Propositional Logic

[ShadyS87]

Hello everybody, could you plesae help analyzing these 2 issues? In the first one I might have an idea but in the second I am really lost. Thanks in advance!

The questions are shown better in the attachment

Question 1
Let BBB be a set with A={n:n∈Z+ and x<16} and p,qp, qp,q and r r r be three propositions concerning an integer nnn in AAA.
ppp means 'n is an odd number'
qqq means 'n is a prime number'
rrr means 'n is less than 8'
a) Find the truth set of the following logical expressions: p∧q; p⊕q; p→q p\rightarrow qp→q and r→q r \rightarrow qr→q
b) Express each of the three following compound propositions using p, q, r and appropriate logical symbols:

• 'n is neither neither odd nor is a prime number'
• 'if n is odd and n less than 8 then n is a prime number '
• 'n is a prime number only if n is odd'
• 'n is a prime number if n is odd'

c) Write in words the contrapositive of the statement given symbolicaly by 'q→pq\rightarrow pq→p'.
Question 2:
Let ppp and qqq be two propositions. Using the laws of propositional logic seen in this topic, show that (p→q)∧p=p∧q.

 

[ShadyS87]

The first question I believe it would be:

- {1, 3, 5, 7, 11, 23}
- { 2, 9, 15}
- everything in p and q except {9, 15}
- everything in r and q except {2, 4, 6}

I am lost in the the difference between "n is a prime number only if n is odd" and "n is a prime number if n is odd"

The second question honestly I am also stuck.

 

[ShadyS87]

Also, I wrote
n is neither odd nor is a prime number = ¬ p V ¬ q
if n is odd and less than 8 then is a prime number = (p V R) → q

 

[pka]

n is neither odd nor is a prime number = ¬ p V ¬ q
if n is odd and less than 8 then is a prime number = (p V R) → q

TO ShadyS87, why did you mess-up a perfectly clear set of questions with all sort of extraneous letters and symbols the make the post unreadable. As of now I think that you need to re-post most of this.
Now here is help on what I can read: n is neither odd nor is a prime number Is \(\displaystyle \neg p \wedge \neg q\). "NEITHER" is an and proposition.
If n is odd and less than 8 then is a prime number: \(\displaystyle (p\wedge r) \to q\)

 

[ShadyS87]

Sorry I didn't mean to mess it up, I simply copied and pasted it and it came out like this. I will re write it down below.
Meanwhile, thanks for the help.

 

[ShadyS87]

Question 1

Let B be a set with A={n:n∈Z+ and x<16} and p, q and r be three propositions concerning an integer n in A.
p means 'n is an odd number'
q means 'n is a prime number'
r means 'n is less than 8'
a) Find the truth set of the following logical expressions: p∧q; p⊕q; p→q and r→q.
b) Express each of the three following compound propositions using p, q, r and appropriate logical symbols:

• 'n is neither neither odd nor is a prime number'
• 'if n is odd and n less than 8 then n is a prime number '
• 'n is a prime number only if n is odd'
• 'n is a prime number if n is odd'

c) Write in words the contrapositive of the statement given symbolicaly by 'q→p'.

Question 2:
Let p and q be two propositions. Using the laws of propositional logic seen in this topic, show that (p→q)∧p=p∧q.

 

[Dr.Peterson]

It appears that the text was copied and pasted from a source that tries to provide expressions in three different forms, accounting for the tripling of the p, for example. If you choose to do this, it is important to proofread what you pasted, and remove parts that are duplicate and/or unreadable. Just posting the image (full-size) is easier.

But even the image has some odd content: "n is neither neither odd ..."??? And why is B mentioned at all?

Anyway, just looking at the image and the answers you gave in post #2,

• You've missed some numbers in p∧q .
• You have p⊕q correct.
• What do you mean by "everything in p and q"? Those aren't sets.

 

[ShadyS87]

p∧q I meant it to be {1, 3, 5, 7, 11, 13}
p⊕q {2, 9, 15}
p→q {1, 3, 5, 7, 11, 13}
r→q {1, 3, 5, 7

Some help in regards to at least the question 2?

 

[Dr.Peterson]

Your answer for p→q implies (no pun intended) that p→q is false for 2. Is it?

For question 2, to show that (p→q)∧p=p∧q, I would start by rewriting p→q. What is that equivalent to, using only conjunction, disjunction, and negation? After you've done that, make at least some attempt at the next step, or else list all the laws you have learned, so we can see what you are allowed to use.

 

[pka]

p∧q I meant it to be {1, 3, 5, 7, 11, 13}
p⊕q {2, 9, 15}
p→q {1, 3, 5, 7, 11, 13}
r→q {1, 3, 5, 7
Question 2: Some help in regards to at least the question 2?
Using the laws of propositional logic seen in this topic, show that (p→q)∧p=p∧q.

\(\displaystyle \begin{gathered}
(P \to Q) \wedge P \hfill \\
(\neg P \vee Q) \wedge P \hfill \\
(\neg P \wedge P) \vee (Q \wedge P) \hfill \\
(f) \vee (P \wedge Q) \hfill \\
\therefore \;(P \wedge Q) \hfill \\
\end{gathered} \)
Your turn to post: SUPPLY THE REASONS>

 

[ShadyS87]

(P→Q)∧P
(¬P∨Q)∧P This equivalence I can reach it only with the truth tables, I don't know the name of the law
(¬P∧P)∨(Q∧P) distributive law
(f)∨(P∧Q) negation law
∴(P∧Q)

If I am not wrong this should be the explanation.

Thanks a lot for your help.

In regards to the first exercise:

• 'n is a prime number only if n is odd'
• 'n is a prime number if n is odd'

Am I wrong to assume that the answers are:

• P→Q
• Q→P

?

 

[pka]

(P→Q)∧P
(¬P∨Q)∧P This equivalence I can reach it only with the truth tables, I don't know the name of the law
(¬P∧P)∨(Q∧P) distributive law
(f)∨(P∧Q) negation law
∴(P∧Q)

If I am not wrong this should be the explanation.
'n is a prime number only if n is odd'
'n is a prime number if n is odd'
[/LIST]
Am I wrong to assume that the answers are:
P→Q
Q→P

\(\displaystyle P\to Q\equiv \neg P\wedge Q\) is known as Material Implication.

\(\displaystyle P\to Q\equiv P \text{ only if }Q\)

 

[ShadyS87]

Thank you very much for your help indeed, it helped me a lot to understand it.