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. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            foldf#_A(x1,x2) = ((0,1),(0,1)) x1 + ((0,1),(0,1)) x2 + (0,1)
            cons_A(x1,x2) = ((1,0),(1,1)) x1 + ((0,0),(1,1)) x2 + (0,22)
            f#_A(x1,x2) = ((0,1),(0,1)) x1 + ((1,1),(0,0)) x2 + (12,6)
            foldf_A(x1,x2) = ((1,1),(0,1)) x1 + ((0,1),(0,1)) x2 + (4,1)
            |f'|#_A(x1,x2) = ((0,1),(0,1)) x1 + ((0,1),(0,0)) x2 + (7,6)
            g_A(x1) = ((1,1),(1,1)) x1 + (13,0)
            triple_A(x1,x2,x3) = ((0,0),(0,1)) x2 + ((0,0),(1,1)) x3
            A_A() = (1,0)
            nil_A() = (0,2)
            |f''|#_A(x1) = ((0,1),(0,1)) x1 + (3,3)
            B_A() = (16,6)
            |f''|_A(x1) = ((0,1),(0,1)) x1 + (7,3)
            |f'|_A(x1,x2) = ((0,1),(0,1)) x1 + x2 + (10,22)
            C_A() = (0,6)
            f_A(x1,x2) = ((0,1),(0,1)) x1 + ((1,1),(1,1)) x2 + (24,22)
    
      precedence:
      
        |f'| > |f''| > foldf > f > g > cons = B = C > f# = |f'|# > |f''|# > foldf# = triple > A > nil
      
      partial status:
    
        pi(foldf#) = []
        pi(cons) = [1]
        pi(f#) = []
        pi(foldf) = [1]
        pi(|f'|#) = []
        pi(g) = []
        pi(triple) = []
        pi(A) = []
        pi(nil) = []
        pi(|f''|#) = []
        pi(B) = []
        pi(|f''|) = []
        pi(|f'|) = [2]
        pi(C) = []
        pi(f) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            foldf#_A(x1,x2) = (3,2)
            cons_A(x1,x2) = (0,0)
            f#_A(x1,x2) = (3,2)
            foldf_A(x1,x2) = ((0,0),(0,1)) x1 + (2,4)
            |f'|#_A(x1,x2) = (3,2)
            g_A(x1) = (0,4)
            triple_A(x1,x2,x3) = (0,3)
            A_A() = (4,2)
            nil_A() = (0,8)
            |f''|#_A(x1) = (3,2)
            B_A() = (0,1)
            |f''|_A(x1) = (5,7)
            |f'|_A(x1,x2) = (6,7)
            C_A() = (0,4)
            f_A(x1,x2) = (1,4)
    
      precedence:
      
        nil > foldf > |f'| = f > g = A > C > B > f# = |f'|# > |f''|# > triple = |f''| > cons > foldf#
      
      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:

  p1, p3, 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: foldf#(x,cons(y,z)) -> foldf#(x,z)
p3: |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, p3}
  {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. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2) = max{4, x1, x2}
          |f'|#_A(x1,x2) = max{3, x1, x2}
          g_A(x1) = x1
          triple_A(x1,x2,x3) = max{x1 + 5, x2 + 5, x3 + 1}
          B_A = 5
          A_A = 0
          C_A = 6
    
      precedence:
      
        f# = |f'|# = g = triple = B = A = C
      
      partial status:
    
        pi(f#) = [1, 2]
        pi(|f'|#) = [1, 2]
        pi(g) = []
        pi(triple) = [1, 2, 3]
        pi(B) = []
        pi(A) = []
        pi(C) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2) = max{x1 + 15, x2 + 6}
          |f'|#_A(x1,x2) = x1 + 12
          g_A(x1) = 5
          triple_A(x1,x2,x3) = max{x1 + 5, x2 + 5, x3 + 5}
          B_A = 1
          A_A = 4
          C_A = 0
    
      precedence:
      
        g > A > B > f# = |f'|# = triple = C
      
      partial status:
    
        pi(f#) = [2]
        pi(|f'|#) = [1]
        pi(g) = []
        pi(triple) = [1]
        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.

-- 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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          foldf#_A(x1,x2) = max{0, x2 - 2}
          cons_A(x1,x2) = max{3, x1, x2 + 1}
    
      precedence:
      
        foldf# = cons
      
      partial status:
    
        pi(foldf#) = []
        pi(cons) = [1, 2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          foldf#_A(x1,x2) = 0
          cons_A(x1,x2) = x2
    
      precedence:
      
        foldf# = cons
      
      partial status:
    
        pi(foldf#) = []
        pi(cons) = [2]
    

The next rules are strictly ordered:

  p1

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