I'm trying to use the following rule:
x[b,c,d]+x[a,c,d]+x[a,b,d]+x[a,b,c]->0
for all distinct integers a,b,c,d to simplify a messy expression with terms like
(x[b,c,d]+x[a,c,d]+x[a,b,d]+x[a,b,c])(x[i,j,k]+x[l,m,n])
but less recognizable.
I hope to find a way to define the rule so that I can use
FullSimplify[expression]/.r
However, it seems like I can only define r with a specific {a,b,c,d}. Is there a way to get a generic r that covers all {a,b,c,d}s?
Any help would be appreciated:D
You can use Subsets and Total as follows:
Total[Subsets[x[a, b, c, d], {3}]] == 0
x[a, b, c] + x[a, b, d] + x[a, c, d] + x[b, c, d] == 0
Update: A function that constructs the desired expression:
ClearAll[f]
f[a_: x] := EqualTo[0] @* Total @* Map[Apply[a]] @* Subsets
Examples:
f[][{a, b, c, d}, {3}]
x[a, b, c] + x[a, b, d] + x[a, c, d] + x[b, c, d] == 0
f[z][{a, b, c, d}, {2}]
z[a, b] + z[a, c] + z[a, d] + z[b, c] + z[b, d] + z[c, d] == 0
f[w][{a, b, c, d}, {1, 2}]
w[a] + w[b] + w[c] + w[d] + w[a, b] + w[a, c] + w[a, d] + w[b, c] + w[b, d] + w[c, d] == 0
Update 2: You can use the expression produced by f as the second argument of FullSimplify:
expression = (x[b, c, d] + x[a, c, d] + x[a, b, d] + x[a, b, c]) (x[i, j, k] + x[l, m, n]);
FullSimplify[expression, f[][{a, b, c, d}, {3}]]
0
Update 3: Similarly you can define a function that generates the desired Rule which can be used with ReplaceAll:
ClearAll[f2]
f2[a_: x] := Rule[#, 0] &@*Total@*Map[Apply[a]]@*Subsets
Examples:
rule = f2[][{a, b, c, d}, {3}]
x[a, b, c] + x[a, b, d] + x[a, c, d] + x[b, c, d] -> 0
expression /. rule
0
