Replacement rule only matches part of expression

Question

I have the result of a calculation (which is too long to mention here) and it gives

-I (t2 - t4) (((Hy vx^2 - Hx vx vy + vz (Hz vy - Hy vz)) Re[Ex] + 
   Hx vz^2 Re[Ey] - Ey vz^2 Re[Hx] + 
   vx (Hy vy Re[Ey] - Hz vz Re[Ey] + Ex vy Re[Hx]) + 
   vy (3 Hx vz Re[Ez] - 3 Ez vz Re[Hx]) + (-Ex vx^2 + Ex vz^2 + 
   vx (-Ey vy + 3 Ez vz)) Re[Hy] - Ex vy vz Re[Hz] + 
   vx (-3 Hy vz Re[Ez] + Ey vz Re[Hz]) + 
   vy^2 Real[I Ey Conjugate[Hx]]))

So, I have the following replacement rule

{f_ x_ + f_ y_ -> f (x + y), f_ x_ - f_ y_ -> f (x - y)}

If I now do

3 Hx Re[Ez] vz - 3 Ez Re[Hx] vz //. {f_ x_ + f_ y_ -> f (x + y), f_ x_ - f_ y_ -> f (x - y)}

This gives me

(3 Hx Re[Ez] - 3 Ez Re[Hx]) vz

If I do the same on the above equation the rule just gives the same thing back to me. What I want is (I have manually manipulated this to give what I want to obtain as requested.)

-I (t2 - t4) (-vx vy Abs[Ex]^2) +
   vz^2 (Hx Re[Ey] - Ey Re[Hx]) + 
   vx vy (Hy Re[Ey] + Ex Re[Hx]) + 
   vy vz 3 (Hx Re[Ez] - Ez Re[Hx]) +
   (-vx^2 + vz^2) (Ex Re[Hy] - Hy Re[Ex])
   - vy vx Abs[Hy]^2
   - vy vz (Ex Re[Hz] + Hz Re[Ex])
   vx vz 3 (Hy Re[Ez] + Ez Re[Hy])
   + vx vz (Ey Re[Hz] - Hz Re[Ey]) + 
   vy^2 Real[I Ey Conjugate[Hx]]

My substitution rules are

 replrule1 = {f_ x_ + f_ y_ -> f (x + y), f_ x_ - f_ y_ -> f (x - y)}
 replrule2 = a_Complex x_ Re[y_] + b_Complex y_ Re[x_] :> 
   Abs[a] Real[I x Conjugate[y]] /; a == -b
 replrule4 = x_ Re[y_] - y_ Re[x_] -> Real[I x Conjugate[y]]

After that substitution I also want to apply

b_Complex y_ Re[x_] + a_Complex x_ Re[y_] :>  Abs[a] Real[I x Conjugate[y]] /; a == -b

Which has been a previous question of mine here.

Answer 1

In reply to the remaining issue addressed in the comments, here is the reason the 3 is not taken out of the brackets.

You have

rule = {f_ x_ + f_ y_ -> f (x + y), f_ x_ - f_ y_ -> f (x - y)};

3 Hx Re[Ez] vz - 3 Ez Re[Hx] vz //. rule

vz (3 Hx Re[Ez] - 3 Ez Re[Hx])

leaving the multiple 3 inside. Taking a simplified case, this works:

3 a + 3 b /. f_ x_ + f_ y_ :> f (x + y)

3 (a + b)

But this does not work, for the reason stated by Mr. Wizard:

3 a - 3 b /. f_ x_ - f_ y_ -> f (x - y)

3 a - 3 b

And neither does this:

3 a - 3 b /. f_ x_ + f_ y_ -> f (x + y)

3 a - 3 b

for reason that FullForm[3 a - 3 b] is

Plus[Times[3, a], Times[-3, b]

so f cannot ever match 3 and -3.

An awkward solution to this simple case is:

3 a - 3 b /. 
 f_ x_ + g_ y_ :> 
  Which[f == g, f (x + y), f == -g, f (x - y), True, f x + g y]

3 (a - b)

But this does not work for more complicated inputs.

Simplify on the other hand works ok:

Simplify[3 a - 3 b]

3 (a - b)

3 Hx Re[Ez] vz - 3 Ez Re[Hx] vz // Simplify

3 vz (Hx Re[Ez] - Ez Re[Hx])

And perhaps rule = f_ x_ + g_ y_ :> Simplify[f x + g y] would work for your original case, depending on what you are specifically seeking.

Answer 2

Try ReplaceRepeated (//.)

On examination

The rule is progressively implemented

rule = {f_ x_ + f_ y_ -> f (x + y), f_ x_ - f_ y_ -> f (x - y)}

on a sample selection

 (Hx vz^2 Re[Ey] - Ey vz^2 Re[Hx] + 
   vx (Hy vy Re[Ey] - Hz vz Re[Ey] + Ex vy Re[Hx]) + 
   vy (3 Hx vz Re[Ez] - 3 Ez vz Re[Hx])) /. rule

Output of which is further modified by % /. rule to a final form.

ReplaceRepeated goes straight to the final form:

 (Hx vz^2 Re[Ey] - Ey vz^2 Re[Hx] + 
   vx (Hy vy Re[Ey] - Hz vz Re[Ey] + Ex vy Re[Hx]) + 
   vy (3 Hx vz Re[Ez] - 3 Ez vz Re[Hx])) //. rule

Simple Case

ReplaceAll matches one match:

a + b + c /. x_ + y_ -> x y

a (b + c)

ReplaceRepeated matches all occurrences

a + b + c //. x_ + y_ -> x y

a b c

Answer 3

I still believe this kind of transformation is often misguided. Nevertheless, here are some tips.

You should usually use RuleDelayed (:>) when using named patterns:

rule = {f_ x_ + f_ y_ :> f (x + y)};

Mathematica represents a - b as a + (-1 b), therefore:

{f x + f y, f x - f y - f z, f x - f y + f z} //. rule

{f (x + y), f (x - y - z), f (x - y + z)}