Proof of Correctness of PeakFindingImproved algorithm

Consider Step $3.$ This will be executed only if A$[$mid $1]>$ A$[$mid$]$ Therefore, any element that is a peak in A$[$x$..$mid $1]$ will be a peak. For indexes $<($mid $1),$ it is obvious; if $($mid $1)$ is a peak in A$[$x$..$mid $1],$ then we know $($mid $1)$ is a peak in A$[$x$..$y$],$ because A$[$mid $1]>=$ A$[$mid $2]($from Step $3$ in Algorithm$)$ and A$[$mid $1]>=$ A$[$mid$]($from Step $2$ in Algorithm$).$ We can similarly prove for Step $5.$ We can also prove Step $6$ is correct, when Step $6$ is executed.