Boolean Algebra: Can't prove that ô.e+ô.ê+o.e=ô+e (the accent means negation).

[haiku11]

Consider a Boolean algebra (B,+,.,negation symbol). 0 is the zero element and 1 is the unit element. + has the lowest precedence and negation has the highest. Show that ô.e+ô.ê+o.e=ô+e for all o and e in B.

This isn't a logic gate algebra because that'd be easy, so I can't use the values 0s and 1s to prove this. I can only use the Boolean algebra laws:
Double Complement law
Idempotent law
Identity law
Domination law
Commutative law
Associative law
Distributive law
De Morgan's law
Absorption law
Unit property
Zero property

This is one of my attempts and I got pretty close but still pretty close isn't the answer.

 ô.e+ô.ê+o.e
=ô.(e+ê)+o.e      cause of distributive law
=ô.1+o.e       cause of unit property
=ô.o+ô+o.e      1=o+ô cause of unit property, I'm unsure as to whether o+ô should be in brackets but I assumed not
=0+ô+o.e      cause of zero property
=o.ô+ô+o.e      0=o.ôcause of zero property
=o.ô+o.e+ô        cause of commutative property
=o.(ô+e)+ô        cause of distributive property

I didn't know what to do at this point.

 

[Valkarie]

Any help would be appreciated. The answer is: ô.e+ô.ê+o.e=ô+(e+ê)=ô+1 (by unit property)=ô (by zero property)