If a subset of a free group $F$ is Nielsen reduced, then it is a basis of $F$. Is the converse statement true? I mean if I take a basis $U$ of $F$, then is it true that it has to be Nielsen reduced?
Nielsen Reduced means this: If $F$ is free group with basis $X$, and $U=(u_1,...)$, an ordered tuple of elements from $F$, we say $U$ is Nielsen reduced if for arbitrary $v_1,v_2,v_3$, where $v_i = u_j^{\pm 1}$, the following conditions hold:
$(N0)$: $v_1 \neq 1$
$(N1)$: $v_1v_2 \neq 1$ implies that $|v_1v_1| \geq |v_1|,|v_2|$
$(N2)$: $v_1v_2 \neq 1$ and $v_2v_3 \neq 1$ implies that $|v_1v_2v_3| > |v_1|-|v_2|+|v_3|.$
A basis does not have to be Nielsen reduced: $(x^{-1},xy)$ is a basis for the free group $\langle x,y \rangle$, but $|x^{-1}xy|=|y|<2=|xy|$ (which contradicts one of the requirements to be Nielsen reduced).
It is true that you can reduce finite sets to Nielsen reduced set, by using Nielsen transformations, and this is discussed in Combinatorial Group Theory by Lyndon and Schupp.