YES

We show the termination of the TRS R:

  a__f(|0|()) -> cons(|0|(),f(s(|0|())))
  a__f(s(|0|())) -> a__f(a__p(s(|0|())))
  a__p(s(X)) -> mark(X)
  mark(f(X)) -> a__f(mark(X))
  mark(p(X)) -> a__p(mark(X))
  mark(|0|()) -> |0|()
  mark(cons(X1,X2)) -> cons(mark(X1),X2)
  mark(s(X)) -> s(mark(X))
  a__f(X) -> f(X)
  a__p(X) -> p(X)

-- SCC decomposition.

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

p1: a__f#(s(|0|())) -> a__f#(a__p(s(|0|())))
p2: a__f#(s(|0|())) -> a__p#(s(|0|()))
p3: a__p#(s(X)) -> mark#(X)
p4: mark#(f(X)) -> a__f#(mark(X))
p5: mark#(f(X)) -> mark#(X)
p6: mark#(p(X)) -> a__p#(mark(X))
p7: mark#(p(X)) -> mark#(X)
p8: mark#(cons(X1,X2)) -> mark#(X1)
p9: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__f(|0|()) -> cons(|0|(),f(s(|0|())))
r2: a__f(s(|0|())) -> a__f(a__p(s(|0|())))
r3: a__p(s(X)) -> mark(X)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(p(X)) -> a__p(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r8: mark(s(X)) -> s(mark(X))
r9: a__f(X) -> f(X)
r10: a__p(X) -> p(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9}


-- Reduction pair.

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

p1: a__f#(s(|0|())) -> a__f#(a__p(s(|0|())))
p2: a__f#(s(|0|())) -> a__p#(s(|0|()))
p3: a__p#(s(X)) -> mark#(X)
p4: mark#(s(X)) -> mark#(X)
p5: mark#(cons(X1,X2)) -> mark#(X1)
p6: mark#(p(X)) -> mark#(X)
p7: mark#(p(X)) -> a__p#(mark(X))
p8: mark#(f(X)) -> mark#(X)
p9: mark#(f(X)) -> a__f#(mark(X))

and R consists of:

r1: a__f(|0|()) -> cons(|0|(),f(s(|0|())))
r2: a__f(s(|0|())) -> a__f(a__p(s(|0|())))
r3: a__p(s(X)) -> mark(X)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(p(X)) -> a__p(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r8: mark(s(X)) -> s(mark(X))
r9: a__f(X) -> f(X)
r10: a__p(X) -> p(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          a__f#_A(x1) = ((1,0),(0,0)) x1 + (2,4)
          s_A(x1) = x1 + (6,2)
          |0|_A() = (1,1)
          a__p_A(x1) = ((1,0),(0,0)) x1 + (0,2)
          a__p#_A(x1) = ((1,0),(0,0)) x1 + (1,2)
          mark#_A(x1) = ((1,0),(0,0)) x1 + (4,5)
          cons_A(x1,x2) = x1 + (1,1)
          p_A(x1) = ((1,0),(0,0)) x1 + (0,1)
          mark_A(x1) = ((1,0),(1,0)) x1 + (2,3)
          f_A(x1) = ((1,0),(0,0)) x1 + (5,6)
          a__f_A(x1) = ((1,0),(0,0)) x1 + (5,7)
  
    precedence:
    
      s = a__p = mark = a__f > f > cons > p > a__f# = |0| = a__p# = mark#
    
    partial status:
  
      pi(a__f#) = []
      pi(s) = []
      pi(|0|) = []
      pi(a__p) = []
      pi(a__p#) = []
      pi(mark#) = []
      pi(cons) = [1]
      pi(p) = []
      pi(mark) = []
      pi(f) = []
      pi(a__f) = []

The next rules are strictly ordered:

  p5

We remove them from the problem.

-- SCC decomposition.

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

p1: a__f#(s(|0|())) -> a__f#(a__p(s(|0|())))
p2: a__f#(s(|0|())) -> a__p#(s(|0|()))
p3: a__p#(s(X)) -> mark#(X)
p4: mark#(s(X)) -> mark#(X)
p5: mark#(p(X)) -> mark#(X)
p6: mark#(p(X)) -> a__p#(mark(X))
p7: mark#(f(X)) -> mark#(X)
p8: mark#(f(X)) -> a__f#(mark(X))

and R consists of:

r1: a__f(|0|()) -> cons(|0|(),f(s(|0|())))
r2: a__f(s(|0|())) -> a__f(a__p(s(|0|())))
r3: a__p(s(X)) -> mark(X)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(p(X)) -> a__p(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r8: mark(s(X)) -> s(mark(X))
r9: a__f(X) -> f(X)
r10: a__p(X) -> p(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8}


-- Reduction pair.

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

p1: a__f#(s(|0|())) -> a__f#(a__p(s(|0|())))
p2: a__f#(s(|0|())) -> a__p#(s(|0|()))
p3: a__p#(s(X)) -> mark#(X)
p4: mark#(f(X)) -> a__f#(mark(X))
p5: mark#(f(X)) -> mark#(X)
p6: mark#(p(X)) -> a__p#(mark(X))
p7: mark#(p(X)) -> mark#(X)
p8: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__f(|0|()) -> cons(|0|(),f(s(|0|())))
r2: a__f(s(|0|())) -> a__f(a__p(s(|0|())))
r3: a__p(s(X)) -> mark(X)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(p(X)) -> a__p(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r8: mark(s(X)) -> s(mark(X))
r9: a__f(X) -> f(X)
r10: a__p(X) -> p(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          a__f#_A(x1) = (8,7)
          s_A(x1) = ((1,0),(0,0)) x1 + (3,4)
          |0|_A() = (2,5)
          a__p_A(x1) = ((1,0),(0,0)) x1 + (0,2)
          a__p#_A(x1) = x1 + (2,2)
          mark#_A(x1) = ((1,0),(0,0)) x1 + (5,5)
          f_A(x1) = ((1,0),(0,0)) x1 + (4,4)
          mark_A(x1) = x1 + (2,2)
          p_A(x1) = ((1,0),(0,0)) x1
          a__f_A(x1) = ((1,0),(0,0)) x1 + (4,5)
          cons_A(x1,x2) = (1,2)
  
