YES

Problem:
 a__f(X,X) -> a__f(a(),b())
 a__b() -> a()
 mark(f(X1,X2)) -> a__f(mark(X1),X2)
 mark(b()) -> a__b()
 mark(a()) -> a()
 a__f(X1,X2) -> f(X1,X2)
 a__b() -> b()

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [f](x0, x1) = x0 + x1 + 5,
   
   [a] = 0,
   
   [a__b] = 0,
   
   [mark](x0) = 2x0,
   
   [a__f](x0, x1) = x0 + x1 + 5,
   
   [b] = 0
  orientation:
   a__f(X,X) = 2X + 5 >= 5 = a__f(a(),b())
   
   a__b() = 0 >= 0 = a()
   
   mark(f(X1,X2)) = 2X1 + 2X2 + 10 >= 2X1 + X2 + 5 = a__f(mark(X1),X2)
   
   mark(b()) = 0 >= 0 = a__b()
   
   mark(a()) = 0 >= 0 = a()
   
   a__f(X1,X2) = X1 + X2 + 5 >= X1 + X2 + 5 = f(X1,X2)
   
   a__b() = 0 >= 0 = b()
  problem:
   a__f(X,X) -> a__f(a(),b())
   a__b() -> a()
   mark(b()) -> a__b()
   mark(a()) -> a()
   a__f(X1,X2) -> f(X1,X2)
   a__b() -> b()
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [f](x0, x1) = 2x0 + 4x1,
    
    [a] = 0,
    
    [a__b] = 4,
    
    [mark](x0) = x0 + 5,
    
    [a__f](x0, x1) = 4x0 + 4x1 + 5,
    
    [b] = 0
   orientation:
    a__f(X,X) = 8X + 5 >= 5 = a__f(a(),b())
    
    a__b() = 4 >= 0 = a()
    
    mark(b()) = 5 >= 4 = a__b()
    
    mark(a()) = 5 >= 0 = a()
    
    a__f(X1,X2) = 4X1 + 4X2 + 5 >= 2X1 + 4X2 = f(X1,X2)
    
    a__b() = 4 >= 0 = b()
   problem:
    a__f(X,X) -> a__f(a(),b())
   DP Processor:
    DPs:
     a__f#(X,X) -> a__f#(a(),b())
    TRS:
     a__f(X,X) -> a__f(a(),b())
    EDG Processor:
     DPs:
      a__f#(X,X) -> a__f#(a(),b())
     TRS:
      a__f(X,X) -> a__f(a(),b())
     graph:
      
     SCC Processor:
      #sccs: 0
      #rules: 0
      #arcs: 0/1