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()
  foldB(t,|0|()) -> t
  foldB(t,s(n)) -> f(foldB(t,n),B())
  foldC(t,|0|()) -> t
  foldC(t,s(n)) -> f(foldC(t,n),C())
  f(t,x) -> |f'|(t,g(x))
  |f'|(triple(a,b,c),C()) -> triple(a,b,s(c))
  |f'|(triple(a,b,c),B()) -> f(triple(a,b,c),A())
  |f'|(triple(a,b,c),A()) -> |f''|(foldB(triple(s(a),|0|(),c),b))
  |f''|(triple(a,b,c)) -> foldC(triple(a,b,|0|()),c)


-- SCC decomposition.

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

p1: foldB#(t,s(n)) -> f#(foldB(t,n),B())
p2: foldB#(t,s(n)) -> foldB#(t,n)
p3: foldC#(t,s(n)) -> f#(foldC(t,n),C())
p4: foldC#(t,s(n)) -> foldC#(t,n)
p5: f#(t,x) -> |f'|#(t,g(x))
p6: f#(t,x) -> g#(x)
p7: |f'|#(triple(a,b,c),B()) -> f#(triple(a,b,c),A())
p8: |f'|#(triple(a,b,c),A()) -> |f''|#(foldB(triple(s(a),|0|(),c),b))
p9: |f'|#(triple(a,b,c),A()) -> foldB#(triple(s(a),|0|(),c),b)
p10: |f''|#(triple(a,b,c)) -> foldC#(triple(a,b,|0|()),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: foldB(t,|0|()) -> t
r8: foldB(t,s(n)) -> f(foldB(t,n),B())
r9: foldC(t,|0|()) -> t
r10: foldC(t,s(n)) -> f(foldC(t,n),C())
r11: f(t,x) -> |f'|(t,g(x))
r12: |f'|(triple(a,b,c),C()) -> triple(a,b,s(c))
r13: |f'|(triple(a,b,c),B()) -> f(triple(a,b,c),A())
r14: |f'|(triple(a,b,c),A()) -> |f''|(foldB(triple(s(a),|0|(),c),b))
r15: |f''|(triple(a,b,c)) -> foldC(triple(a,b,|0|()),c)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p7, p8, p9, p10}


-- Reduction pair.

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

p1: foldB#(t,s(n)) -> f#(foldB(t,n),B())
p2: f#(t,x) -> |f'|#(t,g(x))
p3: |f'|#(triple(a,b,c),A()) -> foldB#(triple(s(a),|0|(),c),b)
p4: foldB#(t,s(n)) -> foldB#(t,n)
p5: |f'|#(triple(a,b,c),A()) -> |f''|#(foldB(triple(s(a),|0|(),c),b))
p6: |f''|#(triple(a,b,c)) -> foldC#(triple(a,b,|0|()),c)
p7: foldC#(t,s(n)) -> foldC#(t,n)
p8: foldC#(t,s(n)) -> f#(foldC(t,n),C())
p9: |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: foldB(t,|0|()) -> t
r8: foldB(t,s(n)) -> f(foldB(t,n),B())
r9: foldC(t,|0|()) -> t
r10: foldC(t,s(n)) -> f(foldC(t,n),C())
r11: f(t,x) -> |f'|(t,g(x))
r12: |f'|(triple(a,b,c),C()) -> triple(a,b,s(c))
r13: |f'|(triple(a,b,c),B()) -> f(triple(a,b,c),A())
r14: |f'|(triple(a,b,c),A()) -> |f''|(foldB(triple(s(a),|0|(),c),b))
r15: |f''|(triple(a,b,c)) -> foldC(triple(a,b,|0|()),c)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15

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:
              foldB#_A(x1,x2) = 0
              s_A(x1) = x1 + 8
              f#_A(x1,x2) = 0
              foldB_A(x1,x2) = x1 + x2 + 1
              B_A() = 3
              |f'|#_A(x1,x2) = 0
              g_A(x1) = 3
              triple_A(x1,x2,x3) = x2 + x3 + 1
              A_A() = 1
              |0|_A() = 1
              |f''|#_A(x1) = 0
              foldC#_A(x1,x2) = 0
              foldC_A(x1,x2) = x1 + x2
              C_A() = 1
              f_A(x1,x2) = x1 + 8
              |f'|_A(x1,x2) = x1 + 8
              |f''|_A(x1) = x1 + 1
              foldB#!_A(x1,x2) = x1 + x2 + 1
              s!_A(x1) = 18
              f#!_A(x1,x2) = x1 + 7
              foldB!_A(x1,x2) = x1 + x2 + 11
              B!_A() = 17
              |f'|#!_A(x1,x2) = x1 + 7
              g!_A(x1) = x1 + 15
              triple!_A(x1,x2,x3) = x2 + 6
              A!_A() = 15
              |0|!_A() = 0
              |f''|#!_A(x1) = x1 + 2
              foldC#!_A(x1,x2) = x1 + x2
              foldC!_A(x1,x2) = x1 + x2 + 8
              C!_A() = 19
              f!_A(x1,x2) = x1 + 13
              |f'|!_A(x1,x2) = x1 + 13
              |f''|!_A(x1) = x1 + 10
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          max/plus interpretation on natural numbers:
          
            foldB#_A(x1,x2) = max{72, x1 - 6}
            s_A(x1) = x1 + 180
            f#_A(x1,x2) = max{72, x2 - 53}
            foldB_A(x1,x2) = max{76, x1 + 70}
            B_A = 125
            |f'|#_A(x1,x2) = 72
            g_A(x1) = max{63, x1 + 10}
            triple_A(x1,x2,x3) = 2
            A_A = 64
            |0|_A = 35
            |f''|#_A(x1) = 72
            foldC#_A(x1,x2) = 72
            foldC_A(x1,x2) = max{75, x1 + 33}
            C_A = 115
            f_A(x1,x2) = max{75, x2 - 54}
            |f'|_A(x1,x2) = max{75, x2 - 116}
            |f''|_A(x1) = max{75, x1 - 1}
            foldB#!_A(x1,x2) = max{x1 + 186, x2 - 68}
            s!_A(x1) = 1
            f#!_A(x1,x2) = max{x1 + 67, x2 + 1}
            foldB!_A(x1,x2) = max{188, x1 + 186}
            B!_A = 189
            |f'|#!_A(x1,x2) = max{66, x2 - 55}
            g!_A(x1) = max{191, x1 - 126}
            triple!_A(x1,x2,x3) = max{x1 + 72, x3 + 72}
            A!_A = 190
            |0|!_A = 109
            |f''|#!_A(x1) = 10
            foldC#!_A(x1,x2) = max{191, x1 + 101}
            foldC!_A(x1,x2) = max{110, x1 - 4, x2 - 70}
            C!_A = 191
            f!_A(x1,x2) = max{x1 + 111, x2 + 9}
            |f'|!_A(x1,x2) = max{110, x2 - 11}
            |f''|!_A(x1) = max{108, x1 - 3}
        
  
    partial status:
  
      pi(foldB#) = []
      pi(s) = [1]
      pi(f#) = []
      pi(foldB) = [1, 2]
      pi(B) = []
      pi(|f'|#) = []
      pi(g) = [1]
      pi(triple) = []
      pi(A) = []
      pi(|0|) = []
      pi(|f''|#) = []
      pi(foldC#) = []
      pi(foldC) = [1]
      pi(C) = []
      pi(f) = []
      pi(|f'|) = []
      pi(|f''|) = []

The following rules are removed:

  p1, p2, p3, p4, p5, p6, p7, p8, p9

Then no dependency pair remains.