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 $f$. 

There is no difficulty in supplying a "Prop. $0$" strictly weaker than Prop. $1$ such that Prop. $0$ and Props. $2$ and $3$ imply the current Prop. $1$.  Just say that for any $x$ there is a $y$ such that $x \sim y$. (Of course that $y$ might be $x$ itself.)  Then Prop. $0$, together with $2$ and $3$, imply Prop. $1$.  Thus Prop. $0$, $2$, and $3$ give an alternate axiomatization.  Perhaps this is the intended answer.  Or not. 
