Question asked:
> I'd like to reduce it to a string that
> would still contain 'n' words but where each word is
> separated by one space.  'repl' isn't exactly working....

I think I know why REPL doesn't work for you:
                        1         2         3
               1234567890123456789012345678901
e.g. setvar a "   abc def:  ghi;jkl ,   mnop  "
     calc     repl(a, "  ", " ")
  result:     "  abc def: ghi;jkl ,  mnop"  (quotes added by me)

so, you need a recursive REPL function that doesn't advance to the
next character position after each replacement operation.  Gavin's
suggestion of:

:SETVAR MYSTR LTRIM(RTRIM(MYSTR))
:WHILE 0 < POS("  ", MYSTR)
:  SETVAR MYSTR REPL(MYSTR, "  ", " ")
:ENDWHILE

is probably the most efficient and concise approach, but I haven't
measured it and compared it to the other choices.

I think it is safe to use WORD as long as you specify the delimiters
as " ".  This will treat all other characters as part of the selected
word/token.  Also, WORD skips blanks before and after the desired
token.  So you can detect when you've hit the end of the string
passed to WORD when WORD returns "" -- valid only when a space is
the sole delimiter.

Here is another possibility, just FYI, showing an empty WHILE
loop and an obscure way of terminating the loop:

parm a
# "a" is a string that needs all spaces folded to a single space.
# Assigns the variable 'result' to the derived string.
setvar strng anyparm(a)
setvar result ""
setvar i 0
while rht(setvar(result,result+ &
                 word(strng," ",setvar(i,i+1))+" "),2) <> "  " do
endwhile
setvar result rtrim(result)

FWIW!

Jeff Vance, CSY