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


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

Goal: $\neg E \to D$

Thank You


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

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

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

  • 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

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 $\neg E$.

  • $\begingroup$ 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

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: $D$.

  • $\begingroup$ 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

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