Tentative – the actual lecture might deviate in some ways from these notes.

Reductio ad Absurdum is, in some ways, a controversial method of proof. Instead of showing, positively, that something is the case, we show, negatively, that something cannot be the case – by showing that it leads to a contradiction. But most scientists (logicians, mathematicians) agree that this reasoning is unproblematic. There are only some dissenters about whether this kind of reasoning can be used to establish a positive conclusion. According to “classical” logic, we can apply Reductio ad Absurdum to a negated premise (e.g. $\neg P$) to derive $\neg\neg P$, and then apply DN-elim to get $P$. According to “intuitionistic” logic, this Reductio only shows that $\neg\neg P$, but fails to establish the positive result that $P$. So intuitionists have a problem with DN-elim, and not actually with RA.

In this course, we will continue to assume DN-elim, and that makes RA a quite powerful method of proof.

An example

Claim: there are irrational numbers $a$ and $b$ such that $a^b$ is rational.

Proof: We assume that ( ) is irrational. If $\sqrt{2}^\sqrt{2}$ is rational, then we are done. If $\sqrt{2}^\sqrt{2}$ is irrational, then we have $$(\sqrt{2}^\sqrt{2}) ^\sqrt{2}=\sqrt{2}^2 = 2,$$ which is rational.

DeMorgan’s laws

Show $\neg (P\vee Q)\vdash \neg P$

1   (1) ¬(P∨Q)         A
2   (2) P              A
2   (3) P∨Q            2 ∨I
1,2 (4) (P∨Q)∧¬(P∨Q)   3,1 ∧I
1   (5) ¬P             2,4 RA

It should be clear that there is a similar proof of $\neg (P\vee Q)\vdash \neg Q$, and putting these together, $\neg (P\vee Q)\vdash \neg P\wedge\neg Q$.

The previous result can be used to show that $\neg (\neg P\vee Q)\vdash \neg (P\to Q)$.

1   (1) ¬(¬P∨Q)         A
2   (2) P→Q             A
1   (3) ¬¬P             1 (copy proof from above)
1   (4) P               3 DN
1,2 (5) Q               2,4 MP
1,2 (6) ¬P∨Q            5 ∨I
1,2 (7) (¬P∨Q)∧¬(¬P∨Q)  6,1 ∧I
1   (8) ¬(P→Q)          2,7 RA

On the homework you are asked to prove $P\to Q\vdash \neg P\vee Q$, which is the contrapositive of the previous result. Note that at line 8, we could have chosen instead to infer $\neg\neg (\neg P\vee Q)$ depending on line $2$. In general, if one has a proof of $\neg \phi ,\psi \vdash \bot$, then one can apply RA to infer either $\neg \phi\vdash \neg \psi$ or $\psi\vdash\neg\neg\phi$. (Here $\bot$ stands for any contradiction.)

Law of Non-Contradiction

1 (1) P∧¬P     A
  (2) ¬(P∧¬P)  1,1 RA

Ex Falso Quodlibet (EFQ)

This means: from the false, everything follows.

1   (1) P      A
2   (2) ¬P     A
3   (3) ¬Q     A
1,2 (4) P∧¬P   1,2 ∧I
1,2 (5) ¬¬Q    3,4 RA
1,2 (6) Q      5 DN

Note that the argument form $P,\neg P\vdash Q$ is valid, but no sound argument can have this form.

Disjunctive Syllogism

EFQ can be used to re-derive disjunctive syllogism $P\vee Q,\neg P\vdash Q$.

1   (1) P∨Q      A
2   (2) ¬P       A
3   (3) P        A
2,3 (4) Q        2,3 EFQ (copy in proof from above)
5   (5) Q        A
1,2 (6) Q        1,3,4,5,5 ∨E

DeMorgan’s Laws

First we show that $\neg P\vee \neg Q\vdash \neg (P\wedge Q)$. We will do it in two steps, starting with $\neg P\vdash \neg (P\wedge Q)$. You have already proven the latter, but we give a different proof using RA.

1   (1) ¬P      A
2   (2) P∧Q     A
2   (3) P       2 ∧E
1,2 (4) P∧¬P    3,1 ∧I
1   (5) ¬(P∧Q)  2,4 RA

If you think about it for a minute, it should be clear that we can change the roles of $P$ and $Q$ in the previous proof and get a proof of $\neg Q\vdash\neg (P\wedge Q)$. (This depends on the fact that $P\wedge Q$ plays exactly the same role vis-a-vis our rules as $Q\wedge P$ plays.) So now we have a proof of $\neg P\vee \neg Q\vdash \neg (P\wedge Q)$.

1 (1) ¬P∨¬Q       A
2 (2) ¬P          A
2 (3) ¬(P∧Q)      2 (copy in proof from above)
4 (4) ¬Q          A
4 (5) ¬(P∧Q)      4 (copy in proof from above)
1 (6) ¬(P∧Q)      1,2,3,4,5 ∨E

Now we show that $¬P,¬Q\:\vdash\:\neg (P\vee Q)$

There are a few different ways that this proof could be approached. The most direct way is to assume $\neg P$ and $\neg Q$, then assume $P\vee Q$ for RA. Arguing from $P$, we get $P\wedge\neg P$, a contradiction. Arguing from $Q$, we get $Q\wedge\neg Q$, which is also a contradiction. But it’s not the same contradiction on both sides, so we would need to do a couple additional steps – which is no problem, because once you have a contradiction, you can derive anything (EFQ).

