$x,y,z\in\mathbb{N}$ with $0$
$(\forall x)(\forall y)(\exists z)(y \geq x \Rightarrow y=x+z)$
Can you help me with trivial thing above? I imagine it is because I am tired, but I can't see if this is true or false.
I always negate things to test if they are true or false, and $(\exists x)(\exists y)(\forall z)(y \ne x+z \Rightarrow y\lt x)$ seemed true, but so did the original statement
From, Clasic Algebra - Cohn.
Pick any $x, y \in \mathbb N_0$ such that $y \geq x$. Then, let $z= y-x$ so that $y = x+z$. Hence, for every pair of natural numbers $(x, y)$ such that $y \geq x$, there does indeed exist a natural $z$ (namely, $z = y-x$, which depends on $x, y$) for which $y = x + z$. Hence, the posted statement is indeed true.
Your negation of the posted statement is incorrect. It should be $$\begin{align} \exists x \exists y \forall z\Big(\lnot(y\geq x \rightarrow y = x+z)\Big) &\equiv \exists x \exists y \forall z \Big(\lnot(y \lt x \lor y = x+z)\Big)\\ \\ &\equiv \exists x \exists y \forall z \Big( y\geq x \land y \neq x+z\Big) \end{align}$$