Let p(x, y) be a propositional function if ꓯyꓱxp(x, y) is true does it necessarily follow that ꓱxꓯyp(x, y) is true? justify your answer or give a counter-example

It is NOT TRUE

Step-by-step explanation:

ꓯyꓱxP(x,y)  

means that for each value of  y there exist x such that P(x,y) is true

whereas

ꓱxꓯyP(x,y)

means that there exists x such that for each value y, P(x,y) is true.

In the second case, the same x must work for every element y.

Counter-example

Consider  

P(x,y) the following proposition

x-y = 0, for x, y integers.

Given an integer y, there is another integer x (namely, x=-y) such that

x-y = 0

so, ꓯyꓱxP(x,y) is TRUE

If ꓱxꓯyP(x,y) were TRUE, then would exist a single unique value of x such that P(x,y) is TRUE for every integer y.

Then P(x,1) and P(x,2) would be both TRUE and

x-1 = 0

x-2 = 0

and we conclude 1=2, which is a contradiction.

So  ꓱxꓯyP(x,y) is NOT TRUE (FALSE)

the answer to your question a or b but i am pretty sure it is b. i hope this : )

step-by-step explanation:

the answer is y < -3 or   y > 3

-3y > 7 + 2

-3 y > 9

y < -3