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How does one prove $C$ from the premises: $A \leftrightarrow (B \leftrightarrow C)$ and $A \leftrightarrow B$ ?

I've tried to prove $C$ by contradiction, using a sub-proof which presumes $\neg C $, but although I can conclude all of the following in the subproof: $\neg A$, $ \neg B$, $ \neg (B \leftrightarrow C)$, I'm unable to find a contradiction this way.

I've been stuck on this for the whole day, and I think I might be over-thinking the problem.

Note: I want to prove this using the basic first-order logic rules (I'm using the First-Order Logic from the Language, Proof and Logic book).

Git Gud
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Qqwy
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    There are several 'basic FOL rules'. I take it you're using the ones from the LPL book because you mentioned it in this question. If this is the case, then I suggest you instead say it's the rules from this book. – Git Gud Sep 20 '14 at 18:34
  • You are indeed correct. I did not know that there were so many different kinds of FOL basics. I have appended it in the question. – Qqwy Sep 20 '14 at 18:49
  • I would try proving that ((A↔(B↔C))$\rightarrow$((A↔B)↔C)) first. I could put up an answer a proof in a different natural deduction system than yours if you'd like. The system I refer to has no negation introduction rule. – Doug Spoonwood Sep 20 '14 at 21:43

5 Answers5

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This basically follows from the associativity of $\leftrightarrow$. But let's pretend that we didn't know that.

We consider two exhaustive, mutually exclusive cases.

Case 1: Suppose that $B$ is true. Then since $A \leftrightarrow B$ is true, we know that $A$ is true. Thus, since $A \leftrightarrow (B \leftrightarrow C)$ is true, we know that $B \leftrightarrow C$ is true. But then since $B$ is true, we know that $C$ is true, as desired.

Case 2: Suppose that $B$ is false. Then since $A \leftrightarrow B$ is true, we know that $A$ is false. Thus, since $A \leftrightarrow (B \leftrightarrow C)$ is true, we know that $B \leftrightarrow C$ is false. But then since $B$ is false, we know that $C$ is true (otherwise, if $C$ was actually false, then $B \leftrightarrow C$ would be true, a contradiction). So we're done!

Adriano
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  • Thanks a lot! The only thing I still am having trouble with is how conclude $C$ from $\neg (B \leftrightarrow C)$ and $\neg B$ – Qqwy Sep 20 '14 at 20:33
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    Use a proof by contradiction. Suppose instead that $C$ is actually false. Now since $\neg B$ is true, we know that $B$ is false. But then since $B$ and $C$ are both false, we know that $B \leftrightarrow C$ is true. But this contradicts the fact that $\neg(B \leftrightarrow C)$ is true. – Adriano Sep 20 '14 at 20:41
  • It is unfortunate that I cannot mark two answers as accepted. This answer was enough for me to arrive at a conclusion. However, I did mark Git Gud's answer because other people who stumble across the same problem might have more use for the complete proof themselves. – Qqwy Sep 21 '14 at 00:31
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$ \newcommand{\calc}{\begin{align} \quad &} \newcommand{\calcop}[2]{\\ #1 \quad & \quad \text{"#2"} \\ \quad & } \newcommand{\endcalc}{\end{align}} $Your proof by contradiction approach is fine, here is how you can complete your proof.

Assume $\;C\;$ is false, then $$\calc A \leftrightarrow (B \leftrightarrow C) \calcop{\leftrightarrow}{using what we know about $\;C\;$} A \leftrightarrow (B \leftrightarrow \text{false}) \calcop{\leftrightarrow}{left hand side: use $\;A \leftrightarrow B\;$; right hand side: simplify} B \leftrightarrow \lnot B \calcop{\leftrightarrow}{logic} \text{false} \endcalc$$

which is a contradiction. Therefore $\;C\;$ is true.

MarnixKlooster ReinstateMonica
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I've tried to prove $C$ by contradiction, using a sub-proof which presumes $\neg C $, but although I can conclude all of the following in the subproof: $\neg A$, $ \neg B$, $ \neg (B \leftrightarrow C)$, I'm unable to find a contradiction this way.

