Expanding Expressions
Ruby, 698 648
O=[A='&amp;',B='+',C='*']
M='Syntax error'
def u(x)c=x.shift;(c.nil?)?S[0]=~/[a-z]/?(S.shift):S[0]=='('?(S.shift;o=u(O*1);S[0]==')'?(S.shift;o):abort(M)):abort(M):(o=[c];loop{o&lt;&lt;u(x*1);break if S[0]!=c;S.shift};o)end
def d(e,a,b)a!=e[0]?e:(i=e.index{|x|x[0]==b};(i.nil?)?e:y([b]+e[i][1..-1].map{|k|z=e*1;z[i]=k;z}))end
def y(e)e.map!{|x|(x.is_a?Array)?(y(x)):(x)};o=[e[0]];e[1..-1].map{|x|x[0]==o[0]?(o+=x[1..-1]):(o&lt;&lt;x)};d(d(d(o.size==2?o[1]:o,C,B),C,A),B,A)end
def j(e)(e.is_a?Array)?(e.map!{|x|j(x)};o=e.shift;e*(o=='*'?'':&quot; #{o} &quot;)):e;end
S=gets.gsub(/ /,'').gsub(/([a-z)])(?=[a-z(])/,'\1*\2').split(//)
k=u(O*1)
puts S!=[&quot;\n&quot;]?(M):(j y k)

This solution consists of a basic recursive descent parser (method u) which builds an syntax tree which is then flattened (y) and transformed (d) according to the distribution axioms. The output (j) simply joins the operands again.
Unfortunately I broke the functionality when I later tried to reduce the code by restructuring. Nevertheless I think there is still some chance for golfing.