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: dim=1
  
  interpretation:
   [f](x0, x1, x2) = x0 + 6x1 + x2,
   
   [c] = 2,
   
   [a__c] = 2,
   
   [mark](x0) = 5x0,
   
   [b] = 2,
   
   [a__f](x0, x1, x2) = 5x0 + 6x1 + x2
  orientation:
   a__f(b(),X,c()) = 6X + 12 >= 6X + 12 = a__f(X,a__c(),X)
   
   a__c() = 2 >= 2 = b()
   
   mark(f(X1,X2,X3)) = 5X1 + 30X2 + 5X3 >= 5X1 + 30X2 + X3 = a__f(X1,mark(X2),X3)
   
   mark(c()) = 10 >= 2 = a__c()
   
   mark(b()) = 10 >= 2 = b()
   
   a__f(X1,X2,X3) = 5X1 + 6X2 + X3 >= X1 + 6X2 + X3 = f(X1,X2,X3)
   
   a__c() = 2 >= 2 = 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: dim=1
   
   interpretation:
    [f](x0, x1, x2) = x0 + 2x1 + x2 + 4,
    
    [c] = 5,
    
    [a__c] = 5,
    
    [mark](x0) = 4x0 + 1,
    
    [b] = 5,
    
    [a__f](x0, x1, x2) = x0 + 2x1 + x2 + 7
   orientation:
    a__f(b(),X,c()) = 2X + 17 >= 2X + 17 = a__f(X,a__c(),X)
    
    a__c() = 5 >= 5 = b()
    
    mark(f(X1,X2,X3)) = 4X1 + 8X2 + 4X3 + 17 >= X1 + 8X2 + X3 + 9 = a__f(X1,mark(X2),X3)
    
    a__f(X1,X2,X3) = X1 + 2X2 + X3 + 7 >= X1 + 2X2 + X3 + 4 = f(X1,X2,X3)
    
    a__c() = 5 >= 5 = 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