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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          foldf#_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (37,37)
          cons_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (55,58)
          f#_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (90,36)
          foldf_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (1,59)
          |f'|#_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (50,1)
          g_A(x1) = x1 + (0,58)
          triple_A(x1,x2,x3) = ((0,0),(1,0)) x1 + ((1,0),(1,0)) x2 + ((1,0),(1,0)) x3 + (6,35)
          A_A() = (0,38)
          nil_A() = (5,0)
          |f''|#_A(x1) = ((1,0),(0,0)) x1 + (43,39)
          B_A() = (40,0)
          |f''|_A(x1) = ((1,0),(1,1)) x1 + (7,19)
          |f'|_A(x1,x2) = x1 + ((1,0),(1,0)) x2 + (55,56)
          C_A() = (41,133)
          f_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (55,57)
  
    precedence:
    
      foldf# = f# = foldf = |f'|# = g = triple = |f''|# = B = |f''| > nil > cons = A = |f'| = C = f
    
    partial status:
  
      pi(foldf#) = []
      pi(cons) = []
      pi(f#) = []
      pi(foldf) = []
      pi(|f'|#) = []
      pi(g) = []
      pi(triple) = []
      pi(A) = []
      pi(nil) = []
      pi(|f''|#) = []
      pi(B) = []
      pi(|f''|) = []
      pi(|f'|) = []
      pi(C) = []
      pi(f) = []

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- 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: f#(t,x) -> |f'|#(t,g(x))
p3: foldf#(x,cons(y,z)) -> foldf#(x,z)
p4: |f'|#(triple(a,b,c),A()) -> |f''|#(foldf(triple(cons(A(),a),nil(),c),b))
p5: |f''|#(triple(a,b,c)) -> foldf#(triple(a,b,nil()),c)
p6: |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, p3, p4, p5, p6}


-- 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),B()) -> f#(triple(a,b,c),A())
p4: |f'|#(triple(a,b,c),A()) -> |f''|#(foldf(triple(cons(A(),a),nil(),c),b))
p5: |f''|#(triple(a,b,c)) -> foldf#(triple(a,b,nil()),c)
p6: foldf#(x,cons(y,z)) -> foldf#(x,z)

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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          foldf#_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (5,9)
          cons_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,1)) x2
          f#_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (3,5)
          foldf_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,0)) x2 + (1,3)
          |f'|#_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (2,4)
          g_A(x1) = ((1,0),(0,0)) x1 + (0,6)
          triple_A(x1,x2,x3) = ((1,0),(0,0)) x2 + ((1,0),(0,0)) x3 + (0,10)
          B_A() = (17,3)
          A_A() = (15,2)
          |f''|#_A(x1) = ((1,0),(0,0)) x1 + (14,13)
          nil_A() = (1,12)
          |f''|_A(x1) = ((1,0),(0,0)) x1 + (12,11)
          |f'|_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,0)) x2 + (0,1)
          C_A() = (18,4)
          f_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,0)) x2 + (0,2)
  
    precedence:
    
      cons = foldf = |f''| = |f'| = f > C > g = B = A > f# > foldf# = |f'|# = triple = |f''|# = nil
    
    partial status:
  
      pi(foldf#) = []
      pi(cons) = []
      pi(f#) = []
      pi(foldf) = []
      pi(|f'|#) = [1]
      pi(g) = []
      pi(triple) = []
      pi(B) = []
      pi(A) = []
      pi(|f''|#) = []
      pi(nil) = []
      pi(|f''|) = []
      pi(|f'|) = []
      pi(C) = []
      pi(f) = []

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- 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: |f'|#(triple(a,b,c),B()) -> f#(triple(a,b,c),A())
p3: |f'|#(triple(a,b,c),A()) -> |f''|#(foldf(triple(cons(A(),a),nil(),c),b))
p4: |f''|#(triple(a,b,c)) -> foldf#(triple(a,b,nil()),c)
p5: foldf#(x,cons(y,z)) -> foldf#(x,z)

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:

  {p5}


-- Reduction pair.

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

p1: foldf#(x,cons(y,z)) -> foldf#(x,z)

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

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          foldf#_A(x1,x2) = ((1,0),(1,0)) x2 + (1,0)
          cons_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(1,0)) x2 + (1,0)
  
    precedence:
    
      cons > foldf#
    
    partial status:
  
      pi(foldf#) = []
      pi(cons) = []

The next rules are strictly ordered:

  p1

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