Question

Let P(x,y) denote the sentence x2 + 1≥ x + 1. What are the truth value of the following where the domain of x and y is the set of all integers?

a. ⱯxⱯyP(x,y)

b. ⱯxƎyP(x,y)

c. ƎxⱯyP(x,y) .

d. ƎxƎyP(x,y)

Answer 1

As i understand, there is a mistake in condition, and P(x,y) denote the sentence $x^2+ 1≥ y + 1$ , otherwise it is independent from y, which doesn't make much sence.

a. ⱯxⱯyP(x,y). For x = 0 and y = 1 we have $0^2+1≥1+1\implies 1≥2$ , which is false

This statement is false

b. ⱯxƎyP(x,y). Let $x=a, a\isin Z$ , then $a^2+1≥y+1\implies y≤a^2$ , which means we can find such y, for exmaple, $y=a^2\in Z$ , that P(x, y) is true

This statement is true

c. ƎxⱯyP(x,y). Let $x=a, a\isin Z$ , then $a^2+1≥y+1\implies y≤a^2$, but we can put, for example,

$y=a^2+1$ , which means P(x, y) would be false

So, this statemnet is false

d. ƎxƎyP(x,y). For x = 0 and y = 0 we have $0^2+1=0+1\implies 1=1$ . P(x, y) is true

This statement is true. Also this statement implies from the true statement (b)