3
$\begingroup$

Hi I'm having trouble solving a Fitch Style Proof and I was hoping someone would be able to help me.

Premises:

$A \land (B \lor C)$
$B \to D$
$C \to E$

Goal: $\neg E \to D$

Thank You

$\endgroup$

closed as off-topic by choco_addicted, Ben Sheller, Kamil Jarosz, Semiclassical, Johannes Kloos Apr 11 '16 at 20:43

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – choco_addicted, Ben Sheller, Kamil Jarosz, Semiclassical, Johannes Kloos
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Do you still need help with this? $\endgroup$ – Git Gud Jun 11 '14 at 18:23
2
$\begingroup$

You should be able to transform the following in a formal proof.

Assume $\neg E$.

Prove $B\lor \neg B$ with the intent to use $\lor$-$\text{Elim}$.

If $B$ holds, then use $\to$-$\text{Elim}$ on the premise $B\to D$ to conclude $D$.

Suppose $\neg B$ holds. Use $\land$-$\text{Elim}$ on the first premise to get $B\lor C$. You will want to use $\lor$-$\text{Elim}$ on $B\lor C$. If $B$ holds you can get $D$ in two ways, choose one of them. If $C$ holds, eliminate $\to$ on the premise $C\to E$ to get $E$ and a consequently a contradiction, thus getting $D$ with $\bot$-$\text{Elim}$. Eliminating the disjunction $B\lor C$ gives you $D$.

Eliminating the disjunction $B\lor \neg B$ yields $D$.

Finish off.

You can find the proof hidden in the grey area below.

Formal proof

$\endgroup$
  • 1
    $\begingroup$ Revision Notes: You don't need to use LEM ($B\lor\lnot B$) because the first premise entails $B\lor C$ and the second premise is $B\to D$. $~$ Since $C\to D$ can be deduced from the third premise under the assumption of $\neg E$ , you may then derive $D$ by or-elimination, and therefore discharge the assumption to deduce $\neg E\to D$. $\endgroup$ – Graham Kemp Jul 17 '18 at 0:25
  • $\begingroup$ @GrahamKemp Thanks. $\endgroup$ – Git Gud Jul 18 '18 at 21:37
1
$\begingroup$

Do a proof by contradiction. Assume $\neg E \wedge \neg D$. Then we have from $A \wedge (B \vee C)$ that $B \vee C$. If $B$, then $D$, which contradicts $\neg D$. If $C$ then $E$ which contradicts $\neg E$.

$\endgroup$
  • $\begingroup$ Please read this comment which I gave in another answer. Though I feel like your suggestion may be a bit closer than the other answer, dealing with the details is where the 'hard' work is. $\endgroup$ – Git Gud Jun 10 '14 at 23:06
  • 1
    $\begingroup$ @GitGud: There is no hard work in this question and I was giving him a strategy instead of a full Fitch proof. Mine does work formally, minus some $\wedge$-Elim's I didn't explicitly mention; there is no "correct" strategy for giving a simple proof like this. $\endgroup$ – Samuel Reid Jun 12 '14 at 2:28
-1
$\begingroup$

To begin with: you have $\neg E$, and combine this with $C \to E$, then you have: $\neg C$, but $B \lor C$ must be true, so this yields $B$, and with $B \to D$, apply Modus Ponen: we have: $D$.

$\endgroup$
  • $\begingroup$ I wasn't gonna say anything, but this came up in the review queue, so now I'm compelled to. I don't think your answer gets the OP closer to solving the problem. It does hint at a way to do it, but not necessarily an easier way. Reiterating: I don't think proving what you suggest is easier than solving the problem in a more standard way. $\endgroup$ – Git Gud Jun 10 '14 at 23:03
  • $\begingroup$ That stupid user is me. I had four options while reviewing: Looks OK (it didn't look OK, it still doesn't look OK, I think you need to review your natural deduction), Edit (It wasn't at all reasonable to transform your answer into something that can actually help the OP), Skip and Delete. Maybe I could have chosen skip and kept an eye on your answer to see what happens, but it's also perfectly reasonable to delete. You took this personally unnecessarily, I have nothing against you. $\endgroup$ – Git Gud Jun 11 '14 at 0:03

Not the answer you're looking for? Browse other questions tagged or ask your own question.