http://scu.edu/cas/religiousstudies/facultystaff/Regular/pinault/index.cfm

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Russell’s theorem (aka russel’s paradox) goes along these lines.

If A implies that B is true if and only if B is not true, then not(A) is true. This is one variation on a theme of proof called proof by contradiction.

In other words, RT (Russell’s theorem) utilizes the following tautology where --> denotes implication and symbolizes the logical terms if/then:

(A -->( B < --> not(B) ) )--> not(A).

< --> denotes if and only if

Russell’s argument uses that tautology with specific statements represented by A and B. Statement A is that there is a set such that all sets are elements of that set. That set can be labeled U = {x : x=x}, assuming every set equals itself. Alternatively, statement A can be a bit more general, for all unary relations P, there is a set U such that for all sets x, x is an element of U if and only if P is true of x.

The conclusion of the theorem is not(A) which is, in other words, there is no set such that all sets are elements of that set.

Statement B is a little more complicated and regards the existence of a set R such that for all sets x, x is an element of R if and only if x is not an element of x (this is not a theorem, it’s a definition of R). If we assume A is true, then there is such a set R. Then someone (named Russell) comes along and asks if R is an element of R and, by definition, R is an element of R if and only if R is not an element of R. Statement B says that R is an element of R.

Thus we have shown that A --> ( B < --> not(B) ), in English: If there is a set with the aforementioned „universal“ property, there is a set that is an element of itself if and only if it is not an element of itself.

Now we invoke the tautology to conclude not(A), i.e., there is no set such that every set is an element of this set.

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