Dependence of Axioms of Equivalence Relation? This question is problem 11(a) in chapter 1 in 'Topics in Algebra' by I.N. Herstein. 
These are the properties of equivalence relation given in this book.


*

*Prop 1 $a \sim a$

*Prop 2 $a \sim b$ implies $b \sim a$.

*Prop 3 $(a \sim b$ and $b \sim c)$ imply $a \sim c$.


Statement Property 2 of an equivalence relation states that if $a \sim b$ then $b \sim a$. By property 3, we have the transitivity i.e. $a \sim b$, and $b \sim c$ then $a \sim c$. What is wrong with following proof that property 2 and 3 imply property 1? Let $a \sim b$; then $b \sim a$, whence by property 3 (using $a = c$), $a \sim a$.

I think I can prove this to be wrong. Without proving equivalence relation first, one can not use $a = c$. Right? After all, equality is equivalent to 'equivalence relation' and 'axiom of substitution' are satisfied. If this is right, then I have trouble with the next part of this problem.

Part 2 Can you suggest an alternative of property 1 which will insure us that prop 2 and prop 3 do imply 1?
Can one give such a formulation without using the idea of '=' or otherwise?

EDIT : Italics are my comments. Rest is as it appeared in the book.
Notion of Equality
I have read in Terry 'Analysis 1' in Appendix A.7 published by Hindustan Book Agency that there are four axioms of 'equality'. First 3 are same as equivalence relation where $\sim $ in replaced by $ = $. The fourth one is known as axiom of substitution. Given any two objects $x$ and $y$ of some type, if $ x = y $, then $f(x) = f(y) $ for all functions or operations $f$. 

For the first part: The "proof" assumes that for $a$ there is a $b$ such that $a\sim b$. This of course is not necessarily given. The empty relation i.e. for no $a,b$ we have $a\sim b$ is transitive, symmetric but not reflexive. (This example seems to many students not very explanatory although it is the simplest example of this situation. See the example of Arturo Magidin if that helps.)
For the second part: I think the book might ask for something like this.
Prop 1': For any $a$, for which there is any $b$ such that $a\sim b$, we also have $a\sim a$.
This might seem to be an weird property, but we could also take "suffices Prop 2 and Prop 3", which wouldn't make much more sense. 
Or as Arturo Magidin suggested
Prop 1'': For any $a$, there is a $b$ such that $a\sim b$
together with Prop 2 and Prop 3 implies Prop 1.
