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Boolean algebra

I'd like to solve the following problem in Mathematica,

Given are the following preliminaries:
(a1 || a2)=True
(a1 && a2) = False

Then, given the following conjunctive normal form:

(b1 || a1) && (b1 || a2)

This can be simpliefied to: b1 || (a1 && a2)

Given the prelininaries I'd like Mathematica to simplify this to: b1.
However, I cant figure out how to do this. I tried the following:

prelims := {(a1 && a2) -> False, (a1 || a2) -> True};
f1 := (b1 || a2) && (b1 || a1);
Simplify[f1] /. prelims

and this seems to work. However

f2 := (b1 || a1 || c1) && (b2 || a2 || a1);
Simplify[f2] /. prelims

evaluates to: (b1 || a1 || c1) && (b2 || a2 || a1) instead of b1 || a1
|| c1.
The reason of this seems to be the order in which a2 and a1 occur in
f2. How can I make sure it works regardless of the order of a1 and a2?

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