Question

PLEASE READ ALL QUESTION BEFORE ANSWERING!
A survey revealed the following statement are true:
a) either Laura goes to Towson University or Rob goes to Harvard, but not both.
b) If Danny goes to Frostburg University, then Laura does not go to Townson University.
c) Rob is not going to Harvard
Let p be the statment:Laura goes to Towson
Let q be the statment: Rob goes to Harvard
Let r be the statmente Danny goes to Frostburg.
Show by DIRECT PROOF that: Danny does not go to Frostburg. (for each statment give a reason such as "by law of detachment"
show by INDIRECT PROOF that the above statetement is true. (also give the reason). Start with the additional assumption that r is true.

Answer 1

The given statements are:

$\newline p:$ Laura goes to Towson$\newline q: $ Rob goes to Harvard$\newline r: $ Danny goes to Frostburg$$

Hence, the survery statements can be written as:

Startement a: $(p\wedge \sim q)\vee (\sim p\wedge q)$ - either Laura goes to Towson or Rob to Harvard but not both so exactly one of the two statement $p$ or $q$ is true

Statement b: $r\to \sim p$ - if Danny goe to Frostburg then Lauran doesn't go to Towson

Statement c: $\sim q$ - it is not true that Rob is going to Harvard

a) We want to show that $\sim r$ - Danny doesn't go to Frostburg

$(p\wedge \sim q)\vee (\sim p\wedge q)$

$\sim q\equiv T\Rightarrow q\equiv F$

Thus, $\sim p\wedge q\equiv F$ as $s\wedge F\equiv F$

Therefore $(p\wedge \sim q)\vee (\sim p\wedge q)\equiv (p\wedge \sim q)\vee F\equiv p\wedge \sim q$

Thus, we must have $p$ as $p \wedge\sim q\Rightarrow p$ and $\sim q$

We are also given $r\Rightarrow \sim p\equiv p\vee \sim r$

Combinig $p\wedge \left[ p\vee \sim r \right ]\Rightarrow \sim r$ (Modus Ponens)

Thus, we have $\sim r$ which follows from the given premises.

b) Suppose $r$ is true

Thus, we have $r\to \sim p$ and so $r\wedge (r\to \sim p)\Rightarrow \sim p$ via (Modus Ponens) and so $\sim p\equiv T\Rightarrow p\equiv F$

Furthermore, $\sim q$ is also true as it is given which means $\sim q\equiv T\Rightarrow q\equiv F$

Therefore, statement a) tells us $p\wedge \sim q\equiv F\wedge T\equiv F$ and $\sim p\wedge q\equiv T\wedge F\equiv F$

Therefore, $(p\wedge\sim q)\vee(\sim p\wedge q)\equiv F\vee F\equiv F$

Thus our assumption that $r$ negates one of the premises

Hence $r\equiv F\Rightarrow \sim r\equiv T$ as required

$\blacksquare$