Using the 20-rule proof system (replacement rules, rules of inference, conditional proof, and reductio ad absurdum) and given these 3 premises:

  1. A -> ~B
  2. ~C -> B
  3. ~A -> ~C

I know that since I'm trying to prove a biconditional, the penultimate step will be (A -> C) * (C -> A), but not sure how to get there... should I be using conditional proof to start off or conditional exchange? Any help would be greatly appreciated!


given that the two main things you need to prove are A -> C and C -> A. As a general strategy, it is often the easiest to do so with conditional proofs. Given the three assumptions you've been given, it's also a successful strategy.

There's quite a few different syntaxes and allowed procedures (it'd be better to give a link or spell out what you can and cannot use rather than just saying "20-rule proof system").

Here's how I'd do it:

 1.|  A -> ~ B             P
 2.|  ~C -> B              P
 3.|_ ~A -> ~C             P
 4.|  |_ A                 A
 5.|  |  ~B                MP 1,4
 6.|  |  ~~C               MT 5,2
 7.|  |  C                 DN 6
 8.|  A -> C               CP 4-7
 9.|  |_ C                 A
10.|  |  ~~C               DN 9
11.|  |  ~~A               MT 10
12.|  |  A                 DN 11
13.|  C -> A               CP 10-12
14.|  (A -> C) & (C -> A)  &I 9,13
15.| A <-> C               BiCond. Int 14

In the above A = assumption, | means we are in a subproof, DN = double negation, MT = modus tollens, MP = modus ponens, CP = conditional proof, &I = conjunction introduction, and BiCond Int = biconditional introduction. For some proof systems, you need to use R to repeat things to use them in subproofs (omitted).

The largest dependency in the above is MT. If you need to avoid it, then the basic pattern is:

 1.|  P -> Q     P
 2.|_ ~Q         P
 3.|  |_ P       A
 4.|  |  Q       MP 1,3
 5.|  |  Q & ~Q  &I 2,4
 6.|  ~P         ~I 3-5

which you would need to substitute for each use of MT.

  • Thank you so much! In the future, I'll specify the syntax, but the syntax you used is essentially the same that's used in my class, so this is perfect!
    – ajq
    Oct 24 '16 at 3:06

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