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I've been struggling with a natural deduction problem for a while now...can anyone perhaps shed some light on where I'm going wrong here?

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As you can see, I need to prove C from the given premises. My natural first move is to assume the negation of C and then prove a contradiction, but I run into a dead end if I start with ¬C as it doesn't flow into any other premises.

I therefore decided to start with a subproof with A as the premise. This leads to a couple of other deductions, and I get to the point where I have (C V D). I know that D is false (as per line 1) but I don't know how to get this over the line (to show that C is true as D is false).

Can someone please put me out of my misery?

Thanks all!

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The idea of the proof of $C \lor D, \neg D \therefore C$ (known under the name "disjunctive syllogism") is the following: We know that one out of C or D must hold, so we consider both cases and start to assume either of the disjuncts, D and C. But we know that it can't be D that holds, because if we assume that it does, this contradicts our premise that D is false, so we derive a contradiction from which we can conclude anything, including C. And in the other case where we assume C to be true, we trivially get C as well, so in any case, we can conclude C, not matter which of the disjuncts C or D is in fact true.

This is precisely the idea of disjunction elimination. And with a disjunction elimination, the proof has the following structure:

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The lines to be cited in the $\lor E$ application are the lines of the major premise $C \lor D$ and the two subproofs $C \vdash C$ and $D \vdash C$. This explains one of the errors you got: The line range for the subproof with premise $C$ was missing. The other error is that the subproof with premise $D$ does not have the conclusion $C$.

With the above skeleton, we just ned to find out how to get from C to C, and from D to C, using the premises in lines 1 and 2.
$C \vdash C$ is easy: We already have $C$ as the premise of the subproof, so we can just reiterate the line.
For $D \vdash C$, you were already on the right track: The premise $D$ of the subproof contradicts the previous assumption $\neg D$, which yields a contradiction. By es falso quodlibet, from $\bot$ we may conclude anything; conventiently, we can choose $C$.
Since from both $C$ and $D$ we derived $C$, we can apply $\lor E$ on the premise $C \lor D$ and conclude $C$, as desired.

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(The rule name $X$ in line 7 is more commonly known as $\bot$, or $\bot E$, or "ex falso quodlibet (EFQL)", or "principle of explosion", depending on which of the hundreds of textbooks out there you're using.)

Then to finish the whole proof, you just need one final $\lor E$ to derive $A \lor C \vdash C$ and you're done:

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  • $\begingroup$ Thank you so much for the concise explanation @lemontree! Just some context: I am working from the "Language Proof And Logic" book by Barwise and Etchemendy. From what I can see, they do not have "X" as a justification rule (as per your line 12 above). Is there another option for X, or can you explain what it means? Oh, and I am assuming that "R" is equivalent to "Reit" in the abovementioned book. Thanks again! $\endgroup$ – Gerhardus Carinus May 1 at 16:56
  • $\begingroup$ As I added as a comment somehwere in my explnation, that X should be a $\bot$ (apparently I accidentally typed "X" instead of "XX" (which is the key combination to produce $\bot$), and it wasn't detected as an error for some reason). I just didn't want to set up the proofs all anew when I realized the error. $\endgroup$ – lemontree May 1 at 16:57
  • $\begingroup$ Yes, R is equivalent to Reit (reiteration). "R" is just the rule name that the online tool uses. $\endgroup$ – lemontree May 1 at 17:01
  • $\begingroup$ Thanks! Okay so I finished the proof by using ⊥E - I'm assuming that's what you meant to illustrate? So just to get my bearings straight: if you have a disjunction and you manage to prove that one side of the disjunction is false, you use ⊥E to show that the other side of the disjunction is true. That's what the solution makes it seem like. Thanks for your help thus far! $\endgroup$ – Gerhardus Carinus May 1 at 21:15
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    $\begingroup$ Yes, $\bot E$ is another name for the rule I mean, if by $\bot E$ you refer to the rule that allows you to conclude any formula $A$ from $\bot$ without closing any assumptions, often referred to as "ex falso quodlibet (EFQL)". (And it turns out that indeed $X$ and not $\bot$ is the symbol used by the ND proof editor, so not a typo.) Naming conventions differ between different books/tools, it can get messy especially with rules involving $\bot$ and $\neg$. $\endgroup$ – lemontree May 1 at 21:21

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