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:
  
      trivial reduction triple (T x T, T x T, Φ)
  
    base order 2:
  
      lexicographic combination of reduction triples:
      
        1. 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) = 0
              cons_A(x1,x2) = x1 + x2 + 17
              f#_A(x1,x2) = 0
              foldf_A(x1,x2) = x1 + x2 + 2
              |f'|#_A(x1,x2) = 0
              g_A(x1) = x1
              triple_A(x1,x2,x3) = x2 + x3 + 5
              A_A() = 7
              nil_A() = 1
              |f''|#_A(x1) = 0
              B_A() = 10
              C_A() = 12
              f_A(x1,x2) = x1 + x2 + 17
              |f'|_A(x1,x2) = x1 + x2 + 17
              |f''|_A(x1) = x1 + 3
              foldf#!_A(x1,x2) = x1 + x2
              cons!_A(x1,x2) = x1 + 35
              f#!_A(x1,x2) = x1 + x2 + 15
              foldf!_A(x1,x2) = x1 + x2 + 18
              |f'|#!_A(x1,x2) = x1 + x2 + 13
              g!_A(x1) = x1 + 15
              triple!_A(x1,x2,x3) = x2 + x3 + 7
              A!_A() = 26
              nil!_A() = 5
              |f''|#!_A(x1) = x1
              B!_A() = 27
              C!_A() = 48
              f!_A(x1,x2) = x1 + 32
              |f'|!_A(x1,x2) = x1 + 32
              |f''|!_A(x1) = x1 + 20
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            foldf#_A(x1,x2) = 1
            cons_A(x1,x2) = max{17, x1 + 15, x2}
            f#_A(x1,x2) = 1
            foldf_A(x1,x2) = max{83, x1, x2 + 2}
            |f'|#_A(x1,x2) = 1
            g_A(x1) = max{11, x1}
            triple_A(x1,x2,x3) = max{x1 + 13, x2 + 25, x3 + 22}
            A_A = 12
            nil_A = 4
            |f''|#_A(x1) = 1
            B_A = 46
            C_A = 46
            f_A(x1,x2) = max{83, x1, x2 + 17}
            |f'|_A(x1,x2) = max{83, x1, x2 + 17}
            |f''|_A(x1) = max{83, x1}
            foldf#!_A(x1,x2) = x2 + 14
            cons!_A(x1,x2) = max{x1 + 12, x2 + 4}
            f#!_A(x1,x2) = max{30, x2 + 14}
            foldf!_A(x1,x2) = max{13, x2 - 4}
            |f'|#!_A(x1,x2) = max{29, x2 - 17}
            g!_A(x1) = x1 + 14
            triple!_A(x1,x2,x3) = max{28, x2 + 27, x3 + 23}
            A!_A = 60
            nil!_A = 1
            |f''|#!_A(x1) = max{2, x1 - 8}
            B!_A = 60
            C!_A = 1
            f!_A(x1,x2) = x2 + 10
            |f'|!_A(x1,x2) = max{9, x2 - 6}
            |f''|!_A(x1) = max{0, x1 - 22}
        
  
    partial status:
  
      pi(foldf#) = []
      pi(cons) = []
      pi(f#) = []
      pi(foldf) = [1]
      pi(|f'|#) = []
      pi(g) = [1]
      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.