Translating into first order predicate expressions

[Id1emind]

Homework Statement

I have been given the following predicates of the domain of real numbers

Homework Equations

P(x): x>0 E(x): x is even D(x): x is exactly divisable by 5

(i) At least one integer is even
(ii) There exists a positive integer that is even
(iii) If x is even then x is not exactly divisible by 5
(iv) There exists an even integer exactly divisable by 5

The Attempt at a Solution

(i) Ax(E(x))
(ii) Ex(P(x) ^ E(x))
(iii) Ax(P(x) ¬ D(x))
(iv) Ax(P(x) U D(x))

Any help would be greatly appreciated, as I am not sure if the above is correct.

Many Thanks
Steve

 

[mighty2000]

Dear Steve,

Your attempt at a solution is correct. Let me explain in more detail why each statement is true:

(i) Ax(E(x)) - This statement means "For all x, x is even." Since we are dealing with the domain of real numbers, there will always be at least one integer that is even, so this statement is true.

(ii) Ex(P(x) ^ E(x)) - This statement means "There exists an x such that x is positive and even." Since we know that there are positive integers and even integers, there will always be a positive even integer in the domain of real numbers, making this statement true.

(iii) Ax(P(x) ¬ D(x)) - This statement means "For all x, if x is even, then x is not exactly divisible by 5." Since we are given the predicate D(x), which means "x is exactly divisible by 5," we know that if x is even, it cannot also be exactly divisible by 5. Therefore, this statement is true.

(iv) Ax(P(x) U D(x)) - This statement means "For all x, x is even or x is exactly divisible by 5." Since we are given the predicates P(x) and D(x), which both deal with integers, we know that there will always be an even integer that is also exactly divisible by 5. Therefore, this statement is true.

I hope this helps clarify your understanding of these statements.


(Scientist)