# Clifford algebra is isomorphic to exterior algebra, proof in Lawson's

Context: We have a homomorphism of graded algebra $$\wedge^*(V) \rightarrow G^*$$ induced on each component from the map $$\wedge^r(V) \rightarrow G^r, \quad v_{i_1} \wedge \cdots \wedge {v_{i_r}} \mapsto [v_{i_1} \wedge \cdots \wedge v_{i_r}]$$ where $G^r$ are components of associated graded algebra of $Cl(V,q):=T(V)/J_q(V)$. where $T(V)$ is tensor algebra, $J_q(V)$ is ideal generated by $\{ v \otimes v + q(v) \rangle \, : \, v \in V \}$.

Question: It suffices to show for a $\varphi \in T^r(V) \cap J_q(V)$ (the $r$ homogenous pieces of elements in $J_q(V)$), $\varphi$ is $0$ in $\wedge^r(V)$, under quotient $T^r(V) \rightarrow \wedge^r(V)$.

[...] Any such $\varphi$ can be written as a finite sum $\varphi = \sum a_i \otimes (v_i \otimes v_i + q(v_i) ) \otimes b_i$ where wlog, $a_i, b_i$ are homogenous and $\deg a_i+\deg b_i \le r-2$.

The last statement "wlog" part is not clear to me. Why cannot $\deg a_i+\deg b_i =r$? This simply means $\sum a_i \otimes v_i \otimes v_i \otimes b_i =0$. and this is possible?