YES

Problem:
 g(A()) -> A()
 g(B()) -> A()
 g(B()) -> B()
 g(C()) -> A()
 g(C()) -> B()
 g(C()) -> C()
 foldf(x,nil()) -> x
 foldf(x,cons(y,z)) -> f(foldf(x,z),y)
 f(t,x) -> f'(t,g(x))
 f'(triple(a,b,c),C()) -> triple(a,b,cons(C(),c))
 f'(triple(a,b,c),B()) -> f(triple(a,b,c),A())
 f'(triple(a,b,c),A()) -> f''(foldf(triple(cons(A(),a),nil(),c),b))
 f''(triple(a,b,c)) -> foldf(triple(a,b,nil()),c)

Proof:
 DP Processor:
  DPs:
   foldf#(x,cons(y,z)) -> foldf#(x,z)
   foldf#(x,cons(y,z)) -> f#(foldf(x,z),y)
   f#(t,x) -> g#(x)
   f#(t,x) -> f'#(t,g(x))
   f'#(triple(a,b,c),B()) -> f#(triple(a,b,c),A())
   f'#(triple(a,b,c),A()) -> foldf#(triple(cons(A(),a),nil(),c),b)
   f'#(triple(a,b,c),A()) -> f''#(foldf(triple(cons(A(),a),nil(),c),b))
   f''#(triple(a,b,c)) -> foldf#(triple(a,b,nil()),c)
  TRS:
   g(A()) -> A()
   g(B()) -> A()
   g(B()) -> B()
   g(C()) -> A()
   g(C()) -> B()
   g(C()) -> C()
   foldf(x,nil()) -> x
   foldf(x,cons(y,z)) -> f(foldf(x,z),y)
   f(t,x) -> f'(t,g(x))
   f'(triple(a,b,c),C()) -> triple(a,b,cons(C(),c))
   f'(triple(a,b,c),B()) -> f(triple(a,b,c),A())
   f'(triple(a,b,c),A()) -> f''(foldf(triple(cons(A(),a),nil(),c),b))
   f''(triple(a,b,c)) -> foldf(triple(a,b,nil()),c)
  Matrix Interpretation Processor: dim=1
   
   usable rules:
    g(A()) -> A()
    g(B()) -> A()
    g(B()) -> B()
    g(C()) -> A()
    g(C()) -> B()
    g(C()) -> C()
    foldf(x,nil()) -> x
    foldf(x,cons(y,z)) -> f(foldf(x,z),y)
    f(t,x) -> f'(t,g(x))
    f'(triple(a,b,c),C()) -> triple(a,b,cons(C(),c))
    f'(triple(a,b,c),B()) -> f(triple(a,b,c),A())
    f'(triple(a,b,c),A()) -> f''(foldf(triple(cons(A(),a),nil(),c),b))
    f''(triple(a,b,c)) -> foldf(triple(a,b,nil()),c)
   interpretation:
    [f''#](x0) = 1/2x0 + 5/2,
    
    [f'#](x0, x1) = 1/2x0 + 3/2x1,
    
    [f#](x0, x1) = 1/2x0 + 3/2x1 + 1/2,
    
    [foldf#](x0, x1) = 1/2x0 + 1/2x1 + 2,
    
    [g#](x0) = 0,
    
    [f''](x0) = x0 + 3/2,
    
    [triple](x0, x1, x2) = x1 + x2 + 2,
    
    [f'](x0, x1) = x0 + 3x1 + 1,
    
    [f](x0, x1) = x0 + 3x1 + 1,
    
    [cons](x0, x1) = 3x0 + x1 + 1,
    
    [foldf](x0, x1) = x0 + x1,
    
    [nil] = 0,
    
    [C] = 3,
    
    [B] = 3,
    
    [g](x0) = x0,
    
    [A] = 2
   orientation:
    foldf#(x,cons(y,z)) = 1/2x + 3/2y + 1/2z + 5/2 >= 1/2x + 1/2z + 2 = foldf#(x,z)
    
    foldf#(x,cons(y,z)) = 1/2x + 3/2y + 1/2z + 5/2 >= 1/2x + 3/2y + 1/2z + 1/2 = f#(foldf(x,z),y)
    
    f#(t,x) = 1/2t + 3/2x + 1/2 >= 0 = g#(x)
    
    f#(t,x) = 1/2t + 3/2x + 1/2 >= 1/2t + 3/2x = f'#(t,g(x))
    
    f'#(triple(a,b,c),B()) = 1/2b + 1/2c + 11/2 >= 1/2b + 1/2c + 9/2 = f#(triple(a,b,c),A())
    
    f'#(triple(a,b,c),A()) = 1/2b + 1/2c + 4 >= 1/2b + 1/2c + 3 = foldf#(triple(cons(A(),a),nil(),c),b)
    
    f'#(triple(a,b,c),A()) = 1/2b + 1/2c + 4 >= 1/2b + 1/2c + 7/2 = f''#(foldf(triple(cons(A(),a),nil(),c),b))
    
    f''#(triple(a,b,c)) = 1/2b + 1/2c + 7/2 >= 1/2b + 1/2c + 3 = foldf#(triple(a,b,nil()),c)
    
    g(A()) = 2 >= 2 = A()
    
    g(B()) = 3 >= 2 = A()
    
    g(B()) = 3 >= 3 = B()
    
    g(C()) = 3 >= 2 = A()
    
    g(C()) = 3 >= 3 = B()
    
    g(C()) = 3 >= 3 = C()
    
    foldf(x,nil()) = x >= x = x
    
    foldf(x,cons(y,z)) = x + 3y + z + 1 >= x + 3y + z + 1 = f(foldf(x,z),y)
    
    f(t,x) = t + 3x + 1 >= t + 3x + 1 = f'(t,g(x))
    
    f'(triple(a,b,c),C()) = b + c + 12 >= b + c + 12 = triple(a,b,cons(C(),c))
    
    f'(triple(a,b,c),B()) = b + c + 12 >= b + c + 9 = f(triple(a,b,c),A())
    
    f'(triple(a,b,c),A()) = b + c + 9 >= b + c + 7/2 = f''(foldf(triple(cons(A(),a),nil(),c),b))
    
    f''(triple(a,b,c)) = b + c + 7/2 >= b + c + 2 = foldf(triple(a,b,nil()),c)
   problem:
    DPs:
     
    TRS:
     g(A()) -> A()
     g(B()) -> A()
     g(B()) -> B()
     g(C()) -> A()
     g(C()) -> B()
     g(C()) -> C()
     foldf(x,nil()) -> x
     foldf(x,cons(y,z)) -> f(foldf(x,z),y)
     f(t,x) -> f'(t,g(x))
     f'(triple(a,b,c),C()) -> triple(a,b,cons(C(),c))
     f'(triple(a,b,c),B()) -> f(triple(a,b,c),A())
     f'(triple(a,b,c),A()) -> f''(foldf(triple(cons(A(),a),nil(),c),b))
     f''(triple(a,b,c)) -> foldf(triple(a,b,nil()),c)
   Qed