Trying to prove that if $F=\alpha\wedge\beta+\sigma\wedge\tau$ then $\alpha,\beta,\sigma,\tau$ are linearly independent. $F$ is a $2$-form, $\alpha,\beta,\sigma,\tau$ are $1$-forms and $n=4$. Here is my attempt:
Suppose $F=\alpha\wedge\beta+\sigma\wedge\tau$, with both $2$-forms. Then suppose for a contradiction that the $1$-forms $\alpha,\beta,\sigma,\tau$ are not linearly independent. Then it is easy to see that $\alpha\wedge\beta\wedge\sigma\wedge\tau=0$, by permuting elements as required. Now, $F\wedge F=2\alpha\wedge\beta\wedge\sigma\wedge\tau=0$, so since $F$ is a $2$-form we conclude $F$ is simple from part (a). But then $F=\gamma\wedge\epsilon=\alpha\wedge\beta+\sigma\wedge\tau$, which I think is a contradiction but I'm not sure.
I'm not sure if why I say it's a contradiction is the contradiction I'm supposed to get. What is wrong with $\gamma\wedge\epsilon=\alpha\wedge\beta+\sigma\wedge\tau$?
Note: (a) was just the proof that if $n=4$ and $F$ is a $2-form$ then $F\wedge F=0$ implies $F$ is simple/decomposable. i.e. $F=v\wedge w$, for two $1$-forms $v,w$.
I think this is the reasoning:
Assume both the $2$-forms in the sum are non-zero as a user commented. Now, wedge both sides of the final equality by $\beta$ from the right. We get $$\gamma\wedge\epsilon\wedge\beta=0+\sigma\wedge\tau\wedge\beta.$$ I believe this implies that $\gamma\wedge\epsilon=\sigma\wedge\tau$. But in this case $\alpha\wedge\beta=0$ contradicting the initial assumption.