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I would like to define a function f(x,y,z) by its attribute, namely by the specific equation f(b,c,d)f(a,b+c,d)f(a,b,c)=f(a+b,c,d)f(a,b,c+d) for all a,b,c,d. Is it pssible to set this specific attribute to f?

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  • $\begingroup$ Will this be used in simplifying expressions? You could perhaps use your constraint (expressed as an equation, using ==) as an Assumption within Simplify. Can you provide some more context? $\endgroup$
    – MarcoB
    Feb 9, 2017 at 17:10

1 Answer 1

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Use UpSet

f /: f[b_, c_, d_] f[a_, b_ + c_, d_] f[a_, b_, c_] = 
  f[a + b, c, d] f[a, b, c + d];

f[b, c, d] f[a, b + c, d] f[a, b, c]

(*  f[a, b, c + d] f[a + b, c, d]  *)

f[y, x, t] f[z, y + x, t] f[z, y, x]

(*  f[z, y, t + x] f[y + z, x, t]  *)

However, this won't work for numeric inputs that cause the additions to be carried out

f[2, 3, t] f[z, 2 + 3, t] f[z, 2, 3]

(*  f[2, 3, t] f[z, 2, 3] f[z, 5, t]  *)
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