Prolog Logic/Einstein Puzzle

Question

The problem is

Brown, Clark, Jones and Smith are four substantial citizens who serve the community as architect, banker, doctor and lawyer, though not necessarily respe

Answer 1

Here's my answer to the problem:

puzzle(Puzzle) :-
    Names = [brown,clark,jones,smith],

    permute(Names,Conservatives),

% Brown is more conservative than Jones.
    ismore(brown,jones,Conservatives),

% Brown is less conservative than Smith.
    isless(brown,smith,Conservatives),

    permute(Names,Golfs),
    permute(Names,Ages),

% Brown is a better golfer than those older than him.
    worsethans(brown,Golfs,WorseAtGolfThanBrown),
    betterthans(brown,Ages,OlderThanBrown),
    members(OlderThanBrown,WorseAtGolfThanBrown),

    permute(Names,Incomes),

% Brown has a higher income than those younger than Clark.
    worsethans(brown,Incomes,WorseIncomeThanBrown),
    worsethans(clark,Ages,YoungerThanClark),
    members(YoungerThanClark,WorseIncomeThanBrown),

    permute([banker,architect,lawyer,doctor],Jobs),

% Banker has a higher income than architect.
    lookup(banker,Jobs,Names,Banker),
    lookup(architect,Jobs,Names,Architect),
    ismore(Banker,Architect,Incomes),

% Banker is neither youngest nor oldest.
    ([_,Banker,_,_]=Ages;[_,_,Banker,_]=Ages),

% Doctor is a worse golfer than lawyer.
    lookup(doctor,Jobs,Names,Doctor),
    lookup(lawyer,Jobs,Names,Lawyer),
    ismore(Lawyer,Doctor,Golfs),

% Doctor is less conservative than architect.
    ismore(Architect,Doctor,Conservatives),

% Oldest is most conservative and has highest income.
    [Oldest,_,_,_]=Ages,
    [Oldest,_,_,_]=Conservatives,
    [Oldest,_,_,_]=Incomes,

% Youngest is the best golfer.
    [_,_,_,Youngest]=Ages,
    [Youngest,_,_,_]=Golfs,

    Puzzle = [Names,Jobs,c(Conservatives),g(Golfs),i(Incomes),a(Ages)].

It needs these supporting predicates:

ismore(X,Y,Zs) :-
    append(Xs,[Y|_],Zs),
    member(X,Xs).

isless(X,Y,Zs) :-
    append(_,[Y|Xs],Zs),
    member(X,Xs).

betterthans(X,Ys,Zs) :-
    append(Zs,[X|_],Ys).

worsethans(X,Ys,Zs) :-
    append(_,[X|Zs],Ys).

%lookup(X,Xs,Ys,Y)
lookup(X,[X|_],[Y|_],Y).
lookup(X,[_|Xs],[_|Ys],Y) :-
    lookup(X,Xs,Ys,Y).

members([], _).
members([X|Xs], Ys) :-
    member(X, Ys),
    members(Xs, Ys).

select([X|Xs], X, Xs).
select([X|Xs], Y, [X|Ys]) :- select(Xs, Y, Ys).

permute([], []).
permute(Xs, [X|Zs]) :-
    select(Xs, X, Ys),
    permute(Ys, Zs).

Now, the only issue I had is that this gives me more than one answer. Unless I've got the logic wrong this is what I got:

[[brown,clark,jones,smith],[banker,architect,doctor,lawyer],c([smith,brown,clark,jones]),g([clark,brown,smith,jones]),i([smith,brown,clark,jones]),a([smith,brown,jones,clark])]
[[brown,clark,jones,smith],[banker,architect,doctor,lawyer],c([smith,brown,clark,jones]),g([clark,brown,smith,jones]),i([smith,brown,jones,clark]),a([smith,brown,jones,clark])]
[[brown,clark,jones,smith],[banker,architect,doctor,lawyer],c([smith,brown,clark,jones]),g([clark,brown,smith,jones]),i([smith,jones,brown,clark]),a([smith,brown,jones,clark])]
[[brown,clark,jones,smith],[banker,architect,doctor,lawyer],c([smith,brown,clark,jones]),g([clark,brown,smith,jones]),i([smith,brown,clark,jones]),a([smith,jones,brown,clark])]
[[brown,clark,jones,smith],[banker,architect,doctor,lawyer],c([smith,brown,clark,jones]),g([clark,brown,smith,jones]),i([smith,brown,jones,clark]),a([smith,jones,brown,clark])]
[[brown,clark,jones,smith],[banker,architect,doctor,lawyer],c([smith,brown,clark,jones]),g([clark,brown,smith,jones]),i([smith,jones,brown,clark]),a([smith,jones,brown,clark])]
[[brown,clark,jones,smith],[banker,doctor,architect,lawyer],c([smith,brown,jones,clark]),g([jones,brown,smith,clark]),i([smith,brown,clark,jones]),a([smith,brown,clark,jones])]
[[brown,clark,jones,smith],[banker,doctor,architect,lawyer],c([smith,brown,jones,clark]),g([jones,brown,smith,clark]),i([smith,brown,jones,clark]),a([smith,brown,clark,jones])]
[[brown,clark,jones,smith],[banker,doctor,architect,lawyer],c([smith,brown,jones,clark]),g([jones,brown,smith,clark]),i([smith,clark,brown,jones]),a([smith,brown,clark,jones])]
[[brown,clark,jones,smith],[banker,architect,doctor,lawyer],c([smith,clark,brown,jones]),g([clark,brown,smith,jones]),i([smith,brown,clark,jones]),a([smith,brown,jones,clark])]
[[brown,clark,jones,smith],[banker,architect,doctor,lawyer],c([smith,clark,brown,jones]),g([clark,brown,smith,jones]),i([smith,brown,jones,clark]),a([smith,brown,jones,clark])]
[[brown,clark,jones,smith],[banker,architect,doctor,lawyer],c([smith,clark,brown,jones]),g([clark,brown,smith,jones]),i([smith,jones,brown,clark]),a([smith,brown,jones,clark])]
[[brown,clark,jones,smith],[banker,architect,doctor,lawyer],c([smith,clark,brown,jones]),g([clark,brown,smith,jones]),i([smith,brown,clark,jones]),a([smith,jones,brown,clark])]
[[brown,clark,jones,smith],[banker,architect,doctor,lawyer],c([smith,clark,brown,jones]),g([clark,brown,smith,jones]),i([smith,brown,jones,clark]),a([smith,jones,brown,clark])]
[[brown,clark,jones,smith],[banker,architect,doctor,lawyer],c([smith,clark,brown,jones]),g([clark,brown,smith,jones]),i([smith,jones,brown,clark]),a([smith,jones,brown,clark])]

I can constrain it to a single solution if I also say that Clark's income is greater than Brown's.

Can anyone confirm if my answer is correct or not and if there should be more constraints?