@digerdiga 

Maple 2016: For infinite sums, Maple is now more careful regarding potentially divergent parametric sums.

@digerdiga 

Mathematically the equality is not true without restrictions. The default option is now formal=false.

@nm 

But e.g. dsolve cannot solve the simple:

dsolve(diff(y(x),x)+Int((y(x)),x=0..x)=x,y(x));

and also

dsolve(diff(y(x),x)+Int((y(x)),x=0..1)=x,y(x));

@acer 

For an ODE, y(x) under Int should be (probably) also rejected.

@Adam Ledger 

1. Any programmer or designer commits mistakes ==> bugs; e,g. one of the many possible cases was not considered. Once discovered, such bugs are fixed.

2. There are theoretical aspects. E.g. there is no general algorithms able to detect whether an expression is 0 or not. If such a zero expression is not detected, a series expansion could be wrong. In this case, a workaround (of the user) could be to rewrite the expression in a simpler form.
These "bugs" cannot be fixed, but maybe a more general (not "universal" though) method could be implemented.

You should post such problems and maybe some user will come with a confirmation/workaround/explanation.

@sand15 

Any name returned from the concatenation operator will be a global variable. See ?||

@Carl Love 

I am not sure whether ad hoc is the right term. If a user asks something about a function f : R --> R,  should the answer refer to the case f : X --> Y,  with X,Y Banach spaces, in order to be non "ad hoc"?

@Carl Love 

My point is that it is not very difficult to write a procedure with a local status for a variable.
 

@Carl Love 

Maybe something like this:

restart;
x:=10:
pl:=proc(f::uneval, r::uneval)
plot(f, r);
end:
pl(sin(x), x=-Pi..Pi);

@Carl Love 

I know, my answer was in the OP's context and it is correct.

@nm 

Because it is global.

@Carl Love 

Heinz'  is not that bad with a small change.

Heinz:= n-> min(select(d-> (d^2>n), Divisors(n))):

@Carl Love 

I wonder how useful is the type  `&+` due to the fact that the ordering of the terms in a sum could be difficult to predict.

@Al86 

This is not true. Take for instance p_0(t) = -t  or  p_0(t) = -t/2.
Not even when p_0 > 0.