# $\mathrm{BPEXP} = \mathrm{BPP} \iff \mathrm{BPEE} = \mathrm{BPE}$

Concerning about a wide variety of complexity classes, I have come up with the above conjecture.

Please, establish the claim in the title formally.

This is true because both sides of the $\iff$ are false.
$\mathrm{BPEXP}\neq \mathrm{BPP}$ because if not, then $\mathrm{BPP} = \mathrm{EXP}$ and by padding, we have $\mathrm{BPEXP} = \mathrm{EEXP} \neq \mathrm{EXP} = \mathrm{BPP}$
$\mathrm{BPEE}\neq \mathrm{BPE}$ because if not, then by padding, we have $\mathrm{BPEEXP} = \mathrm{BPEXP}$, this implies that $\mathrm{BPEXP} = \mathrm{EEXP}$, and by padding again, we have $\mathrm{BPEEXP} = \mathrm{EEEXP} \neq \mathrm{EEXP} = \mathrm{BPEXP}$