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:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          foldB#_A(x1,x2) = x1 + x2
          s_A(x1) = x1 + 15
          f#_A(x1,x2) = x1 + x2 + 8
          foldB_A(x1,x2) = x1 + x2 + 1
          B_A() = 4
          |f'|#_A(x1,x2) = x1 + x2 + 6
          g_A(x1) = x1 + 1
          triple_A(x1,x2,x3) = x2 + x3 + 16
          A_A() = 1
          |0|_A() = 2
          |f''|#_A(x1) = x1 + 3
          foldC#_A(x1,x2) = x1 + x2
          foldC_A(x1,x2) = x1 + x2 + 1
          C_A() = 5
          f_A(x1,x2) = x1 + 15
          |f'|_A(x1,x2) = x1 + 15
          |f''|_A(x1) = x1 + 4
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          foldB#_A(x1,x2) = max{83, x1 - 35}
          s_A(x1) = x1
          f#_A(x1,x2) = max{83, x1 - 35, x2 - 23}
          foldB_A(x1,x2) = max{36, x1}
          B_A = 1
          |f'|#_A(x1,x2) = max{83, x1 - 35}
          g_A(x1) = 25
          triple_A(x1,x2,x3) = max{x1 + 117, x2 - 75, x3 + 46}
          A_A = 22
          |0|_A = 46
          |f''|#_A(x1) = max{83, x1 - 35}
          foldC#_A(x1,x2) = max{83, x1 - 35, x2 + 1}
          foldC_A(x1,x2) = max{50, x1, x2 - 1}
          C_A = 25
          f_A(x1,x2) = max{x1, x2 + 25}
          |f'|_A(x1,x2) = max{25, x1, x2}
          |f''|_A(x1) = max{50, x1}
          foldB#!_A(x1,x2) = max{0, x1 - 46}
          s!_A(x1) = x1 + 72
          f#!_A(x1,x2) = max{128, x1 + 75, x2 - 26}
          foldB!_A(x1,x2) = x1 + 108
          B!_A = 107
          |f'|#!_A(x1,x2) = max{3, x1 - 122}
          g!_A(x1) = max{129, x1 + 105}
          triple!_A(x1,x2,x3) = 80
          A!_A = 1
          |0|!_A = 3
          |f''|#!_A(x1) = max{80, x1 - 113}
          foldC#!_A(x1,x2) = max{x1 + 74, x2 + 127}
          foldC!_A(x1,x2) = x2 + 107
          C!_A = 130
          f!_A(x1,x2) = 106
          |f'|!_A(x1,x2) = max{106, x2 + 80}
          |f''|!_A(x1) = max{81, x1 - 46}
  
    partial status:
  
      pi(foldB#) = [2]
      pi(s) = [1]
      pi(f#) = []
      pi(foldB) = []
      pi(B) = []
      pi(|f'|#) = [2]
      pi(g) = []
      pi(triple) = []
      pi(A) = []
      pi(|0|) = []
      pi(|f''|#) = [1]
      pi(foldC#) = [1]
      pi(foldC) = [2]
      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.