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!):
        
          max/plus interpretation on natural numbers:
          
            foldB#_A(x1,x2) = 16
            s_A(x1) = x1
            f#_A(x1,x2) = max{16, x2 + 1}
            foldB_A(x1,x2) = max{6, x1}
            B_A = 15
            |f'|#_A(x1,x2) = 16
            g_A(x1) = 16
            triple_A(x1,x2,x3) = max{x1 + 43, x2 + 40, x3 + 39}
            A_A = 15
            |0|_A = 3
            |f''|#_A(x1) = 16
            foldC#_A(x1,x2) = 16
            foldC_A(x1,x2) = x1
            C_A = 6
            f_A(x1,x2) = max{x1, x2 - 16}
            |f'|_A(x1,x2) = x1
            |f''|_A(x1) = max{41, x1}
            foldB#!_A(x1,x2) = max{78, x1 + 35, x2 + 75}
            s!_A(x1) = 8
            f#!_A(x1,x2) = max{x1 + 35, x2 + 63}
            foldB!_A(x1,x2) = max{33, x1 + 9}
            B!_A = 13
            |f'|#!_A(x1,x2) = max{x1 + 35, x2 + 25}
            g!_A(x1) = max{14, x1 - 1}
            triple!_A(x1,x2,x3) = x3 + 42
            A!_A = 12
            |0|!_A = 47
            |f''|#!_A(x1) = x1 + 35
            foldC#!_A(x1,x2) = max{x1 + 35, x2 + 73}
            foldC!_A(x1,x2) = max{36, x1 + 3}
            C!_A = 1
            f!_A(x1,x2) = max{32, x1 + 3, x2 + 1}
            |f'|!_A(x1,x2) = max{x1 + 3, x2 + 16}
            |f''|!_A(x1) = max{37, x1 + 3}
        
        2. 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 + 21
              f#_A(x1,x2) = 0
              foldB_A(x1,x2) = x1 + x2
              B_A() = 4
              |f'|#_A(x1,x2) = 0
              g_A(x1) = x1 + 1
              triple_A(x1,x2,x3) = x2 + x3 + 13
              A_A() = 1
              |0|_A() = 1
              |f''|#_A(x1) = 0
              foldC#_A(x1,x2) = 0
              foldC_A(x1,x2) = x1 + x2 + 1
              C_A() = 3
              f_A(x1,x2) = x1 + 21
              |f'|_A(x1,x2) = x1 + 21
              |f''|_A(x1) = x1 + 2
              foldB#!_A(x1,x2) = x1 + x2 + 4
              s!_A(x1) = x1 + 1
              f#!_A(x1,x2) = x1 + 21
              foldB!_A(x1,x2) = 33
              B!_A() = 0
              |f'|#!_A(x1,x2) = x1 + 21
              g!_A(x1) = 20
              triple!_A(x1,x2,x3) = x2 + 34
              A!_A() = 19
              |0|!_A() = 16
              |f''|#!_A(x1) = x1 + 4
              foldC#!_A(x1,x2) = x1 + x2 + 2
              foldC!_A(x1,x2) = x1 + x2 + 23
              C!_A() = 18
              f!_A(x1,x2) = x1 + x2 + 28
              |f'|!_A(x1,x2) = x1 + x2 + 26
              |f''|!_A(x1) = x1 + 25
        
  
    partial status:
  
      pi(foldB#) = []
      pi(s) = [1]
      pi(f#) = []
      pi(foldB) = [1]
      pi(B) = []
      pi(|f'|#) = []
      pi(g) = []
      pi(triple) = []
      pi(A) = []
      pi(|0|) = []
      pi(|f''|#) = [1]
      pi(foldC#) = []
      pi(foldC) = [1]
      pi(C) = []
      pi(f) = []
      pi(|f'|) = []
      pi(|f''|) = []

The following rules are removed:

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



-- SCC decomposition.

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

p1: f#(t,x) -> |f'|#(t,g(x))
p2: |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 estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: f#(t,x) -> |f'|#(t,g(x))
p2: |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

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!):
        
          max/plus interpretation on natural numbers:
          
            f#_A(x1,x2) = max{0, x1 - 6, x2 - 7}
            |f'|#_A(x1,x2) = max{0, x1 - 6, x2 - 111}
            g_A(x1) = max{106, x1 + 44}
            triple_A(x1,x2,x3) = max{4, x1 - 1, x2 - 101, x3 - 6}
            B_A = 108
            A_A = 6
            C_A = 64
            f#!_A(x1,x2) = max{20, x2 + 2}
            |f'|#!_A(x1,x2) = max{1, x2 - 87}
            g!_A(x1) = max{19, x1 + 1}
            triple!_A(x1,x2,x3) = max{x1 + 104, x2 + 16, x3 + 9}
            B!_A = 64
            A!_A = 2
            C!_A = 13
        
        2. marked-algebra reduction triple (>=A, >=A!, >A!):
        
          matrix interpretation with standard point-wise order:
          
            carrier: N^1
            order: standard order
            interpretations:
              f#_A(x1,x2) = x1
              |f'|#_A(x1,x2) = x1
              g_A(x1) = x1 + 1
              triple_A(x1,x2,x3) = 6
              B_A() = 10
              A_A() = 1
              C_A() = 10
              f#!_A(x1,x2) = x1 + x2 + 5
              |f'|#!_A(x1,x2) = x1 + x2
              g!_A(x1) = x1 + 4
              triple!_A(x1,x2,x3) = x1 + 17
              B!_A() = 13
              A!_A() = 1
              C!_A() = 16
        
  
    partial status:
  
      pi(f#) = []
      pi(|f'|#) = [2]
      pi(g) = []
      pi(triple) = [3]
      pi(B) = []
      pi(A) = []
      pi(C) = []

The following rules are removed:

  p1, p2

Then no dependency pair remains.