Natural Deduction stuck

Question

$$\vdash (( p → q) → p) → p$$ how would I go about solving this with no premises. to get the implication i know i need $$(p → q) → p$$ as my first assumption but im stuck to where to go from there

Answer 1

I just did it with a truth table. Set p to each of 0 and 1 (false and true), and set q to each of 0 and 1 as well. That gives you 4 combinations of the two. Then, work your way from the inside out.

Case 1: $p = 0$, $q = 0$

$p→q = 1$

$1→p=0$

$0→p=1$

Case 2: $p = 0$, $q = 1$

$p→q = 1$

$1→p=0$

$0→p=1$

Case 3: $p = 1$, $q = 0$

$p→q = 0$

$0→p=1$

$1→p=1$

Case 4: $p = 1$, $q = 1$

$p→q = 1$

$1→p=1$

$1→p=1$

Therefore the statement holds true for every case.

Answer 2

First hint: Prove ($\lnot$p$\rightarrow$(p$\rightarrow$q)) as a lemma.

More detailed analysis:

1. What is the goal?

The goal is to produce to (((p $\rightarrow$ q) $\rightarrow$ p) $\rightarrow$ p).

2. What evidence do you have so far?

You have evidence that if you assume ((p$\rightarrow$q)$\rightarrow$p) and produce p, that a solution can get constructed.

3. If you assume something, what is the structure of that assumption?

The assumption ((p$\rightarrow$q)$\rightarrow$p), has the structure such that (p$\rightarrow$q) is the antecedent and p is the consequent. Thus, if you can produce (p$\rightarrow$q), by detachment ("modus ponens"), you can deduce p.

4. Can you assume anything and get rid of that assumption?

Suppose you assume ((p$\rightarrow$q)$\rightarrow$p). You want 'p'. But, let's say your enemy says that obtaining 'p' is impossible. So, let's assume your enemy correct, and thus assume that $\lnot$p makes sense. Can you then obtain 'p' under that assumption, with the analysis of 3. in mind?