# Negating $w_t \cdot x < w_t\cdot x_t \text{ for some } t \text{ s.t. } y_t> y$, does the rightmost inequality become $\leq$?

Negating $w_t \cdot x < w_t\cdot x_t \text{ for some } t \text{ s.t. } y_t> y$, does the rightmost inequality become $\leq$?

So I believe I can write this as: $$\exists t \ni y_t>y, x\cdot w_t < x_t \cdot w_t$$ for which I believe the negation is $$\tag{1} \forall t \ni y_t>y, x\cdot w_t \geq x_t \cdot w_t$$ What I am asking in this question is whether $(1)$ is the correct negation, or if the correct negation is $$\tag{2} \forall t \ni y_t\leq y, x\cdot w_t \geq x_t \cdot w_t$$ (or something else). The reason I am a bit confused is because I think where I have commas I should have ands ($\wedge$), which would then become ors, in which case I think that $(2)$ may actually be the correct negation. (but I tried writing the truth tables and having an and ($\wedge$) with the $\forall$ doesn't make sense).

Thanks.

So your initial statement is "there exists $t$ with $y_t>y$ such that $w_t\cdot x<w_t\cdot x_t$".
The negation is "there DOESN'T EXIST $t$ with $y_t>y$ such that $w_t\cdot x<w_t\cdot x_t$".
Therefore, "for any $t$ with $y_t>y$ that you try, you will have $w_t\cdot x\geq w_t\cdot x_t$".
So, formally, "$\forall t$ s.t. $y_t>y$, $w_t\cdot x\geq w_t\cdot x_t$".
The statement is not saying anything about $t$ in which $y_t\leq y$.