Use R ::= Q[x,y,z];
F := x+y+z;
Eval(F,[2,1]); -- let x=2 and y=1 in F
z + 3
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Subst(F,[[x,2],[y,1]]); -- the same thing using Subst
z + 3
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Subst(F,y,1); -- the syntax is easier when substituting for a single
-- indeterminate
x + z + 1
-------------------------------
Subst(F,[[y,x-y],[z,2]]); -- substitute x-y for y and 2 for z
2x - y + 2
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