# Section 2.1: Problem 5 Solution

Working problems is a crucial part of learning mathematics. No one can learn... merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.
James R. Munkres
In 3–8, translate each English sentence into the first-order language specified. (You may want to carry out the translation in several steps, as in some of the examples.) Make full use of the notational conventions and abbreviations to make the end result as readable as possible.
(a) You can fool some of the people all of the time. (b) You can fool all of the people some of the time. (c) You can’t fool all of the people all of the time. ($\forall$ , for all things; $P$ , is a person; $T$ , is a time; $Fxy$ , you can fool $x$ at $y$ . One or more of the above may be ambiguous, in which case you will need more than one translation.)
(a) Here ambiguity is in that it either says that there are some (fixed) people you can fool all time, or says that at every moment there are (some, not fixed) people you can fool, i.e. either $\exists x(Px\wedge\forall y(Ty\rightarrow Fxy))$ or $\forall y(Ty\rightarrow\exists x(Px\wedge Fxy))$ . Note, how the two translations differ in the order of quantifiers, essentially, making the first version stronger than the second one.
(b) Here ambiguity is in that it either says that you can fool each person at some time (times can be different for different people), or says that at some (fixed) time you can fool everyone (at that specific time): $\forall x(Px\rightarrow\exists y(Ty\wedge Fxy))$ or $\exists y(Ty\wedge\forall x(Px\rightarrow Fxy))$ . Again, we have two translations that differ in the order of quantifiers, making the second version stronger than the first one.
(c) $\neg\forall x(Px\rightarrow\forall y(Ty\rightarrow Fxy))$ . Note, that we can put the two quantifiers together and even reorder them in this case, for example, $\neg\forall y\forall x(Px\rightarrow Ty\rightarrow Fxy)$ or $\neg\forall y\forall x(Ty\wedge Px\rightarrow Fxy)$ etc. Alternatively, using that $\neg\forall x(Ax\rightarrow\alpha)$ is “logically equivalent” to $\exists x(Ax\wedge\neg\alpha)$ , we have $\exists x\exists y(Px\wedge Ty\wedge\neg Fxy)$ (but again, similar to Exercise 3, this is a translation of another “logically equivalent” English sentence).