Laws of Thought

In his 1912 book The Problems of Philosophy, Bertrand Russell wrote the following three laws of thought, which he considered “self-evident logical principles”.

  1. Law of Identity: Whatever is, is.  Symbolically:  \mathbf{A} \equiv \mathbf{A}
  2. Law of Noncontradiction: Nothing can both be, and not be.  Symbolically: \neg \big(\mathbf{A}\wedge \neg \mathbf{A}\big)
  3. Law of Excluded Middle: Everything must either be, or not be.  Symbolically: \mathbf{A}\vee\neg\mathbf{A}

These all, at first glance, seem self-evident.  However, upon closer inspection it turns out that all three are wrong (or at least, are limited in scope).  In most common cases, Russell is correct, and all three apply.  Any time a mathematician does a proof by reductio ad absurdum

\big(\mathbf{A}\Rightarrow\neg\mathbf{A}\big)\Rightarrow\neg\mathbf{A}
he is applying the second and third assumptions (#2 states that if \mathbf{A}\Rightarrow\neg\mathbf{A} then \mathbf{A} cannot be true, but #3 states that if \mathbf{A} is not true, then \neg\mathbf{A} must be true).

1.  Law of Identity: Whatever is, is.  \mathbf{A} \equiv \mathbf{A}

The law of identity holds if we only restrict ourselves to equivalence relations (it is, in fact, one of the three properties in the definition of equivalence relations, the other two being symmetry and transitivity).  If we consider all possible relations it may or may not hold.  It is trivial to find an example which does not hold:  the “less than” operator.  5<5 is clearly not true.  Relations which have this property are called reflexive relations, but they certainly do not make up all possible relations.  When considering more complex relations than “equals” or “less than”, it is worthwhile to check if you are dealing with a reflexive relation or not, rather than to assume that all relations are reflexive.

2.  Law of Noncontradiction: Nothing can both be, and not be.  \neg \big(\mathbf{A}\wedge \neg \mathbf{A}\big)

The law of noncontradiction is actually an assumption on the theory you are working within, the assumption of consistency.  Unfortunately, thanks to Gödel, given a sufficiently complex logical theory, it is impossible to prove that it is consistent (or rather, if you can prove that it is consistent, then it is inconsistent).  Further, not all logical systems are consistent, so this clearly doesn’t hold in all cases.

3.  Law of Excluded Middle: Everything must either be, or not be.  \mathbf{A}\vee\neg\mathbf{A}

For the law of excluded middle, we will look at Russel’s paradox, given a consistent logical system.  Russel’s paradox is roughly this:  Define the set of all sets which do not contain themselves, and then ask if this set is a member of itself.  Formally:

\mathbf{A} = \big\{\mathbf{X}|\mathbf{X}\not\in\mathbf{X}\big\}
  What is the truth value of the statement \mathbf{P} = \big(\mathbf{A}\in\mathbf{A}\big)?

Assume for a moment that \mathbf{P} holds, that is, \mathbf{A} is a member of itself.  In that case, by the definition of \mathbf{A}, \mathbf{A} is not a member of itself, \neg\mathbf{P}.  As well, if \neg\mathbf{P} holds, then \mathbf{A} is not a member of itself.  In that case, by the definition of \mathbf{A}, \mathbf{A} is a member of itself, \mathbf{P}.

If we assume that #2 holds, that is, we cannot have \mathbf{P}\wedge\neg\mathbf{P}, then we must have neither \mathbf{P} nor, \neg\mathbf{P}.

If on the other hand, we assume that #2 doesn’t hold, then the theory is inconsistent, and everything can be proven true and not true.  If 2 doesn’t hold, then there exists at least one \mathbf{X} such that \mathbf{X}\wedge\neg\mathbf{X}.  Given \mathbf{X}, we know that the statement \mathbf{P} = \big(\mathbf{X}\vee\mathbf{Y}\big) must be true for any \mathbf{Y}.  But given \neg\mathbf{X} and \mathbf{P} we conclude that \mathbf{Y}.  The same holds for \neg\mathbf{Y}, therefore, letting \mathbf{Y} = \big(\mathbf{A}\vee\neg\mathbf{A}\big), \neg\big(\mathbf{A}\vee\neg\mathbf{A}\big), and #3 doesn’t hold.

Notes:

I am aware that my treatement of #3 uses the assumption of #3 in it’s proof (that is, I assumed either #2 or not #2) this is perfectly acceptable, since if the assumption is wrong, we are already done.

A treatment of #3, which is less interesting, can be seen with the statement “This statement is false”.  Symbolically: \mathbf{P} = \neg\mathbf{P}.

This is meant to be an interesting discussion of formal logic, rather than a thorough treatment of the subject, which would take several university courses, and even then you wouldn’t cover it all.  As such, it is far from rigorous, intentionally so.

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