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()
  foldf(x,nil()) -> x
  foldf(x,cons(y,z)) -> f(foldf(x,z),y)
  f(t,x) -> |f'|(t,g(x))
  |f'|(triple(a,b,c),C()) -> triple(a,b,cons(C(),c))
  |f'|(triple(a,b,c),B()) -> f(triple(a,b,c),A())
  |f'|(triple(a,b,c),A()) -> |f''|(foldf(triple(cons(A(),a),nil(),c),b))
  |f''|(triple(a,b,c)) -> foldf(triple(a,b,nil()),c)

-- SCC decomposition.

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

p1: foldf#(x,cons(y,z)) -> f#(foldf(x,z),y)
p2: foldf#(x,cons(y,z)) -> foldf#(x,z)
p3: f#(t,x) -> |f'|#(t,g(x))
p4: f#(t,x) -> g#(x)
p5: |f'|#(triple(a,b,c),B()) -> f#(triple(a,b,c),A())
p6: |f'|#(triple(a,b,c),A()) -> |f''|#(foldf(triple(cons(A(),a),nil(),c),b))
p7: |f'|#(triple(a,b,c),A()) -> foldf#(triple(cons(A(),a),nil(),c),b)
p8: |f''|#(triple(a,b,c)) -> foldf#(triple(a,b,nil()),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: foldf(x,nil()) -> x
r8: foldf(x,cons(y,z)) -> f(foldf(x,z),y)
r9: f(t,x) -> |f'|(t,g(x))
r10: |f'|(triple(a,b,c),C()) -> triple(a,b,cons(C(),c))
r11: |f'|(triple(a,b,c),B()) -> f(triple(a,b,c),A())
r12: |f'|(triple(a,b,c),A()) -> |f''|(foldf(triple(cons(A(),a),nil(),c),b))
r13: |f''|(triple(a,b,c)) -> foldf(triple(a,b,nil()),c)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p5, p6, p7, p8}


-- Reduction pair.

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

p1: foldf#(x,cons(y,z)) -> f#(foldf(x,z),y)
p2: f#(t,x) -> |f'|#(t,g(x))
p3: |f'|#(triple(a,b,c),A()) -> foldf#(triple(cons(A(),a),nil(),c),b)
p4: foldf#(x,cons(y,z)) -> foldf#(x,z)
p5: |f'|#(triple(a,b,c),A()) -> |f''|#(foldf(triple(cons(A(),a),nil(),c),b))
p6: |f''|#(triple(a,b,c)) -> foldf#(triple(a,b,nil()),c)
p7: |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: foldf(x,nil()) -> x
r8: foldf(x,cons(y,z)) -> f(foldf(x,z),y)
r9: f(t,x) -> |f'|(t,g(x))
r10: |f'|(triple(a,b,c),C()) -> triple(a,b,cons(C(),c))
r11: |f'|(triple(a,b,c),B()) -> f(triple(a,b,c),A())
r12: |f'|(triple(a,b,c),A()) -> |f''|(foldf(triple(cons(A(),a),nil(),c),b))
r13: |f''|(triple(a,b,c)) -> foldf(triple(a,b,nil()),c)

The set of usable rules consists of

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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^1
      order: standard order
      interpretations:
        foldf#_A(x1,x2) = x1 + x2
        cons_A(x1,x2) = x2 + 7
        f#_A(x1,x2) = x1 + 5
        foldf_A(x1,x2) = x1 + x2 + 1
        |f'|#_A(x1,x2) = x1 + 5
        g_A(x1) = x1 + 1
        triple_A(x1,x2,x3) = x2 + x3 + 1
        A_A() = 2
        nil_A() = 1
        |f''|#_A(x1) = x1 + 2
        B_A() = 1
        |f''|_A(x1) = x1 + 3
        |f'|_A(x1,x2) = x1 + 7
        C_A() = 1
        f_A(x1,x2) = x1 + 7
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        g > f > A > triple > B > cons > C > nil > foldf > |f''| > |f'| > f# > |f''|# > |f'|# > foldf#
      
      argument filter:
    
        pi(foldf#) = 2
        pi(cons) = 2
        pi(f#) = 1
        pi(foldf) = 2
        pi(|f'|#) = 1
        pi(g) = [1]
        pi(triple) = [2, 3]
        pi(A) = []
        pi(nil) = []
        pi(|f''|#) = [1]
        pi(B) = []
        pi(|f''|) = []
        pi(|f'|) = 1
        pi(C) = []
        pi(f) = 1
    

The next rules are strictly ordered:

  p1, p3, p4, p5, p6

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: foldf(x,nil()) -> x
r8: foldf(x,cons(y,z)) -> f(foldf(x,z),y)
r9: f(t,x) -> |f'|(t,g(x))
r10: |f'|(triple(a,b,c),C()) -> triple(a,b,cons(C(),c))
r11: |f'|(triple(a,b,c),B()) -> f(triple(a,b,c),A())
r12: |f'|(triple(a,b,c),A()) -> |f''|(foldf(triple(cons(A(),a),nil(),c),b))
r13: |f''|(triple(a,b,c)) -> foldf(triple(a,b,nil()),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: foldf(x,nil()) -> x
r8: foldf(x,cons(y,z)) -> f(foldf(x,z),y)
r9: f(t,x) -> |f'|(t,g(x))
r10: |f'|(triple(a,b,c),C()) -> triple(a,b,cons(C(),c))
r11: |f'|(triple(a,b,c),B()) -> f(triple(a,b,c),A())
r12: |f'|(triple(a,b,c),A()) -> |f''|(foldf(triple(cons(A(),a),nil(),c),b))
r13: |f''|(triple(a,b,c)) -> foldf(triple(a,b,nil()),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^1
      order: standard order
      interpretations:
        f#_A(x1,x2) = x1 + x2 + 2
        |f'|#_A(x1,x2) = x1 + x2
        g_A(x1) = x1 + 1
        triple_A(x1,x2,x3) = x1 + x2 + x3 + 1
        B_A() = 4
        A_A() = 1
        C_A() = 4
    
    2. lexicographic path order with precedence:
    
      precedence:
      
        A > B > C > |f'|# > triple > g > f#
      
      argument filter:
    
        pi(f#) = [1, 2]
        pi(|f'|#) = [1]
        pi(g) = [1]
        pi(triple) = [1, 3]
        pi(B) = []
        pi(A) = []
        pi(C) = []
    

The next rules are strictly ordered:

  p1, p2

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