Definition of Fermion Recently, I encounter a problem about the definition of Fermion operator. In our standard textbooks, the Fermions are defined by their exchange/braiding property, that is, if a minus sign appears by exchanging two Fermions, we say that they are Fermions. Bosonic particles do not have this sign. However, mathematically in the textbooks, the Fermion operators are defined in the following way.
\begin{equation}
\{c_i, c_j\} = \{c_i^\dagger, c_j^\dagger\} =0, \quad \{c_i, c_j^\dagger\} = \delta_{ij}
\end{equation}
The first two equations follow exactly the property of braiding. So my question is, why we still need the second equation to fully define a Fermion? OF course, the exact the same condition happens to the definition of boson operator. This is a basic question in quantum mechanics, but it seems that the textbooks do not give a detailed discussion for this issue. In the Fock space, of course, the second equation seems to be redundant. 
I have this question because of the following paper, 
http://dao.mit.edu/~wen/pub/edgere.pdf
by prof. xiaogang wen. In Eq. 2.10, Prof. Wen said that the wavefunction in Eq. 2.9 (in the above link) is fermionic only when $1/\nu = m$ is an odd number. In his discussion, we do not need to discuss the second equation. 
I fact in the above paper,Prof. Wen did not check that his defined wave function respect $\{\psi(x), \psi^\dagger(x')\} = \delta(x-x')$.
OF course, I can find this discussion in other refs.
I think my question maybe formally asked in a straightforward way. If I can define a operator satisfying the following condition
\begin{equation}
\{c_i, c_j\} = \{c_i^\dagger, c_j^\dagger\} =0,
\end{equation}
then are $c_i$ Fermion particles? The potential controversial in this new definition is that the creation and destruction operators are not well-defined.
 A: You may wish to look at http://en.wikipedia.org/wiki/Fermion#Composite_fermions , in particular: "The number of bosons within a composite particle made up of simple particles bound with a potential has no effect on whether it is a boson or a fermion.
Fermionic or bosonic behavior of a composite particle (or system) is only seen at large (compared to size of the system) distances. At proximity, where spatial structure begins to be important, a composite particle (or system) behaves according to its constituent makeup."
and
"The quasiparticles of the fractional quantum Hall effect are also known as composite fermions, which are electrons with an even number of quantized vortices attached to them."
The commutation relations for composite particles only approximately coincide with the "ideal" commutation relations that you use (see, e.g., the book "Quantum Mechanics" by Lipkin, where commutation relations for composite particles containing two fermions are derived). 
A: I think these equations are not the definition of fermionic operators but property we can derive from the definition of boson and fermion.
we can define the creation and annihilation operators of a particle first, and according to the exchange symmetry we can derive the commutation relations of bosonic and fermionic operators. I guess this is the way to understand this.  
