The truth value of each(Domain set of values for both x and y is the set of real numbers):
a) $\neg(\forall x \exists y \forall z(x+y=z))$
(b)$\forall x(1+2+.....x=\frac{x(x+1)}{2})$
(c) $\forall x(1^2+2^2+.....x^2=\frac{x(x+1)(2x+1)}{6})$
(d)$\exists x \forall y(1+2+3...+x=y)$
my attempt:
(b) is true
(c) is true
(d) is false
i am confusing with (a) can any one help me with (a) please ..thanks
I suspect the intended domain is the natural numbers, rather than the real numbers, so let's assume this is about the natiral numbers.
As such, b) and c) are well-known (and thus indeed true) statements about the sum of the first $x$ numbers and the sum of squares of the first $x$ numbers respectively.
And you're also right that d) is false: for whatever $x$ you take, the sum of the first $x$ numbers is a specific number, so it doesn't make sense it would equal all numbers.
Finally, for a), it is easier to first think of the statement without the negation, i.e. $\forall x \exists y \forall z \ (x + y =z)$. That statement has the basically the same problem as d), since whatever the values of $x$ and $y$ are, their sum is of course a specific number, and hence is not equal to all numbers. So, without the negation, the statement is false. But that means that with the negation, the statement is true.