Question about simplification



Also:

(%i1) ? lratsubst;
 -- Function: lratsubst (<L>, <expr>)
     is analogous to `subst (<L>, <expr>)' except that it uses
     `ratsubst' instead of `subst'.

     < rest deleted>


--Barton
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/* ratsubst with multiple equation input -- warning: does no error checking on input */
eratsubst(eqs,val) := (while eqs # [] do (val: ratsubst(rhs(first(eqs)),lhs(first(eqs)),val), eqs: rest(eqs)), val)$

            -s