Writing proposition with connectives and laws of logic Question 1): Pei Ann has been dealt two cards from a standard 52 card deck. She holds one in her left hand and one in her right.
Let $p$ be the proposition "The card in Pei Ann's left hand is an ace".
Let $q$ be the proposition "The card in Pei Ann's right hand is an ace"
Let $r$ be the proposition "The card in Pei Ann's left hand is a club".
Let $s$ be the proposition "The card in Pei Ann's right hand is a club".
Write propositions (using just $p, q, r, s$ and logical connectives) corresponding to the following sentences.


*

*Pei Ann doesn't have two aces.

*Pei Ann has at least one club.

*Pei Ann has the ace of clubs and another club.


Question :2) Show that $(\neg p \lor \neg q) \rightarrow \neg q \equiv p \lor \neg q$ using the laws of logic.
These are two of the questions for my quiz that I had trouble doing.
For question number (1) I am not sure what a standard 52 card deck is, not a cards fan. 
I could solve it for the first sentence "Pei Ann doesn't have two aces" I think. I came with $\neg (p \land q)$. Is it correct? For the second sentence I am not exactly sure. Could it be $(p \rightarrow r) \lor (q \rightarrow s)$?
For the third sentence I have no idea how to even begin.
For question number (2) I got to this result $\neg q \lor p \land q$ after applying various logic laws. I could not get it to $p \lor \neg q$.
Any help will be appreciated. 
 A: Q1.
-(sentence 1) Pei Ann doesn't have two aces. : $\neg p \land \neg q$
-(sentence 2) Pei Ann has at least one club. : $r \lor s$
-(sentecne 3) Pei Ann has the ace of clubs and another club. : $ (p \land s) \lor (q \land r)$
Q2.
show that (¬p∨¬q)→ ¬q ≡ p∨¬q
(¬p∨¬q)→ ¬q  ≡ ¬(p $\land$q) → ¬q ≡ (p $\land$q) $\lor$ $\neg q$ ≡ (p $\lor$ ¬q) $\land$ (q $\lor$ ¬q) ≡ (p $\lor$ ¬q)
A: For question 1, I have the following.


*

*Your answer is right; it is 


$$\neg p \lor \neg q$$


*If Ann has at least one club, we do not know which hand the club(s) is in. Thus we can say that "Either Ann has a club in her left hand or Ann has a club in her right hand," which becomes


$$r \lor s$$


*I will answer this question in pieces: First, we know that Ann has the ace of club, but we do not know which hand has the ace of clubs, so we will have to say, "Either the ace of clubs is in Ann's left hand or the ace of clubs is in her right hand." But we do not have a proposition $\psi$ which literally says "Ann has the ace of clubs," but we do have propositions $p$, $q$, $r$ and $s$ that can be combined to say "Ann has the ace of clubs." So we make the following translations


$$\text{Ann has an ace in her left hand and Ann has a club in her left hand} \implies p \land r \tag{1}$$
Proposition (1) can be translated simply into "Ann has an ace of clubs in her left hand."
$$\text{Ann has an ace in her right hand and Ann has a club in her right hand} \implies q \land s \tag{2}$$
Proposition (2) can be translated simply into "Ann has an ace of clubs in her right hand."
Because we do not know in which hand Ann's ace of clubs is, we will have to combine (1) and (2) using the disjunctive ($\lor$) operator
$$\text{Ann has an ace of clubs in her left or in her right hand} \implies (p \land r) \lor (q \land s) \tag{3}$$
Finally, we also know that she holds another club, and this answer has been provided in part 2.  But because this fact is being combined with the other facts that we know about her in translation (3), we will join the proposition about the club with (3) using the conjunction ($\land$) operator:
$$\text{Ann has the ace of clubs and another club} \implies [(p \land r) \lor (q \land s)] \land (r \lor s) \tag{4}$$
For question 2, I will make reference to the following equivalance:
$$p \rightarrow q \equiv \neg p \lor q \tag{1}$$
Here is a proof that $(\neg p \lor \neg q) \implies \neg q \equiv p \lor \neg q$.
Proof
$$\begin{align}(\neg p \, \lor \neg q) \implies\neg q \, &\equiv \neg(p \land q) \implies \neg q \qquad \color{blue}{\text{: De Morgan's Law}}\ \\ &\equiv \neg(\neg(p \land q)) \lor \neg q \, \qquad \color{blue}{\text{: Identiy 1.}}\ \\ &\equiv \neg q \,  \lor \, (p \land q) \hspace{39 pt} \color{blue}{\text{: Commutative and Negation Laws }}\ \\ &\equiv (\neg q \lor p) \land (\neg q \lor q) \hspace{13 pt} \color{blue}{\text{: Distributive Law}}\ \\ &\equiv (p \lor \neg q) \land T \hspace{41 pt} \color{blue}{\text{: Negation Law}}\ \\ &\equiv p \lor \neg q \hspace{67 pt} \color{blue}{\text{: Identity Law}}\ \end{align}   $$
$\blacksquare$
Here is a table of logical equivalences utilized in the proof, except for equivalence (1).

