[FOM] Odd Thought About Identity

[Joel I. Friedman]

This is my first participation in the FOM discussions.  I'm glad to be 
aboard.

As I remember way back then , 40 to 50 years ago at UCLA, when I was a 
student taking logic courses from Montague, Kaplan, Carnap, Abraham 
Robinson, etc., we sometimes used two axioms for identity, namely,

x = x      (reflexivity)
x = y --> [F(x) <--> F(y)]      (Leibniz' Law of Substitutivity of Identity)

Then from these two axioms, one can prove:

x = y --> y = x  (symmetry)
(x = y & y = z) --> x = z  (transitivity)

For example, to show symmetry, take as an instance of "F(z)" the 
following: 
"z = x".

Then we have:

F(x) has as instance:  "x = x"
F(y) has as instance:  "y = x"

Then using Leibniz' Law, we have,

x = y --> [x = x <--> y = x].

Using the axiom of reflexivity, we eliminate the clause [x = x], to obtain,

x = y --> y = x  (symmetry)

A similar kind of argument yields transitivity.

Therefore, the three usual laws of identity (namely, reflexivity, 
symmetry, and transitivity) do follow from reflexivity and Leibniz' Law 
alone.  Is this what you had in mind?  It seems too simple.


Joel I. Friedman, Professor Emeritus
University of California
Davis, California 95616




Richard Heck wrote:
> This came up in my logic final. There was a deduction in which one got 
> to here:
>     Rxy . ~Ryx
> and needed to get to here:
>     ~(x = y)
> What a lot of students did was this:
>     (x)(y)(x = y --> Rxy <--> Ryx)
> This does not, of course, accord with the usual way we state the laws of 
> identity, but it struck me that it is, in fact, every bit as intuitive 
> as the usual statement. Which, of course, is why they did it that way.
>
> It wouldn't be difficult to formulate a version of the law of identity 
> that allowed this sort of thing. But I take it that it would not be 
> "schematic", in the usual sense, or in the strict sense that Vaught 
> uses. I wonder, therefore, if a logic that had a collection of axioms of 
> this sort might not yield an interesting example somewhere. Or if there 
> isn't a similar phenomenon somewhere else.
>
> Anyone have any thoughts about this?
>
> Richard
>
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