T(x,y) = x has taken class y.
Statement: Every student has taken at least one math class besides Math 101.
My initial thought was the following:
- $$\forall x \exists y : (y\neq Math 101) \rightarrow T(x,y)$$
but discussion has lead to the following:
- $$\forall x \exists y: (y\neq Math101) \land T(x,y)$$
From my understanding 2) would be incorrect because it is implying that no student took Math 101 where the statement says nothing about whether or not a student took math 101. Does this make sense.
I could also see 1) being incorrect as the class not being Math 101 may not necessarily mean that the student took the class?
That 2) doesn't say that no student took Math 101; for $y=Math101$, $(y\ne Math101)$ is false and the result of $T(x,y)$ is irrelevant.
Then 2) would mean there exists a different class $y$ such that $(y\ne Math101)\wedge T(x,y)$. $(y\ne Math101)$ is true for the different class $y$, and $T(x,y)$ has to be true for that $y$ too.
That 1) is also incorrect in the way that, if $Math101$ is among the classes that $y$ can be, then $y=Math101$ makes the implication $(y\ne Math101)\to T(x,y)$ true, regardless of whether $T(x,Math101)$. Then $y=Math101$ satisfies the $\exists y$, and other classes $y$ becomes irrelevant.