Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to prove that $$\{\neg P_1\to P_2\} \vdash (\neg P_2\to P_1)$$ without using the Deduction Theorem.

I'm not sure how to proceed. The class notes are all we have to work from, no text to work on similar proofs.

share|cite|improve this question
What axioms and deduction rules are you working from? – Qiaochu Yuan Apr 28 '11 at 0:47
only modus ponems – Daniel Lopez Apr 28 '11 at 1:26
and my 3 axiom schemas which are 1. B->(A->B) 2. (~A->B)->((~A->~B)->A) 3. (A->(B->C))->((A->B)->(A->C)) – Daniel Lopez Apr 28 '11 at 1:27
Please try to put the question in the body of your message; would you require a reader of a book to look up important information on the spine? – Arturo Magidin Apr 28 '11 at 2:14
I was asked a similar question once, and I responded by going through the proof of the deduction theorem and writing down the steps it gives for this particular case. This is a good exercise in understanding how the deduction theorem works, and technically you have not cited the deduction theorem to do it. – Qiaochu Yuan Apr 28 '11 at 3:27

I will use the axioms you wrote on the comment, namely:

  1. $B\to(A\to B)$
  2. $(\lnot A\to B)\to((\lnot A\to\lnot B)\to A)$
  3. $(A\to(B\to C))\to((A\to B)\to(A\to C))$

Please add them in your original question to make it complete.

We want to prove $\{\lnot P_1\to P_2\}\vdash\lnot P_2\to P_1$. A proof is a finite sequence of sentences that each of them is either an assumption, or one of the logical axioms or it's a sentence that can be deduced through modus ponens using previous sentences of the proof. I will write a number next to every sentence to denote the step in which I am.

The first thing to do is to write down your assumptions:

1.$\lnot P_1\to P_2$ (assumption)

What we want to do now is to find an axiom that looks like the assumption so we can use the modus ponens. Axiom 2 fits perfectly here so our second step can be

2.$(\lnot P_1\to P_2)\to((\lnot P_1\to\lnot P_2)\to P_1)$ (axiom 2)

Using modus ponens on 1 and 2 we can derive the following

3.$(\lnot P_1\to\lnot P_2)\to P_1$ (modus ponens 1,2)

The "right hand side" of the consequence we want to prove is there (namely $P_1$) but the "left hand side" is different. So what we want is some means to replace that $\lnot P_1\to\lnot P_2$ with $\lnot P_2$. Intuitively, what we need is to show that from $\lnot P_2$ we can derive $\lnot P_1\to\lnot P_2$. Looking at the axioms this is exactly what the first one gives:

4.$\lnot P_2\to(\lnot P_1\to\lnot P_2)$ (axiom 1)

We have something of the form $A\to B$ and $B\to C$ and we want to prove $A\to C$. If we can do that then our prove will be complete.

So now what we want to prove is $\{A\to B, B\to C\}\vdash A\to C$.

Again let's begin with our assumptions:

1.$A\to B$ (assumption)

2.$B\to C$ (assumption)

What we want to do is to create somewhere $A\to C$. Looking at the axioms that you have (and since $A\to B$ is an assumption) a good idea is to use the third axiom

3.$(A\to (B\to C))\to((A\to B)\to(A\to C))$ (axiom 3)

Since $B\to C$ is an assumption we should use the first axiom to create that $(A\to (B\to C))$ we have

4.$(B\to C)\to(A\to (B\to C))$ (axiom 1)

Now we have everything we want. We just need to apply modus ponens to derive our result:

5.$A\to(B\to C)$ (modus ponens 2,4)

6.$(A\to B)\to(A\to C)$ (modus ponens 3,5)

7.$A\to C$ (modus ponens 1,6)

So writing it down a bit more formally we have:

A. $\{A\to B, B\to C\}\vdash A\to C$:

  1. $A\to B$ (assumption)
  2. $B\to C$ (assumption)
  3. $(A\to (B\to C))\to((A\to B)\to(A\to C))$ (axiom 3)
  4. $(B\to C)\to(A\to (B\to C))$ (axiom 1)
  5. $A\to(B\to C)$ (modus ponens 2,4)
  6. $(A\to B)\to(A\to C)$ (modus ponens 3,5)
  7. $A\to C$ (modus ponens 1,6)

B. $\{\lnot P_1\to P_2\}\vdash\lnot P_2\to P_1$:

  1. $\lnot P_1\to P_2$ (assumption)
  2. $(\lnot P_1\to P_2)\to((\lnot P_1\to\lnot P_2)\to P_1)$ (axiom 2)
  3. $(\lnot P_1\to\lnot P_2)\to P_1$ (modus ponens 1,2)
  4. $\lnot P_2\to(\lnot P_1\to\lnot P_2)$ (axiom 1)
  5. $\lnot P_2\to P_1$ (using A with 3,4)

I hope that was helpful.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.