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.
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.
I will use the axioms you wrote on the comment, namely:
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. $\{A\to B, B\to C\}\vdash A\to C$:
B. $\{\lnot P_1\to P_2\}\vdash\lnot P_2\to P_1$:
I hope that was helpful.