This is a wee bit late, but if a contradiction may be derived from assuming $\neg C$, then you should also be able to derive $\def\too{\leftrightarrow}(B\too C)$ too, and from that you can obviously derive $A$, $B$, and $C$ in turn.

$$\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}}\fitch{~~1.~~A\too(B\too C)\\~~2.~~A\too B}{~~3.~~A\to(B\too C)\hspace{10ex}\too\!\mathsf e~1\\~~4.~~(B\too C)\to A\hspace{10ex}\too\!\mathsf e~1\\~~5.~~A\to B\hspace{17ex}\too\!\mathsf e~2\\~~6.~~B\to A\hspace{17ex}\too\!\mathsf e~2\\\fitch{~~7.~~\lnot C\hspace{18ex}\mathsf a}{\fitch{~~8.~~B\hspace{16ex}\mathsf a}{~~9.~~A\hspace{15ex}\to\!\mathsf e~8,6\\10.~~B\too C\hspace{10ex}\to\!\mathsf e~9,3\\11.~~B\to C\hspace{10ex}\too\!\mathsf e~10\\12.~~C\hspace{15.5ex}\to\!\mathsf e~8,12}\\13.~~B\to C\hspace{13.5ex}\to\!\mathsf i~8{-}12\\\fitch {14.~~C\hspace{16ex}\mathsf a}{15.~~\bot\hspace{16ex}\neg~\mathsf e\,14,17\\16.~~B\hspace{16ex}\mathsf x~15}\\17.~~C\to B\hspace{13ex}\to\!\mathsf i~14{-}16\\18.~~B\too C\hspace{13ex}\too\!\mathsf i~13,17\\19.~~A\hspace{18ex}\to\!\mathsf e~18,4\\20.~~B\hspace{18ex}\to\!\mathsf e~19,5\\21.~~C\hspace{18ex}\to\!\mathsf e~20,13\\22.~~\bot\hspace{18ex}\neg\,\mathsf e~21,7}\\23.~~\neg\neg C\hspace{18ex}\neg~\mathsf i~7{-}22\\24.~~C\hspace{21ex}\neg\neg\,\mathsf e~23}$$

Graham Kemp
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Due to the transitivity of $\leftrightarrow$ and due to the fact that $A$ comes up on both premises 'at the same level', I find it natural to focus on $A$ and let it act as a pivot of sorts.

Start by proving $A\lor \neg A$ and perform $\lor$-$\text{Elim}$ on this disjunction.

In the first case just use $\leftrightarrow$-$\text{Elim}$ successively on the premises to get $C$.

In the second case (where one starts a subproof with the premise $\neg A$), use the premise $A\leftrightarrow B$ to get $\neg B$ and the premise $A\leftrightarrow (B \leftrightarrow C)$ to get $\neg(B\leftrightarrow C)$ (in both cases by negation introduction).
Now assume $\neg C$, prove $\neg B\leftrightarrow \neg C$ and from this last statement get $B\leftrightarrow C$.
At this point you can find a contradiction allowing you to conclude $C$ in the subproof whose premise is $\neg A$.

I leave the proof below.

Formal proof

Git Gud
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    The software that comes with the book doesn't seem to deal well with big proofs.This was the best possible presentation I could get. One can note that in justification of the steps there are some missing steps, the software itself did that. – Git Gud Sep 20 '14 at 22:18
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I would use the algebraic machinery:

$A\leftrightarrow B\quad$ iff $\quad 1\oplus A\oplus B$

and get $(A\leftrightarrow (B\leftrightarrow C))\wedge (A\leftrightarrow B)\quad$ iff $\quad (1\oplus A\oplus 1\oplus B\oplus C)(1\oplus A\oplus B)=$ $=(A\oplus B\oplus C)(1\oplus A\oplus B)$ $=A\oplus B\oplus C\oplus A\oplus AB\oplus AC\oplus AB\oplus B\oplus BC=$ $=C\oplus AC\oplus BC=C(1\oplus A\oplus B)\quad$ iff $\quad C\wedge(A\leftrightarrow B)$, which imply C.

Which considering the note may not be as relevant, but...

Lehs
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