Is the elimination reaction between 1 1 dibromo cyclopropane and phenyl lithium feasible in alkaline medium (and some heat) I found this question in one of my question papers for the JEE , 
My sir gave me an explanation that due to resonance between lone pairs of the bromide and π bond formed.

This reasoning doesn't seem to be satisfying because of the cyclopropyl ring strain which is not lost even at the end of reaction.
Please give an alternate pathway of the above product is not correct.
There are two pathways that phenyllithium could follow - it could deprotonate, but there really not any reactive protons, or it could do lithium-halogen exchange with one of gem-bromines. This second pathway will lead to a carbene by loss of $\ce{Br-}$ and the carbene (a highly reactive intermediate) will rearrange to give allene $\ce{H2C=C=CH2}$.

This method is documented:


In Page 275: H. Maskill, "Chapter 1: Three- and Four-Membered Carbocyclic Rings," In Specialist Periodical Reports - Aliphatic, Alicyclic and Saturated Heterocyclic Chemistry, Volume 1, Part II: Three‐ and four‐membered rings (carbocyclic and saturated heterocyclic); W. Parker, Ed. (Senior Reporter); The Chemical Society: Burlington House, London, England, 1973, pp. 3-292 (ISBN: 0-85186-502-X). 
In Pages 1492-1493: Armin de Meijere, Editor, Carbocyclic Three-Membered Ring Compounds, Houben-Weyl Methods of Organic Chemistry, Volume E 17 b - Cyclopropanes: Synthesis; Georg Thieme Verlag: Stuttgart, Germany, 1997,  p. 1492 (ISBN: 978-3-13-798204-3).