@Anthrazit This gives a True answer.  This post should really be converted to a question.

if is(op(1, b) < op(1, a)) then
    "True";
else
    "False";
end if

Edit:-
I purposely kept in "is" because it forces evaluation.

if 1 < sqrt(2) then
    "True";
else
    "False";
end if;
Error, cannot determine if this expression is true or false: 1 < 2^(1/2)
if is(1 < sqrt(2)) then
    "True";
else
    "False";
end if;

You can see the problem  by looking at op(A), it has two parts when A is non  zero i.e. value,unit but only one part when A is zero i.e. value.

Try if is(a<b) then "True" else ""False" end if

@acer  Just out of interest. Are there other situations outside of trig. where solve misses solutions? I mean towards relatively simple type equations like what is presented above. I realise that might be hard to define.

@acer  Thank you. Nice to read the history and orign of such things.

@tomleslie Am using this version. Thank you.

@vv     @tomleslie explained the cause of my problem well in the first reply. I have since converted the formula.

@tomleslie  Thank you. I understand now. Had never read the details on || or cat. Also checking over replies to my series of questions this year some of the answers have become clearer. The m's formula is extendible. So I just could build it around the size of the list L which I would probably call M. So for example  (1 + m2 + 2*m3 + 3*m4 + 4*m5 + 5*m6 + 6*m7)!  would become (1+add(i * M[i], i=1..nops(M)))! , assuming my typing is correct here.

@acer I should have taken more care. Basicially I was looking for 

 5*c^3+ 21*c^2*d + 28*c*d^2 + 12*d^3 

as the 3rd degree form. I was able to get things working properly overall. 

I have combined you answer with  the answer to a previous question I asked  https://www.mapleprimes.com/questions/232082-Is-It-Possible-To-Addsum-By-Mapping#comment278942 

Download 25_Polynomial_General_Expansion.mw

@acer Yes all the table entries are copy and pasted from other equations. I was trying to make the document look similar to 

what was being done in Scientific Workplace. Basicially I am formatting the documents to print them as I find it easier to flick through printed pages than  open half a dozen Maple documents or videos to find something.

A link to one of the videos. It about developing a power series type solution to polynomials. Solving Polynomial Equations 2: The Quadratic case | Exploring Research Level Maths | Wild Egg Maths | N J Wildberger on Patreon

You might find it interesting.

@acer All the answers on this are very helpful. II missed that part of the help. I use the the help pages as I like to have something to show when I ask a question. Sometimes rushing I mass important details as  in "%".

@acer Thank you. That works well. But I dont understand the syntax. I didnt realise % made operations inert.  I haven't seen that in the help or at least I have found that.

InertForm:-Display~(M,inert=false):  I tried it with inert=true that made no difference and I left out inert altogether and that made no difference either. What am I missing here?

It would be nice to know how to get to display 2! 2!.  I couldn't get  `%*` to work.   

@Carl Love Thank you. When I first asked the question I hadn't bothered about the sign of the terms. which for all cases is (-1)^(m3+1). I got it to work abe modifying the itterator expression. It just means that P=2 doesnt work but easly work around is  coeff(%,c3,0) for P=3

restart;

Iterator:-CartesianProduct([1,1]): #Force compilation


P:= 2..3: #suffixes of m and c variables
N:= [$0..20]: #evaluation values of m variables

B:= subs(
    {_C= [$P], _V= [c||P]}, 
    proc(M)
    local r:= add(M*~_C), s:= 1+r, t:= s-add(M);
        r!*c0^t*mul(_V^~M)/t!/mul(M!~)/c1^s
    end proc
):
CodeTools:-Usage(
    seq[reduce= `+`]((-1)^(v[2]+1)*B(v), v= Iterator:-CartesianProduct(N$(rhs(P)-1)))
)

@Carl Love  to answer both replies here. Yes I had wondered about the For loop when I red your code yesterday but didn't get to try it then.

On the Iterator:-CartesianProduct there is quite a time delay when it is first run after a restart. Then it is really fast. Would adding this into the start up code region (not that I have ever used it) get around this? 

@tomleslie I seperated out the code onto different lines to help me understand what is happening.

The Iterator:-Cartesian product then gave an error. So I had to change $=0..4 to $ 0..4  Why?

As A an B use the arrow operator, why add(A~...   instead of  add(A(m)~......  which actually gives incorrect output.

I see the Iterator:-CartesianProduct is very fast. I gather  this auto compiles which causes some bit of a delay 1st time. Question here. Coud auto compile type routines be added to start up code so restart doesn't affect this?

Download modified_Sums.mw

Both are very useful answers.