Let’s try a more subtle approach. We already know how to prove $P\vee Q,\neg P\vdash Q$, a result called disjunctive syllogism. But that means that $P\vee Q,\neg P,\neg Q$ entails both $\neg Q$ and $Q$, and so $\neg P,\neg Q\vdash \neg (P\vee Q)$ by RA. In the following proof, $P\vee Q,\neg P\vdash Q$ occupies lines 1 through 9. Since we already assumed $\neg Q$ in that proof, we can reuse it to derive $P\vee Q,\neg P,\neg Q\vdash Q\wedge\neg Q$.

1     (1) P∨Q       A
2     (2) ¬P        A
3     (3) P         A
4     (4) ¬Q        A
2,3   (5) P∧¬P      3,2 ∧I
2,3   (6) ¬¬Q       4,5 RA
2,3   (7) Q         6 DN
8     (8) Q         A
1,2   (9) Q         1,3,7,8,8 ∨E
1,2,4 (10) Q∧¬Q     9,4 ∧I
2,4   (11) ¬(P∨Q)   1,10 RA

Note for more advanced readers: this proof is intuitionistically valid (since EFQ can be used instead of RA and DN).

Excluded Middle

It was already possible to prove Excluded Middle with the previous rules. But RA makes the proof more intelligible and more efficient.

1   (1) ¬(P∨¬P)          A
2   (2) P                A
2   (3) P∨¬P             2 ∨I
1,2 (4) (P∨¬P)∧¬(P∨¬P)   3,1 ∧I
1   (5) ¬P               2,4 RA
1   (6) P∨¬P             5 ∨I
1   (7) (P∨¬P)∧¬(P∨¬P)   6,1 ∧I
    (8) ¬¬(P∨¬P)         1,7 RA
    (9) P∨¬P             8 DN

Redundancies in our system

We have kept the rules almost to a minimum, but there are few different reductions that could be made.

Without MT and DN-Intro

Now that we have RA, we could actually eliminate MT and DN-Intro. He’s a way to “simulate” MT using RA.

1     (1) P→Q      A
2     (2) ¬Q       A
3     (3) P        A
1,3   (4) Q        1,3 MP
1,2,3 (5) Q∧¬Q     4,2 ∧I
1,2   (6) ¬P       3,5 RA

And here’s a way to simulate DN-Intro:

1   (1) P      A
2   (2) ¬P     A
1,2 (3) P∧¬P   1,2 ∧I
1   (4) ¬¬P    2,3 RA

Is it possible to show that that the system without MT and DN-Intro is minimal, i.e. that none of the remaining rules can be simulated by others. I think that the answer to that question is yes, and I think it might be shown by altering the truth tables of the connectives in suitable ways. For example, if we altered the truth table of $\wedge$ so that $P\wedge Q$ is true in all circumstances, then (conjecture) all of the other rules are still truth-preserving, but $\wedge$-elim is not.

Without Reductio ad Absurdum

We now show that Reductio ad Absurdum can be reduced to the other rules. Suppose that $\Gamma ,P\vdash Q\wedge \neg Q$. Then $\Gamma \vdash P\to Q$ and $\Gamma \vdash P\to \neg Q$. By contraposition $\Gamma \vdash \neg Q\to \neg P$, and so $\Gamma \vdash P\to\neg P$. By a previous result, $P\to\neg P\vdash\neg P$, and so $\Gamma\vdash \neg P$.

However, RA is, in some important ways, a more natural rule than Modus Tollens. Logical purists do not like MT because it requires the presence of two connectives ($\wedge$ and $\neg$) instead of just one. To restore the purity and symmetry of our rules, we could take RA to be the intro rule for negation, and DN-elimination to be the elim rule for negation.

Long and nasty proofs

To show: $\vdash (P\to Q)\vee (Q\to P)$

This one is not obvious. You could just assume its negation and try to derive a contradiction. From $\neg ((P\to Q)\vee (Q\to P))$, you can use DeMorgans to get $\neg (P\to Q)$ and $\neg (Q\to P)$. Then you can use material conditional to get $P$ from the first, and $\neg P$ from the second. Easier said than done, of course.

Another approach is to start by deriving excluded middle: $P\vee \neg P$. Then from $P$, use positive paradox to get $Q\to P$. And from $\neg P$ use negative paradox to get $P\to Q$. Use $\vee$-intro on both sides to get $(P\to Q)\vee (Q\to P)$, and then finish off by applying $\vee$-elim to $P\vee\neg P$.

To show: $P\to (Q\vee R)\:\vdash\: (P\to Q)\vee (P\to R)$

There are at least two methods that you could use on this one. The first is to assume the negation of the conclusion and apply DeMorgans. That will yield $\neg (P\to Q)$ and $\neg (P\to R)$. Then material conditional will give $P,\neg Q,\neg R$, from which you can derive a contradiction with the premise.

The second method is to derive $P\vee \neg P$. In the case that $P$, the premise gives $Q\vee R$, and in either case we have $(P\to Q)\vee (P\to R)$. In the case that $\neg P$, we have $P\to Q$ and hence $(P\to Q)\vee (P\to R)$.

Important sequents

DeMorgans $\neg (\phi\vee\psi )\vdash \neg\phi\wedge\neg\psi$

Material Conditional $\neg (\phi\to\psi )\vdash \phi\wedge\neg \psi$

Excluded Middle $\vdash \phi\vee\neg\phi$

Disjunctive Syllogism $\phi\vee\psi ,\neg \phi \vdash \psi$.