Proof by Contradiction with Multiple Axioms

Looking at proofs by contradiction and it seems I've run into something that does not sit well with me. I am fine with the law of the excluded middle (thus not an intuitionist) and more fundamentally the Principle of Explosion seems reasonable.

The standard form is:

$(P \wedge \neg Q \implies \bot) \implies (P \implies Q)$

However I've seen a number which claim to be reductio ad absurdem but follow the following format:

$(\neg P \implies \bot) \implies P$

Which seems to be not entirely robust when people use it in a similar way to as follows. Let:

$P = A \wedge B \wedge C$

Then through reductio ad absurdem they find that P is true. Thus any of A, B or C is true.

Think of the infinite primes proof with the original P statement, "If there are infinite primes and cats are plants". I'm concerned about this use. Thanks.

• In classical logic : $((P \land \lnot Q) \rightarrow \bot) \rightarrow (P \rightarrow Q)$ is simply : $\lnot (P \land \lnot Q) \rightarrow (P \rightarrow Q)$ which amount to : $(P \rightarrow Q) \rightarrow (P \rightarrow Q)$. So, what are you meaning with : "standard form" ? Sep 2 '14 at 8:25
• I was merely talking about the standard way in which I have seen these proofs being done. My logic and mathematics background is restricted to third year university mathematics and a Coursera logic course.
– AER
Sep 2 '14 at 9:48
• Your "second format" is a particular case of your "standard form". Note indeed that (i) $\neg Q$ is equivalent to $\top\land\neg Q$, and that (ii) $\top\Rightarrow Q$ is equivalent to $Q$. Thus, your "standard form" $((P\land\neg Q)\Rightarrow \bot)\Rightarrow(P\Rightarrow Q)$ instantiates into $((\top\land\neg Q)\Rightarrow \bot)\Rightarrow(\top\Rightarrow Q)$, and this is turn equivalent, given (i) and (ii), to $(\neg Q\Rightarrow \bot)\Rightarrow Q$, a slight variation of your "second format". Sep 2 '14 at 17:25

The RAA principle :

$(¬P → \bot) → P$

can be equivalently rewritten as :

$\lnot \lnot P → P$

which is Double Negation.

If you agree with it, you are using classical logic.

Thus, when you apply it to a "complex" statement like : $P := A \land B \land C$, you have simply :

$\lnot \lnot (A \land B \land C) \rightarrow (A \land B \land C)$.

If we consider the original formulation : $(¬P → \bot) → P$, with $P := A \land B \land C$, the denial of $P$ amounts to :

$\lnot (A \land B \land C)$

which, again in classical logic, is :

$\lnot A \lor \lnot B \lor \lnot C$.

Proving that this assumption implies the falsum (i.e. $\bot$) amounts to saying that no one between $\lnot A,\lnot B,\lnot C$ is true, and thus that $A,B,C$ are all true.

Note : also the above use of disjunction is not intuitionistically allowed ...

In your example, you are trying to derive a contradiction (⊥) from the denial of :

Primes are infinite and Cats are plants

which is :

Primes are not infinite or Cats are not plants.

How can we do this ?

By the ∨-elim rule :

if P⊢A and Q⊢A, then P∨Q⊢A.

Assuming a "standard" theory of numbers, according to which we can prove that "Primes are infinite", we can derive ⊥ from the assumption : "Primes are not infinite" (the first disjunct).

But what about the second disjunct : "Cats are not plants" ?

• You are right. Thank you for spelling it out so clearly. I've had a number of lecturers do this very sloppily and effectively shown a contradiction and selected one of the many axioms used to be the one that is contradicted. Which is as you have pointed out sloppy logic. Fears allayed; logic stands as always.
– AER
Sep 2 '14 at 9:56
• @AER - I think that your "lecturers" mix-up different rules all naming them "proof by contradiction". It may help to see my answer to this post. As you can see, the rule we are discussing is RAA, which is classically equivalent to Excluded Middle and Double Negation. We have another rule, which is intuitionistically valid, call the Absurdity rule or Ex Falso Quodlibet : "from $\bot$, derive $\varphi$". Is this rule which licences us, in presence of a contradiction ($\bot$) ... 1/2 Sep 2 '14 at 10:18
• ... to introduce a formula $\varphi$ whatever. There is also another rule, called $\lnot$-introduction : "if $\varphi \vdash \bot$, then $\lnot \varphi$". Of course, if you derive $\bot$ in a proof with more than one assumption, you can choose the assumption to be "blamed" for the contradiction , deriving its denial. 2/2 Sep 2 '14 at 10:27

The statement $P=$"there are infinite primes and cats are plants" cannot be proven by contradiction.

Since $\neg P$ is "There are finite primes or cats are not plants" is a true statement, the statement $\neg P\implies\bot$ is not true, so you cannot conclude that $P$ is true.

• Do you mean "or cats are not plants"? What I'm saying is though you can use the first statement alone (by expanding the "and").
– AER
Sep 2 '14 at 9:51
• The negation of $A\wedge B$ is $\neg A \vee \neg B$, so yes.
– 5xum
Sep 2 '14 at 9:53