Does 1-ethylidene-2-(propan-2-ylidene)cyclobutane show geometrical isomerism?

[![(1E)-1-ethylidene-2-(propan-2-ylidene)cyclobutane](https://i.stack.imgur.com/aRK6O.jpg)](https://i.stack.imgur.com/aRK6O.jpg)


I think it should be yes, but my book says no.
I think the two geometrical isomers are 


**Z-isomer**


[![(1Z)-1-ethylidene-2-(propan-2-ylidene)cyclobutane](https://i.stack.imgur.com/7s6CQ.jpg)](https://i.stack.imgur.com/7s6CQ.jpg)


**E-isomer**


[![(1E)-1-ethylidene-2-(propan-2-ylidene)cyclobutane](https://i.stack.imgur.com/rjU8l.jpg)](https://i.stack.imgur.com/rjU8l.jpg)




The two isomers are distinct substances, differing only in the orientation of a double bond, so they must be geometrical isomers. However, as a practical matter, they may interconvert under mild conditions and/or one may be much less stable than the other.


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