# Logic Question - Why is This an Implication?

I have a question about predicate logic. Suppose we have the following predicates:

$\text{Study}(x,y)$: x studies y

$\text{Comp}(x)$: x is a computing student

I want to encode the following sentence in predicate logic: "Some, but not all computer students study logic."

A potential answer is:

$$\exists x(\text{Comp}(x)\land \text{Study}(x,l))\land\neg \forall x(\text{Comp}(x)\implies \text{Study}(x,l))$$

Why is there an $\implies$ and not a $\land$? Is this formulation correct?

• A useful thing to remember is that usually the $\exists$ is using a $\land$ where the $\forall$ uses a $\rightarrow$. There is a philosopher who drinks vs. all philosophers drink: $\exists x P(x) \land D(x)$ vs $\forall x P(x) \rightarrow D(x)$. Note that $\forall x P(x) \land D(x)$ means that everybody are drinking philosophers whereas $\exists P(x) \rightarrow D(x)$ is true if there are no philosophers! – Pål GD Feb 11 '14 at 12:04

## 1 Answer

Because $\neg\forall x\,(\text{Comp}(x) \wedge \text{Study}(x,l))$ means "It is not true that every student is both a computing student and studying logic." In particular, that would be true if there is at least one student who is not a computing student, regardless of whether all computing students do or do not study logic.

• Thank you, but how would the first segment differ in meaning if there was an implication instead of an and? Eg if the first segment said ∃x(Comp(x) → Study(x,l)) – Natu Myers Feb 11 '14 at 1:05
• I think you need to revise the definition of implication ($X\rightarrow Y$ means that either $X$ is false or $Y$ is true or both.) – David Richerby Feb 11 '14 at 1:07