YES

We show the termination of the TRS R:

  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)

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: foldf#(x,cons(y,z)) -> f#(foldf(x,z),y)
p2: foldf#(x,cons(y,z)) -> foldf#(x,z)
p3: f#(t,x) -> |f'|#(t,g(x))
p4: f#(t,x) -> g#(x)
p5: |f'|#(triple(a,b,c),B()) -> f#(triple(a,b,c),A())
p6: |f'|#(triple(a,b,c),A()) -> |f''|#(foldf(triple(cons(A(),a),nil(),c),b))
p7: |f'|#(triple(a,b,c),A()) -> foldf#(triple(cons(A(),a),nil(),c),b)
p8: |f''|#(triple(a,b,c)) -> foldf#(triple(a,b,nil()),c)

and R consists of:

r1: g(A()) -> A()
r2: g(B()) -> A()
r3: g(B()) -> B()
r4: g(C()) -> A()
r5: g(C()) -> B()
r6: g(C()) -> C()
r7: foldf(x,nil()) -> x
r8: foldf(x,cons(y,z)) -> f(foldf(x,z),y)
r9: f(t,x) -> |f'|(t,g(x))
r10: |f'|(triple(a,b,c),C()) -> triple(a,b,cons(C(),c))
r11: |f'|(triple(a,b,c),B()) -> f(triple(a,b,c),A())
r12: |f'|(triple(a,b,c),A()) -> |f''|(foldf(triple(cons(A(),a),nil(),c),b))
r13: |f''|(triple(a,b,c)) -> foldf(triple(a,b,nil()),c)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p5, p6, p7, p8}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: foldf#(x,cons(y,z)) -> f#(foldf(x,z),y)
p2: f#(t,x) -> |f'|#(t,g(x))
p3: |f'|#(triple(a,b,c),A()) -> foldf#(triple(cons(A(),a),nil(),c),b)
p4: foldf#(x,cons(y,z)) -> foldf#(x,z)
p5: |f'|#(triple(a,b,c),A()) -> |f''|#(foldf(triple(cons(A(),a),nil(),c),b))
p6: |f''|#(triple(a,b,c)) -> foldf#(triple(a,b,nil()),c)
p7: |f'|#(triple(a,b,c),B()) -> f#(triple(a,b,c),A())

and R consists of:

r1: g(A()) -> A()
r2: g(B()) -> A()
r3: g(B()) -> B()
r4: g(C()) -> A()
r5: g(C()) -> B()
r6: g(C()) -> C()
r7: foldf(x,nil()) -> x
r8: foldf(x,cons(y,z)) -> f(foldf(x,z),y)
r9: f(t,x) -> |f'|(t,g(x))
r10: |f'|(triple(a,b,c),C()) -> triple(a,b,cons(C(),c))
r11: |f'|(triple(a,b,c),B()) -> f(triple(a,b,c),A())
r12: |f'|(triple(a,b,c),A()) -> |f''|(foldf(triple(cons(A(),a),nil(),c),b))
r13: |f''|(triple(a,b,c)) -> foldf(triple(a,b,nil()),c)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        foldf#_A(x1,x2) = ((1,1),(1,0)) x1 + ((1,1),(0,1)) x2
        cons_A(x1,x2) = ((1,1),(0,0)) x1 + ((0,0),(1,1)) x2 + (1,10)
        f#_A(x1,x2) = ((1,1),(1,0)) x1 + x2 + (9,7)
        foldf_A(x1,x2) = x1 + ((1,1),(0,0)) x2 + (1,0)
        |f'|#_A(x1,x2) = ((1,1),(1,0)) x1 + x2 + (8,7)
        g_A(x1) = ((1,0),(1,1)) x1 + (0,1)
        triple_A(x1,x2,x3) = x1 + ((1,1),(0,0)) x2 + ((1,1),(0,0)) x3 + (0,1)
        A_A() = (1,2)
        nil_A() = (1,0)
        |f''|#_A(x1) = ((1,0),(1,0)) x1 + (3,1)
        B_A() = (3,1)
        |f''|_A(x1) = x1 + (3,1)
        |f'|_A(x1,x2) = x1 + x2 + (11,0)
        C_A() = (4,0)
        f_A(x1,x2) = x1 + ((1,1),(0,0)) x2 + (11,0)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        foldf#_A(x1,x2) = (62,21)
        cons_A(x1,x2) = ((0,0),(1,1)) x1 + (1,1)
        f#_A(x1,x2) = x1 + x2 + (22,22)
        foldf_A(x1,x2) = ((1,1),(0,0)) x2 + (39,7)
        |f'|#_A(x1,x2) = ((0,1),(1,1)) x1 + ((0,1),(0,1)) x2
        g_A(x1) = ((0,0),(1,0)) x1 + (4,23)
        triple_A(x1,x2,x3) = ((1,1),(0,0)) x3 + (6,1)
        A_A() = (1,12)
        nil_A() = (1,1)
        |f''|#_A(x1) = (61,20)
        B_A() = (3,26)
        |f''|_A(x1) = ((0,0),(1,0)) x1
        |f'|_A(x1,x2) = ((0,1),(0,0)) x1 + ((0,1),(0,0)) x2 + (9,1)
        C_A() = (1,1)
        f_A(x1,x2) = ((0,1),(0,0)) x1 + x2 + (33,1)
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6, p7

We remove them from the problem.  Then no dependency pair remains.