Q:

A skew symmetric map $\phi$ if $\phi(x_1, ... ,x_p)= 0$ when $x_i = x_j$, then $\phi(x_1, ... ,x_p)= 0$ when $x_1, ... ,x_p$ are lin.  dependent

In the book of Linear Algebra By Werner Greub at page 100, it is given that;
Let $\phi$ a p-linear map from E to F.Then the following statements are equivalent. 
...
(i) $\phi$ is skew symmetric
(ii) $\phi(x_1, ... ,x_p)= 0$ whenever $x_i = x_j$ for some pair $i\not = j$
(iii) $\phi(x_1, ... ,x_p)= 0$ whenever the vectors $x_1, ... ,x_p$ are linearly dependent. 
My question is if we assume that $(ii)$ true, the vectors $x_1, ... ,x_p$ have be linearly dependent since $x_i = 1_E x_j$, so (iii) is a direct conclusion of $(ii)$, but in the book, it chooses $p$ without lost of generality and writes it as a linear combination of the vectors $x_1, ... ,x_{p-1}$, but I didn't get it.  I mean there is nothing wrong in that proof, but I think it is unnecessary because there is no way that (ii) is true and the vector $x_1, ... ,x_p$ are linearly independent. 
Edit: 
What I didn't get it is that when $ii$ is assumed, is there any way that $iii$ can be false ? 

A:

The problem is that the statements (ii) and (iii) have quantifiers in them : the word "whenever" must be understood as "for all $x_1,\dots,x_p$ such that...". 
Condition (ii) tells you only what happens when two vectors in $x_1,\dots,x_p$ are equal, and doesn't tell you anything about what happens when $x_1,\dots,x_p$ are only linearly dependent; (ii) is only a special case of (iii), and that's why you need a proof. 

Let us make this more abstract.  Let us give name to some affirmations :

$A $ will denote the affirmation "$\phi(x_1,\dots ,x_p)=0$"
$B$ will denote the affirmation "$x_i=x_j$ for some $i\neq j$"
$C$ will denote the affirmation "$x_1,\dots ,x_p$ are linearly dependent". 

Then $(ii)$ is the affirmation
$$\forall x_1,\dots ,x_p\, (B\Rightarrow A)$$
and $(iii)$ is the affirmation
$$\forall x_1,\dots ,x_p \, (C\Rightarrow A).$$
Moreover, $B\Rightarrow C$; so if $(iii)$ holds then we have for all $x_1,\dots ,x_p$
$$B\Rightarrow C\Rightarrow A,$$
and thus $(ii)$ holds (this is why I said that $(ii)$ was a special case of $(iii)$).  So the proof of $(iii)\Rightarrow (ii)$ is simple. 
But if you only know $(ii)$, then you have $B\Rightarrow A$ and $B\Rightarrow C$, but you can't deduce that $C\Rightarrow A$.  Doing so would just be a bad syllogism, which is a frequent logical mistake.