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:

  generalized weighted path order (>=MG, >G)
  
    base order 1:
  
      matrix interpretation with standard point-wise order:
      
        carrier: N^1
        order: standard order
        interpretations:
          foldf#_A(x1,x2) = x1 + x2
          cons_A(x1,x2) = x1 + x2 + 3
          f#_A(x1,x2) = x1 + x2 + 2
          foldf_A(x1,x2) = x1 + x2 + 1
          |f'|#_A(x1,x2) = x1 + x2
          g_A(x1) = x1
          triple_A(x1,x2,x3) = x2 + x3 + 2
          A_A() = 8
          nil_A() = 1
          |f''|#_A(x1) = x1 + 2
          B_A() = 10
          C_A() = 11
          f_A(x1,x2) = x1 + x2 + 3
          |f'|_A(x1,x2) = x1 + x2 + 3
          |f''|_A(x1) = x1 + 3
  
    base order 2:
  
      marked-algebra reduction triple (>=A, >=A!, >A!):
      
        max/plus interpretation on natural numbers:
        
          foldf#_A(x1,x2) = 1
          cons_A(x1,x2) = max{0, x1 - 3}
          f#_A(x1,x2) = 1
          foldf_A(x1,x2) = max{10, x1}
          |f'|#_A(x1,x2) = max{1, x2 - 38}
          g_A(x1) = 39
          triple_A(x1,x2,x3) = max{x1 + 7, x2 + 26, x3 + 53}
          A_A = 35
          nil_A = 0
          |f''|#_A(x1) = 1
          B_A = 37
          C_A = 1
          f_A(x1,x2) = max{10, x1}
          |f'|_A(x1,x2) = max{7, x1}
          |f''|_A(x1) = max{27, x1}
          foldf#!_A(x1,x2) = 0
          cons!_A(x1,x2) = max{44, x1 - 41, x2 + 13}
          f#!_A(x1,x2) = 0
          foldf!_A(x1,x2) = max{15, x1 + 3}
          |f'|#!_A(x1,x2) = max{13, x2 - 40}
          g!_A(x1) = max{45, x1 - 38}
          triple!_A(x1,x2,x3) = 12
          A!_A = 43
          nil!_A = 25
          |f''|#!_A(x1) = 24
          B!_A = 44
          C!_A = 12
          f!_A(x1,x2) = max{14, x1 + 2}
          |f'|!_A(x1,x2) = max{13, x1 - 10}
          |f''|!_A(x1) = max{43, x1 - 27}
  
    partial status:
  
      pi(foldf#) = [1, 2]
      pi(cons) = [1]
      pi(f#) = []
      pi(foldf) = []
      pi(|f'|#) = []
      pi(g) = [1]
      pi(triple) = []
      pi(A) = []
      pi(nil) = []
      pi(|f''|#) = []
      pi(B) = []
      pi(C) = []
      pi(f) = []
      pi(|f'|) = []
      pi(|f''|) = []

The following rules are removed:

  p1, p2, p3, p4, p5, p6, p7

Then no dependency pair remains.