Question

Help with symbolic logic.
Simplify
[~(p→q)→~(q→p)]∧(p∨q)

Where:
~ means negation
∧ means conjuction
∨ means disjuction
→ is the condicional.


Seems easy , but i'm missing some rule.

Answer 1

sorry friend, I am lost!!

Answer 2

@primeralph

Answer 3

Expand?

Answer 4

Try that.

Answer 5

I'm not sure how to do that.
There´re two formulas in my notes that says:
~(p→q) = p∧~q and
p→q = ~p∨q
So i get that
[~(p→q)→~(q→p)]∧(p∨q)
[(p∧~q)→(q∧~p)]∧(p∨q)
[~(p∧~q)∨(q∧~p)]∧(p∨q)

And then i don't know what to do

Answer 6

boolean!! hand off! take a lot of time to do but not sure about the answer

Answer 7

@Luigi0210

Answer 8

What is this?

Answer 9

@Luigi0210 boolean algebra

Answer 10

I hope you called me here to learn cause I have not seen this yet

Answer 11

Truth tables out the wazoo!

Answer 12

truth table just give out the truth not simple form and the question is simplify

Answer 13

@Loser66, ah right, I thought I saw "identity" somewhere in the question...

Answer 14

Very miserable :(

Answer 15

empathy

Answer 16

hey, you can check my stuff, find out the mistake, it's ok friend, If mine is not wrong, it's right, right? hahahah.. I go around as boolean

Answer 17

Okay, so after some work on paper (using some truth tables), I managed to show that
\[\bigg[\sim(p\to q)\to\sim(q\to p)\bigg]\equiv p\to q\]
Does that help?

Answer 18

Then using one of the given formulas, you have
\[(p\to q)\wedge(p\vee q)\\
(\sim p\vee q)\wedge(p\vee q)\\
q\]