I need a work check please.

Question

Section 3.5 Symbolic Arguments
In exercise 33-48 (a) translate the argument into symbolic form and (b) determine if the argument is valid or invalid. You may compare the argument to a standard form or use a truth table.
44. If the canteen is full, then we can go for a walk.
We can go for a walk and we will not get thirsty.
∴ If we go for a walk, then the canteen is not full.
p: The canteen is full.
q: We can go for a walk.
r: We will not get thirsty
p → q
q → r
~ q → ~p
This is a valid statement

Answer 1

Hi, Cynthia,

I don't think that the conclusion "if we go for a walk, then the canteen is not full" is valid, and hopefully I can show you why.

I like most of your translations of the statements into symbolic form:

p: The canteen is full
q: We can go for a walk
r: We will not get thirsty

Using those definitions, the statements that are given would each then be:

    "If the canteen is full, then we can go for a walk"
    if "p" then "q"
    p → q

 

    "We can go for a walk and we will not get thirsty"

    p AND q

    p ∩ q

 

    Thus, "if we go for a walk, then the canteen is not full"

    if "q" then "not p"

    q → ~p

 

I'll make a truth table to cover all of the possible truth values of p, q, and r. In our table, we will need to include p, q, r, ~p, p→q, p ∩ q, and q → ~p:

 

p    q    r    ~p    p→q    p∩q    q→~p

T    T    T    F        T        T          F

T    T    F    F        T        T          F

T    F    T    F        F        F          T

T    F    F    F        F        F          T

F    T    T    T        T        F          T

F    T    F    T        T        F          T

F    F    T    T        T        F          T

F    F    F    T        T        F          T

 

If you have any questions about how I came up with this truth table, please let me know!

 

We know, from the problem, that "p∩q" is TRUE, so let's look back at the table to determine, when "p∩q" is TRUE, if " q→~p" could also be true:

 

p    q    r    ~p    p→q    p∩q    q→~p
T    T    T     F       T        T          F
T    T    F     F       T        T          F
T    F    T     F       F        F          T
T    F    F     F       F        F          T
F    T    T     T       T        F          T
F    T    F     T       T        F          T
F    F    T     T       T        F          T
F    F    F     T       T        F          T

 

We can see that, there are only two circumstances where "p∩q" is TRUE, and in both of those circumstances, " q→~p" is FALSE. So, the conclusion is NOT valid.

 

I hope this helps! Please let me know if I can help any further!

 

Andy