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Alright, so I've been staring at this problem for two hours trying to figure out what exactly is wrong. Is there a typo? Am I missing something? Here is the problem:

  1. $A$ /∴ $B \rightarrow (\lnot A \rightarrow C)$

My professor said these were supposed to be easy. Now I am new to symbolic logic so I could very well be missing something. But I don't understand how constructing this proof is possible. Could anyone help, using the 8 basic inference rules? Or with Conditional Proof/Indirect proof? Thanks.

Mauro ALLEGRANZA
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T. J.
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3 Answers3

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You don't say which system of rules you are using. But in a natural deduction framework, the proof will go like this:

Given premiss is $A$.

Assume $B$.

Then assume $\lnot A$.

By explosion rule (ex false quodlibet), from the contradictory first and third lines you can infer anything, including $C$.

Discharge second assumption to get $\lnot A \rightarrow C$.

Discharge first assumption to get $B \rightarrow (\lnot A \rightarrow C)$.

(Note standard natural deduction systems allow this second kind of inference where we discharge an assumption which is not actually used.)

Community
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Peter Smith
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If you need some intuitive explanation then $B \implies (\lnot A \implies C)$ is equivalent to $B \land \lnot A \implies C$ because $A$ is true then $B \land \lnot A $ is false and you can infer anything from false statement so $C$ holds.

Trismegistos
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I do not know the logical rules you are using (please, in this type of questions, refer to the textbook you are using, or type the rules you have on your lectures notes) but, in general, you need :

$P \rightarrow Q$ is equivalent to $\lnot P \lor Q$

so that, using Double Negation :

$\lnot A \rightarrow C$ will be $A \lor C$

and you must have a rule (called $\lor$-introduction) that allows you to derive $A \lor C$ form $A$, i.e.

$A \vdash A \lor C$.

Again, $B \rightarrow (\lnot A \rightarrow C)$ is $\lnot B \lor (\lnot A \rightarrow C)$, so that, by the same rule, you "introduce" also $\lnot B$ getting :

$A \vdash \lnot B \lor (\lnot A \rightarrow C)$ that is your $B \rightarrow (\lnot A \rightarrow C)$.

Mauro ALLEGRANZA
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  • As I said before, I've no way to trace the names of the rules you are using... But I think it's correct. – Mauro ALLEGRANZA Feb 06 '14 at 14:54
  • Ok, I checked in Wikipedia List of rules of inference : Addition is "my" Disjunction Introduction (i.e. $\lor$-introduction); and Conditional Exchange is the equivalence between ... I've used. Fine. – Mauro ALLEGRANZA Feb 06 '14 at 15:15
  • (Whoops, my above comment wasn't finished when I sent it) Thank you! Here's what I did:
    1. A /∴ B-->(~A-->C)
    2. A v C (using Addition from premise 1)
    3. ~A-->C (Conditional Exchange, 2)
    4. ~B v (~A-->C) (Addition, 3)
    5. B-->(~A-->C) (Conditional Exchange, 4)

    Does that sound right? I wasn't aware Conditional Exchange worked in that manner. Obviously I am very new at this! Thank you for your time. (The names are from Understanding Symbolic Logic by Virgina Klenk)

     – T. J. Feb 06 '14 at 15:30