Question

Proof question. Help please!

Answer 1

Suppose

\[P \rightarrow Q, R \rightarrow \not Q \]

Answer 2

Prove then

\[P \rightarrow \not R \]

Answer 3

prove if P then not R... what are we doing? Are we using truth tables? boolean algebra

Answer 4

no truth tables

Answer 5

boolean algebra

Answer 6

so boolean algebra? that's going to be tought because I learned it the =truth table way omg... sorry man

Answer 7

but I do know that you can use truth tables to verify an answer. so that's a good check.

Answer 8

I just don't get what to do with if then statements: do I change the form or is it already in logical form?

Answer 9

is there another way to write if then statements

Answer 10

You either have to really prove that the statement is true, or you have to disprove it meaning prove the negation of it which is just if not p then r.

Answer 11

err Q ---> not R

Answer 12

why would you want to do that... you can't change theorems.

Answer 13

Well because we would assume P and therefore Q which means Q ---> not R which is what I want

Answer 14

because if p implies Q and Q implies not R, then P implies not R, right?

Answer 15

now I'm lost D:. I'm sorry I'm a newb at proofs unless it's number theory.

Answer 16

okay well thanks anyway

Answer 17

assume P -> R
then P -> not Q
but contradiction
hence assumption must be false, not (P -> R) must be true
hence (P -> not R) is true

Answer 18

your way of doing it is also correct and more rigorous:
(R -> not Q) = (Q -> not R)
hence, P ->Q leads to P -> not R

Answer 19

basically, (if A then B), is equivalent as (if not B then not A)

Answer 20

okay, I just wasn't sure if I could do that. I'm not sure if I can apply a negation to an if then

Answer 21

u can always do that

Answer 22

and what happens if I do

Answer 23

so does the conditional flip with a negation?

Answer 24

yes exactly

Answer 25

Oh okay, so what is the name of that function

Answer 26

I have to list all of my steps

Answer 27

(R -> not Q) is equivalent as (Q -> not R)
hence, P -> R leads to P -> Q

Answer 28

this is as rigorous as it gets

Answer 29

but aren't we assuming the negation function is true without stating it?

Answer 30

Well thanks for the help PHP

Answer 31

np