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I'm really stuck on how to prove ¬C → B , C → ¬B ⊢ ¬B ↔ C. I know I have $C \implies \sim B$ but in order to introduce the biconditional I have to prove $\sim B \implies C$ and I have no idea how. Any help is greatly appreciated. So far this is what I have:

  1. $\quad \sim C \implies B :PR$

  2. $\quad C \implies \sim B :PR$

  3. $\quad \sim B :AS$

  4. $\quad ?????$

  5. $\quad \sim B \implies C :\implies I\;(???)$

  6. $\quad \sim B \iff C : \iff I2,5$

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greenpotatoes
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    What you need is just the contrapositive of the first implication. Is that something you have access to? – Milten Oct 17 '21 at 20:19
  • @Milten unfortunately no... to be totally honest I don't know what that is. In my course we only have access to the basic rules of truth-functional logic so far (introduction and elimination of ~, V, +, ->, and <->.) – greenpotatoes Oct 17 '21 at 20:35
  • Okay. Do you know that $P\rightarrow Q$ is equivalent to $\neg P \lor Q$? That would help you a lot. – Milten Oct 17 '21 at 20:39
  • @Milten I think I understand that but I don't really know how that helps? Sorry I am super new to natural deduction and honestly just need someone to explain. Even if I can prove ~C -> B is the same as C / B I'll still have to somehow get rid of the disjunction and then what? – greenpotatoes Oct 17 '21 at 20:52
  • I don't actually know the rules of natural deduction myself, so hopefully someone else will step in here. But I would basically do this: $$\neg C\to B \equiv C\lor B \equiv B\lor C \equiv \neg B\to C.$$ Hopefully you can determine yourself how much of that is legal in your case. – Milten Oct 17 '21 at 20:58
  • Also, contraposition (or taking the contrapositive) is the very useful equivalence $P\to Q \equiv \neg Q\to \neg P$. Extremely useful outside of logic too when setting up a proof for example. Somewhat related to proof by contradiction. My previous comment is just a proof that contraposition works, really. – Milten Oct 17 '21 at 21:01
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    There are many different systems of formal logic, each with their own set of specific rules. Even when you say that you have Intro and Elim rules for the ‘typical’ connectives, that does not tell us exactly what your rules look like: The Intro and Elin rules in one system of logic can be different from those in another system. So: you’ll have to tell us exactly what the rules of your system look like! – Bram28 Oct 17 '21 at 21:07
  • @Bram28 the system is called truth-functional logic. The margin of this website has all the rules. I can only use the basic rules in my proof (not including the last three): https://proofs.openlogicproject.org/ – greenpotatoes Oct 17 '21 at 21:48
  • @Milten I see...I have tried every which way to prove those steps but I'm really new at this (and not great at it) so I haven't been able to get there through those steps. But I will keep trying. Thank you. – greenpotatoes Oct 17 '21 at 22:08
  • @greenpotatoes Thanks for the link! Eh, I don't see any rules regarding biconditionals ($\leftrightarrow$). Given as you are trying to prove a binconditional, that would seem to be imperative ... – Bram28 Oct 17 '21 at 22:39
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    @Bram28 Sorry, didn't realize they weren't in there! Biconditional rules are the exact same as conditionals in this system. But I managed to finally figure this proof out and solve it on my own. Thank you anyways!! – greenpotatoes Oct 17 '21 at 22:45
  • @greenpotatoes Excellent! :) – Bram28 Oct 17 '21 at 23:24
  • Yes; having proved $C \to \lnot B$, you have to prove $\lnot B \to C$. Assume $\lnot B$ and $\lnot C$; with this you have a contradiction and from it $\lnot \lnot C$. The conclusion follows by Double Negation. – Mauro ALLEGRANZA Oct 18 '21 at 07:33

2 Answers2

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Using TFL's fundamental rules, here's a derivation for $C\to\neg B\vdash_{\tiny\!\! TFL} B\to\neg C$ : $$\def\fitch#1#2{~~~~\begin{array}{|l}#1\\\hline#2\end{array}}\fitch{~~1.~C\to\neg B}{\fitch{~~2.~B}{\fitch{~~3.~C}{~~4.~\neg B\qquad{\to}E~1,3\\~~5.~\bot~~\qquad\neg E~4,2}\\~~6.~\neg C\qquad\neg I~3{-}5}\\~~7.~B\to\neg C\qquad{\to}I ~2{-}6}$$

The derivation you want is similar.

$$\fitch{~~1.~\neg C\to B}{\fitch{~~2.~\neg B}{~~~~\vdots\\~~7.~C}\\~~8.~\neg B\to C\qquad{\to}I ~2{-}7}$$

Graham Kemp
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Heres one potential way to do it,

$$\begin{align} \neg C \to B , C \to \neg B, C &\vdash C &&\text{($\in$)}\\ \neg C \to B , C \to \neg B, C &\vdash C \to \neg B &&\text{($\in$)}\\ \neg C \to B , C \to \neg B, C &\vdash \neg B &&\text{($\to -$), (1), (2)}\\ \neg C \to B , C \to \neg B, \neg B, \neg C &\vdash \neg B &&\text{($\in$)}\\ \neg C \to B , C \to \neg B, \neg B, \neg C &\vdash \neg C &&\text{($\in$)}\\ \neg C \to B , C \to \neg B, \neg B, \neg C &\vdash \neg C \to B &&\text{($\in$)}\\ \neg C \to B , C \to \neg B, \neg B, \neg C &\vdash B &&\text{($\to -$), (5), (6)}\\ \neg C \to B , C \to \neg B, \neg B &\vdash C &&\text{($\neg -$), (4), (7)}\\ \neg C \to B , C \to \neg B &\vdash \neg B \leftrightarrow C &&\text{($\leftrightarrow +$), (3), (8)} \end{align}$$

2k of awsome
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