactly two apples.

C. There is atmost one apple.

D. There is exactly one apple.

List-II

1. $\forall x \forall y \forall z ((Apple (x) \wedge Apple (y) \wedge Apple (z)) \rightarrow (x=y \vee x=z \vee y=z))$

2. $\forall x \forall y ((Apple (x) \wedge Apple (y)) \rightarrow (x=y \vee y=x))$

3. $\exists x \exists y (Apple (x) \wedge Apple (y) \wedge (x \neq y) \wedge \forall z (Apple (z) \rightarrow ((z=x) \vee (z=y))))$

4. $\exists x (Apple (x) \wedge \forall y (Apple (y) \rightarrow (x=y)))$

Codes:

        A    B   C   D

(a)    1     2    3   4

(b)    3     2    1   4

(c)    1     3    2   4

(d)    3     1    2   4

1. $a$
2. $b$
3. $c$
4. $d$

Answer 1

option C should be the correct answer.

Apple(p): p is an apple.

1) For all x, y & z, if (x is an apple and y is an apple and z is an apple), then (either x is same as y or x is same as z or y is same as z).It means we have at most 2 apples.

2) For all x & y if (x is an apple and y is an apple) then (either x is same as y or y is same as x).It means we have at most 1 apple. here x & y need not to apple always but "if" both of them are apples then they are same apples, it says that we may or may not have apples but if we have then we can not have more than 1 apple.

3) There exists an x & y such that x is an apple and y is an apple and x & y are different apples and for every z if z is an apple then either z is same as x or z is same as y.It means we have exactly 2 apples.

4) There exists an x such that x is an apple and for every y if y is an apple then x & y are same. It means we have exactly 1 apple.

Comments

wow.. Thank you radha.:) :)

Answer 2

option C

Comments

c is correct...