# FullSimplify versus Reduce applied to set of inequalities

I have a complicated expression Expr describing a bunch of inequalities and I am trying to reduce it to a manageable subset that facilitates ready interpretation. I am trying to understand the difference between using Reduce[] and FullSimplify[] to accomplish this.

e.g.,

Expr = Lambda > 0 && beta > 0 && omega > 0 && (-1 + aRS aSR) (aSR kR - kS) > 0 && rS > 0 && (-1 + aRS aSR) (-kR + aRS kS) > 0 && ((0 < kS <= (2 (aSR kR - kS))/(-1 + aRS aSR) && dS > 0 && aSR > 0 && aRS > 0 && (kR - aRS kS) ((-1 + aRS aSR) dS kS + (aSR kR - kS) rS) (aSR kR rS - kS (dS + rS)) < 0 && rR > 0) || ((-1 + aRS aSR) (-2 aSR kR + kS + aRS aSR kS) > 0 && ((0 < dS < ((-2 aSR kR + kS + aRS aSR kS) rS)/((-1 + aRS aSR) kS) && ((0 < aSR <= ((-1 + aRS aSR) dS kS - (-2 aSR kR + kS + aRS aSR kS) rS)/((kR - aRS kS) rS) && aRS > 0 && kR > 0 && rR > 0) || ((-1 + aRS aSR)^2 (-kR + aRS kS) rS ((-1 + aRS aSR) dS kS + (aSR kR - kS) rS) > 0 && rR > 0 && (((kR - aRS kS) ((-1 + aRS aSR) dS kS + (aSR kR - kS) rS) (aSR kR rS - kS (dS + rS)) < 0 && 0 < aRS < (2 (kR - aRS kS) ((-1 + aRS aSR) dS kS + (aSR kR - kS) rS))/((aSR kR - kS) ((-1 + aRS aSR) dS kS - (-2 aSR kR + kS + aRS aSR kS) rS))) || (kR > 0 && aRS >= (2 (kR - aRS kS) ((-1 + aRS aSR) dS kS + (aSR kR - kS) rS))/((aSR kR - kS) ((-1 + aRS aSR) dS kS - (-2 aSR kR + kS + aRS aSR kS) rS))))))) || (dS >= ((-2 aSR kR + kS + aRS aSR kS) rS)/((-1 + aRS aSR) kS) && aSR > 0 && aRS > 0 && (kR - aRS kS) ((-1 + aRS aSR) dS kS + (aSR kR - kS) rS) (aSR kR rS - kS (dS + rS)) < 0 && rR > 0))));

Would anyone know why the output of FullSimplify[Expr] and Reduce[Expr] are quite different?