NO

Problem:
 a__f(b(),X,c()) -> a__f(X,a__c(),X)
 a__c() -> b()
 mark(f(X1,X2,X3)) -> a__f(X1,mark(X2),X3)
 mark(c()) -> a__c()
 mark(b()) -> b()
 a__f(X1,X2,X3) -> f(X1,X2,X3)
 a__c() -> c()

Proof:
 Matrix Interpretation Processor:
  dimension: 1
  interpretation:
   [mark](x0) = 2x0 + 1,
   
   [f](x0, x1, x2) = x0 + 6x1 + 2x2 + 5,
   
   [a__c] = 0,
   
   [a__f](x0, x1, x2) = 2x0 + 6x1 + 4x2 + 5,
   
   [c] = 0,
   
   [b] = 0
  orientation:
   a__f(b(),X,c()) = 6X + 5 >= 6X + 5 = a__f(X,a__c(),X)
   
   a__c() = 0 >= 0 = b()
   
   mark(f(X1,X2,X3)) = 2X1 + 12X2 + 4X3 + 11 >= 2X1 + 12X2 + 4X3 + 11 = a__f(X1,mark(X2),X3)
   
   mark(c()) = 1 >= 0 = a__c()
   
   mark(b()) = 1 >= 0 = b()
   
   a__f(X1,X2,X3) = 2X1 + 6X2 + 4X3 + 5 >= X1 + 6X2 + 2X3 + 5 = f(X1,X2,X3)
   
   a__c() = 0 >= 0 = c()
  problem:
   a__f(b(),X,c()) -> a__f(X,a__c(),X)
   a__c() -> b()
   mark(f(X1,X2,X3)) -> a__f(X1,mark(X2),X3)
   a__f(X1,X2,X3) -> f(X1,X2,X3)
   a__c() -> c()
  Matrix Interpretation Processor:
   dimension: 1
   interpretation:
    [mark](x0) = 4x0 + 2,
    
    [f](x0, x1, x2) = x0 + 5x1 + x2 + 4,
    
    [a__c] = 0,
    
    [a__f](x0, x1, x2) = 4x0 + 5x1 + x2 + 4,
    
    [c] = 0,
    
    [b] = 0
   orientation:
    a__f(b(),X,c()) = 5X + 4 >= 5X + 4 = a__f(X,a__c(),X)
    
    a__c() = 0 >= 0 = b()
    
    mark(f(X1,X2,X3)) = 4X1 + 20X2 + 4X3 + 18 >= 4X1 + 20X2 + X3 + 14 = a__f(X1,mark(X2),X3)
    
    a__f(X1,X2,X3) = 4X1 + 5X2 + X3 + 4 >= X1 + 5X2 + X3 + 4 = f(X1,X2,X3)
    
    a__c() = 0 >= 0 = c()
   problem:
    a__f(b(),X,c()) -> a__f(X,a__c(),X)
    a__c() -> b()
    a__f(X1,X2,X3) -> f(X1,X2,X3)
    a__c() -> c()
   Matrix Interpretation Processor:
    dimension: 1
    interpretation:
     [f](x0, x1, x2) = x0 + 2x1 + 4x2,
     
     [a__c] = 0,
     
     [a__f](x0, x1, x2) = 2x0 + 6x1 + 4x2 + 1,
     
     [c] = 0,
     
     [b] = 0
    orientation:
     a__f(b(),X,c()) = 6X + 1 >= 6X + 1 = a__f(X,a__c(),X)
     
     a__c() = 0 >= 0 = b()
     
     a__f(X1,X2,X3) = 2X1 + 6X2 + 4X3 + 1 >= X1 + 2X2 + 4X3 = f(X1,X2,X3)
     
     a__c() = 0 >= 0 = c()
    problem:
     a__f(b(),X,c()) -> a__f(X,a__c(),X)
     a__c() -> b()
     a__c() -> c()
    Unfolding Processor:
     loop length: 3
     terms:
      a__f(a__c(),a__c(),a__c())
      a__f(a__c(),a__c(),c())
      a__f(b(),a__c(),c())
     context: []
     substitution:
      
     Qed