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?

 

[matt grime]

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.)

 

[MathematicalPhysicist]

let's try conditional proof:
1.(p->(qvp))->r premise
2. p hypothesis
3. qvp 2addition
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

 

[MathematicalPhysicist]

another method which doesn't 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..(-: