The free group of rank two is nonamenable

By [L1], write $F_2=⟨a,b⟩$. For $x\in \{a,a^{-1},b,b^{-1}\}$, let $W(x)$ be the set of nonempty reduced words whose first letter is $x$. Put $P=\{a^n:n\ge 0\}$ and $P_{+}=\{a^n:n\ge 1\}$, and define $A_1=W(a)\setminus P_{+}$, $A_2=W(a^{-1})\cup P$, $B_1=W(b)$, and $B_2=W(b^{-1})$.

The four sets are pairwise disjoint and partition $F_2$: the usual five first-letter classes partition $F_2$, and $P_{+}$ has merely been moved from $W(a)$ into the piece containing the identity.

Left multiplication gives $aW(a^{-1})=F_2\setminus W(a)$ and $aP=P_{+}$, hence $F_2=A_1\bigsqcup a{A}_2$. Likewise $bW(b^{-1})=F_2\setminus W(b)$, hence $F_2=B_1\bigsqcup b{B}_2$. Thus the pieces of step 1.1, with translators $e,a,e,b$, satisfy both partition equalities in [L3].

Steps 1.1-3.1 give a paradoxical decomposition of $F_2$, so [L2] implies that $F_2$ admits no left-invariant mean and is therefore nonamenable.