Formalise the following statements Let $P$ stand for the set of people and let $p \in P$. $C(p)$ is a propositional function that is true when person $p$ plays cricket; $R(p)$ is a propositional function that is true when $p$ plays rugby; and $F(p)$ is true when $p$ plays football. Formalise the following statements:

*

*Someone plays football and rugby.

*No-one plays rugby and cricket.

*If someone plays cricket, then that person also plays either
football or rugby.

*Everyone either plays cricket and football, or they play no sport at all.

1st try:

*

*$p \rightarrow (F(p) \wedge R(p)$

*$(R(p) \wedge C(p))'$

*$C(p) \rightarrow (F(p) \vee R(p))$

*$(C(p) \wedge F(p)) \rightarrow (C(p) \wedge F(p) \wedge R(p))$
2nd try:

*

*$\exists p \in P, F(p) \wedge R(p)$


*$\forall p \in P, (R(p) \wedge C(p))’$
 A: Let us study this question more deeply, in order to make explicit the kind of reasoning that it involves. In this way, you will be able to apply the same argument to similar questions in your further studying.
Let $P$ be the set of all people and $C(p)$, $R(p)$ and $F(p)$ be the following conditions:

*

*$C(p)$: “$p$ plays cricket.”

*$R(p)$: “$p$ plays rugby.”

*$F(p)$: “$p$ plays football.”

The first statement states “Someone plays football and rugby.”. So, we immediately conclude that there is a person, say $p_0$, that plays both football and rugby, i.e. $F(p_0) \wedge R(p_0)$. This clearly involves the use of the existential quantifier, $\exists$ (in fact, we can even [informally] says that this is its definition, in the sense that states that some object exists). Hence, to say that exists a person that plays both football and rugby we would write, symbolically, $\exists p \in P (F(p) \wedge R(p))$.
The second statement is “No one plays rugby and cricket.”. Let’s pick an arbitrary person from $P$, say $p_1$. We know, by our statement, that $p_1$ doesn’t play rugby and doesn’t play cricket, i.e. $\neg R(p_1) \wedge \neg C(p_1)$. Note that we pick an arbitrary person $p_1$, and that our reasoning applies to any other person from $P$ (since we are here dealing with the universal quantifier, $\forall$). In short, $\neg R(p’) \wedge \neg C(p’)$ would happen for any person $p’ \in P$ that we pick. Therefore, we write this statement, symbolically, as $\forall p (\neg R(p) \wedge \neg C(p))$.
I believe that this discussion is sufficient for you to try the statements $3$ and $4$. Also, note that the $3$ and $4$ involved implication and that “implication presupposes universal quantification”. Let me illustrate this with an example. Let $C$ be the set of all cats and let $O(c)$ and $J(c)$ be the following conditions:

*

*$O(c)$: “$c$ is orange.”

*$J(c)$: “$c$ jumps a lot.”

Remark. Suppose that we say $\forall c (O(c) \rightarrow J(c))$. This means that “Given any cat, if its orange then it jumps a lot.”. But we could also as well say that “If a cat is orange, then it jumps a lot.”. Note that the last statement would be written symbolically as $O(c) \rightarrow J(c)$. Informally, and without loss of rigour for the purpose of this problem, we can avoid the use of $\forall$. From your post, I can see that this is exactly what you have done, but I wrote this in order to make you aware of this details.
I hope this helps you in your study.
