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. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        foldB#_A(x1,x2) = ((0,1),(0,0)) x1 + ((1,1),(0,0)) x2 + (3,0)
        s_A(x1) = ((1,1),(0,0)) x1 + (9,5)
        f#_A(x1,x2) = ((0,1),(0,0)) x1 + (3,0)
        foldB_A(x1,x2) = ((1,1),(0,1)) x1 + ((1,0),(1,1)) x2 + (0,1)
        B_A() = (3,4)
        |f'|#_A(x1,x2) = ((0,1),(0,0)) x1 + (3,0)
        g_A(x1) = ((0,1),(1,0)) x1 + (0,1)
        triple_A(x1,x2,x3) = ((0,0),(1,1)) x2 + ((1,0),(1,1)) x3 + (1,0)
        A_A() = (1,2)
        |0|_A() = (0,0)
        |f''|#_A(x1) = ((0,1),(0,0)) x1 + (2,0)
        foldC#_A(x1,x2) = ((1,1),(0,0)) x1 + x2
        foldC_A(x1,x2) = ((1,1),(0,1)) x1 + ((1,1),(1,1)) x2 + (0,1)
        C_A() = (8,9)
        |f''|_A(x1) = ((0,1),(0,1)) x1 + (2,1)
        |f'|_A(x1,x2) = ((0,1),(0,1)) x1 + ((0,1),(0,0)) x2 + (2,14)
        f_A(x1,x2) = ((0,1),(0,1)) x1 + x2 + (4,14)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        foldB#_A(x1,x2) = (1,3)
        s_A(x1) = x1 + (3,1)
        f#_A(x1,x2) = (2,4)
        foldB_A(x1,x2) = ((1,1),(0,1)) x1 + x2
        B_A() = (0,2)
        |f'|#_A(x1,x2) = (2,4)
        g_A(x1) = (2,1)
        triple_A(x1,x2,x3) = ((1,1),(1,1)) x3 + (1,1)
        A_A() = (1,2)
        |0|_A() = (1,1)
        |f''|#_A(x1) = (1,0)
        foldC#_A(x1,x2) = ((1,0),(1,0)) x2 + (0,1)
        foldC_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (3,1)
        C_A() = (2,2)
        |f''|_A(x1) = (5,0)
        |f'|_A(x1,x2) = (5,4)
        f_A(x1,x2) = (7,6)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        foldB#_A(x1,x2) = (0,1)
        s_A(x1) = (1,2)
        f#_A(x1,x2) = (1,0)
        foldB_A(x1,x2) = ((0,1),(0,0)) x1 + ((1,0),(1,0)) x2
        B_A() = (1,2)
        |f'|#_A(x1,x2) = (1,0)
        g_A(x1) = (2,1)
        triple_A(x1,x2,x3) = ((0,0),(1,1)) x3 + (1,1)
        A_A() = (1,1)
        |0|_A() = (1,1)
        |f''|#_A(x1) = (2,1)
        foldC#_A(x1,x2) = (3,2)
        foldC_A(x1,x2) = ((0,0),(1,0)) x1 + ((0,1),(0,0)) x2
        C_A() = (3,2)
        |f''|_A(x1) = (0,1)
        |f'|_A(x1,x2) = (4,3)
        f_A(x1,x2) = (3,2)
    

The next rules are strictly ordered:

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

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),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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1,x2) = ((1,1),(1,1)) x1 + ((1,1),(0,0)) x2 + (2,0)
        |f'|#_A(x1,x2) = ((1,1),(1,1)) x1 + x2
        g_A(x1) = ((1,1),(0,0)) x1 + (1,1)
        triple_A(x1,x2,x3) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + ((1,1),(1,1)) x3 + (1,1)
        B_A() = (5,1)
        A_A() = (1,1)
        C_A() = (4,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1,x2) = ((1,1),(1,1)) x1 + (1,0)
        |f'|#_A(x1,x2) = x1 + ((1,1),(0,1)) x2
        g_A(x1) = x1 + (1,1)
        triple_A(x1,x2,x3) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + ((1,1),(1,1)) x3 + (4,4)
        B_A() = (1,3)
        A_A() = (1,5)
        C_A() = (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1,x2) = ((1,1),(0,1)) x1
        |f'|#_A(x1,x2) = x2 + (3,0)
        g_A(x1) = x1 + (2,1)
        triple_A(x1,x2,x3) = ((0,1),(0,1)) x1 + ((0,1),(0,1)) x2 + ((1,0),(1,0)) x3 + (1,4)
        B_A() = (1,3)
        A_A() = (1,1)
        C_A() = (1,1)
    

The next rules are strictly ordered:

  p1, p2

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