# How do I prove this propositional logic

millani
How do I prove this? (propositional logic)

## Homework Statement

How to prove this
$(p \rightarrow (q \vee p)) \rightarrow r \vdash \neg p \vee (q \vee r)$
using only the natural deduction rules in propositional logic?

## Homework Equations

http://en.wikipedia.org/wiki/Propositional_logic
(natural deduction rules only)

## The Attempt at a Solution

$1: (p \rightarrow (q \vee p)) \rightarrow r$ premise
<start of hypothesis 0> ; I tried to make a box here but failed miserably
$2: \neg (\neg p \vee (q \vee r))$ hypothesis
<start of hypothesis 1>
$3: \neg p$ hypothesis
$4: \neg p \vee (q \vee r)$ conjunction introduction 3
$5: \bot$ negation introduction 2,4
<end of hypothesis 1>
$6: \neg \neg p$ negation introduction 3-5
$7: p$ double negative elimination 6
$8: ?$
<end of hypothesis 0>

As you can see, I don't exactly know how to use TeX.

PS: How do I put a box around the hypothesis in TeX?

Last edited:

Homework Helper
There are many many latex guides easily googlable. However, it is not guaranteed that the code will work in the forum. Still. Searching is you r best bet if you really want to put a box round something. (Though I'd say there was no need to do that at all.)

Gold Member
let's try conditional proof:
1.(p->(qvp))->r premise
2. p hypothesis
4. p->(qvp) 2-3conditional proof
5. r 1,4 modes ponens
6.?
7.?
now i leave you to fill 6 and 7. (hint: look at 3).

millani
Thanks for the help, loop quantum gravity!

1. $(p \rightarrow (q \vee p)) \rightarrow r$ premise
2. $p$ hypothesis
3. $q \vee p$ 2addition
4. $p \rightarrow (q \vee p)$ 2-3conditional proof
5. $r$ 1,4 modes ponens
6. $q \vee r$ disjunction introduction 5
7. $\neg p \vee (q \vee r)$ disjunction introduction 6

Gold Member
another method which doesnt employ having an extra hypothesis is:
use material conditional on the premise as far as you get to ~qvr, and afterwards use addition or disjunctive intro, to get ~qvrvq which is r and then the same as in the first approach.
the differnece is that we trully have (p->(qvp))->r |- ~pvqvr
while in the first approach we have (p->(qvp))->r,p|- ~pvqvr
which by the deudction theorem is the sam as: (p->(qvp))->|-p->(~pvqvr) but by the mc it's again the same result.
so here you have both methods to prove this.
and even more..(-: