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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            foldB#_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,1)) x2 + (20,1)
            s_A(x1) = ((1,0),(1,1)) x1 + (34,24)
            f#_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(0,0)) x2 + (2,20)
            foldB_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,1)) x2 + (5,60)
            B_A() = (33,119)
            |f'|#_A(x1,x2) = x1 + x2
            g_A(x1) = ((1,0),(1,1)) x1 + (1,1)
            triple_A(x1,x2,x3) = ((1,0),(1,1)) x2 + ((1,0),(0,0)) x3 + (4,23)
            A_A() = (26,39)
            |0|_A() = (5,63)
            |f''|#_A(x1) = x1 + (15,1)
            foldC#_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (9,22)
            foldC_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (1,23)
            C_A() = (39,25)
            |f''|_A(x1) = ((1,0),(0,0)) x1 + (7,64)
            |f'|_A(x1,x2) = x1 + ((0,0),(1,0)) x2 + (34,16)
            f_A(x1,x2) = x1 + ((0,0),(1,0)) x2 + (34,18)
    
      precedence:
      
        |f'|# > s > B = g > triple = |f''|# > foldB > |f'| > |f''| > foldB# = A = |0| = C > foldC# = foldC > f# > f
      
      partial status:
    
        pi(foldB#) = []
        pi(s) = [1]
        pi(f#) = []
        pi(foldB) = [2]
        pi(B) = []
        pi(|f'|#) = []
        pi(g) = [1]
        pi(triple) = []
        pi(A) = []
        pi(|0|) = []
        pi(|f''|#) = [1]
        pi(foldC#) = []
        pi(foldC) = [2]
        pi(C) = []
        pi(|f''|) = []
        pi(|f'|) = [1]
        pi(f) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            foldB#_A(x1,x2) = (1,2)
            s_A(x1) = (9,9)
            f#_A(x1,x2) = (8,1)
            foldB_A(x1,x2) = x2 + (22,11)
            B_A() = (6,5)
            |f'|#_A(x1,x2) = (3,3)
            g_A(x1) = (7,6)
            triple_A(x1,x2,x3) = (21,22)
            A_A() = (4,4)
            |0|_A() = (1,2)
            |f''|#_A(x1) = (2,9)
            foldC#_A(x1,x2) = (22,8)
            foldC_A(x1,x2) = x2 + (10,0)
            C_A() = (5,7)
            |f''|_A(x1) = (2,3)
            |f'|_A(x1,x2) = (10,10)
            f_A(x1,x2) = (20,21)
    
      precedence:
      
        g > f# > s > foldC > foldB > |f'|# = triple = |0| = |f''|# = foldC# > |f''| > |f'| > f > foldB# = B > A > C
      
      partial status:
    
        pi(foldB#) = []
        pi(s) = []
        pi(f#) = []
        pi(foldB) = []
        pi(B) = []
        pi(|f'|#) = []
        pi(g) = []
        pi(triple) = []
        pi(A) = []
        pi(|0|) = []
        pi(|f''|#) = []
        pi(foldC#) = []
        pi(foldC) = []
        pi(C) = []
        pi(|f''|) = []
        pi(|f'|) = []
        pi(f) = []
    

The next rules are strictly ordered:

  p1, p3, p4, p6, p8, p9

We remove them from the problem.

-- 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),A()) -> |f''|#(foldB(triple(s(a),|0|(),c),b))
p3: foldC#(t,s(n)) -> foldC#(t,n)

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:

  {p3}


-- Reduction pair.

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

p1: foldC#(t,s(n)) -> foldC#(t,n)

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

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            foldC#_A(x1,x2) = ((0,0),(1,0)) x2
            s_A(x1) = ((1,0),(1,0)) x1 + (1,1)
    
      precedence:
      
        s > foldC#
      
      partial status:
    
        pi(foldC#) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            foldC#_A(x1,x2) = (0,0)
            s_A(x1) = (1,1)
    
      precedence:
      
        s > foldC#
      
      partial status:
    
        pi(foldC#) = []
        pi(s) = []
    

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.