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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      max/plus interpretation on natural numbers:
      
        foldf#_A(x1,x2) = max{96, x1 + 55, x2 + 78}
        cons_A(x1,x2) = max{x1 + 21, x2}
        f#_A(x1,x2) = max{98, x1 + 55}
        foldf_A(x1,x2) = max{29, x1, x2 + 7}
        |f'|#_A(x1,x2) = max{97, x1 + 55}
        g_A(x1) = 27
        triple_A(x1,x2,x3) = max{x1 + 44, x2 + 85, x3 + 40}
        A_A = 11
        nil_A = 0
        |f''|#_A(x1) = max{94, x1 + 55}
        B_A = 26
        C_A = 24
        f_A(x1,x2) = max{x1, x2 + 28}
        |f'|_A(x1,x2) = max{28, x1}
        |f''|_A(x1) = max{85, x1}
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        matrix interpretation with standard point-wise order:
        
          carrier: N^1
          order: standard order
          interpretations:
            foldf#_A(x1,x2) = x1 + x2
            cons_A(x1,x2) = x1 + x2 + 17
            f#_A(x1,x2) = x1 + 3
            foldf_A(x1,x2) = x1 + x2 + 1
            |f'|#_A(x1,x2) = x1 + 3
            g_A(x1) = x1
            triple_A(x1,x2,x3) = x2 + x3 + 1
            A_A() = 1
            nil_A() = 1
            |f''|#_A(x1) = x1 + 1
            B_A() = 9
            C_A() = 9
            f_A(x1,x2) = x1 + x2 + 17
            |f'|_A(x1,x2) = x1 + x2 + 17
            |f''|_A(x1) = x1 + 2
            foldf#!_A(x1,x2) = x1 + x2 + 6
            cons!_A(x1,x2) = 16
            f#!_A(x1,x2) = x1 + x2 + 20
            foldf!_A(x1,x2) = x1 + x2 + 1
            |f'|#!_A(x1,x2) = x1 + x2 + 13
            g!_A(x1) = x1
            triple!_A(x1,x2,x3) = 7
            A!_A() = 9
            nil!_A() = 10
            |f''|#!_A(x1) = x1 + 8
            B!_A() = 21
            C!_A() = 20
            f!_A(x1,x2) = x1 + x2 + 16
            |f'|!_A(x1,x2) = x1 + x2 + 9
            |f''|!_A(x1) = x1 + 3
  
    partial status:
  
      pi(foldf#) = []
      pi(cons) = [1, 2]
      pi(f#) = [1]
      pi(foldf) = [1]
      pi(|f'|#) = []
      pi(g) = []
      pi(triple) = []
      pi(A) = []
      pi(nil) = []
      pi(|f''|#) = [1]
      pi(B) = []
      pi(C) = []
      pi(f) = []
      pi(|f'|) = []
      pi(|f''|) = [1]

The following rules are removed:

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

Then no dependency pair remains.