    precedence:
    
      a__f# = a__p = a__p# = mark# = mark = p > f = a__f > cons > s = |0|
    
    partial status:
  
      pi(a__f#) = []
      pi(s) = []
      pi(|0|) = []
      pi(a__p) = []
      pi(a__p#) = []
      pi(mark#) = []
      pi(f) = []
      pi(mark) = []
      pi(p) = []
      pi(a__f) = []
      pi(cons) = []

The next rules are strictly ordered:

  p8

We remove them from the problem.

-- SCC decomposition.

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

p1: a__f#(s(|0|())) -> a__f#(a__p(s(|0|())))
p2: a__f#(s(|0|())) -> a__p#(s(|0|()))
p3: a__p#(s(X)) -> mark#(X)
p4: mark#(f(X)) -> a__f#(mark(X))
p5: mark#(f(X)) -> mark#(X)
p6: mark#(p(X)) -> a__p#(mark(X))
p7: mark#(p(X)) -> mark#(X)

and R consists of:

r1: a__f(|0|()) -> cons(|0|(),f(s(|0|())))
r2: a__f(s(|0|())) -> a__f(a__p(s(|0|())))
r3: a__p(s(X)) -> mark(X)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(p(X)) -> a__p(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r8: mark(s(X)) -> s(mark(X))
r9: a__f(X) -> f(X)
r10: a__p(X) -> p(X)

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: a__f#(s(|0|())) -> a__f#(a__p(s(|0|())))
p2: a__f#(s(|0|())) -> a__p#(s(|0|()))
p3: a__p#(s(X)) -> mark#(X)
p4: mark#(p(X)) -> mark#(X)
p5: mark#(p(X)) -> a__p#(mark(X))
p6: mark#(f(X)) -> mark#(X)
p7: mark#(f(X)) -> a__f#(mark(X))

and R consists of:

r1: a__f(|0|()) -> cons(|0|(),f(s(|0|())))
r2: a__f(s(|0|())) -> a__f(a__p(s(|0|())))
r3: a__p(s(X)) -> mark(X)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(p(X)) -> a__p(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r8: mark(s(X)) -> s(mark(X))
r9: a__f(X) -> f(X)
r10: a__p(X) -> p(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          a__f#_A(x1) = x1 + (2,1)
          s_A(x1) = x1 + (6,0)
          |0|_A() = (2,21)
          a__p_A(x1) = ((1,0),(0,0)) x1 + (0,9)
          a__p#_A(x1) = ((1,0),(1,0)) x1 + (1,12)
          mark#_A(x1) = ((1,0),(0,0)) x1 + (5,13)
          p_A(x1) = ((1,0),(0,0)) x1 + (0,8)
          mark_A(x1) = ((1,0),(0,0)) x1 + (2,11)
          f_A(x1) = ((1,0),(0,0)) x1 + (0,9)
          a__f_A(x1) = ((1,0),(0,0)) x1 + (0,10)
          cons_A(x1,x2) = ((0,0),(1,0)) x2 + (1,1)
  
    precedence:
    
      mark = a__f > f > a__f# = s = a__p# = mark# > |0| = a__p > p > cons
    
    partial status:
  
      pi(a__f#) = []
      pi(s) = [1]
      pi(|0|) = []
      pi(a__p) = []
      pi(a__p#) = []
      pi(mark#) = []
      pi(p) = []
      pi(mark) = []
      pi(f) = []
      pi(a__f) = []
      pi(cons) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: a__f#(s(|0|())) -> a__p#(s(|0|()))
p2: a__p#(s(X)) -> mark#(X)
p3: mark#(p(X)) -> mark#(X)
p4: mark#(p(X)) -> a__p#(mark(X))
p5: mark#(f(X)) -> mark#(X)
p6: mark#(f(X)) -> a__f#(mark(X))

and R consists of:

r1: a__f(|0|()) -> cons(|0|(),f(s(|0|())))
r2: a__f(s(|0|())) -> a__f(a__p(s(|0|())))
r3: a__p(s(X)) -> mark(X)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(p(X)) -> a__p(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r8: mark(s(X)) -> s(mark(X))
r9: a__f(X) -> f(X)
r10: a__p(X) -> p(X)

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: a__f#(s(|0|())) -> a__p#(s(|0|()))
p2: a__p#(s(X)) -> mark#(X)
p3: mark#(f(X)) -> a__f#(mark(X))
p4: mark#(f(X)) -> mark#(X)
p5: mark#(p(X)) -> a__p#(mark(X))
p6: mark#(p(X)) -> mark#(X)

and R consists of:

r1: a__f(|0|()) -> cons(|0|(),f(s(|0|())))
r2: a__f(s(|0|())) -> a__f(a__p(s(|0|())))
r3: a__p(s(X)) -> mark(X)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(p(X)) -> a__p(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r8: mark(s(X)) -> s(mark(X))
r9: a__f(X) -> f(X)
r10: a__p(X) -> p(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          a__f#_A(x1) = ((1,0),(0,0)) x1 + (3,18)
          s_A(x1) = ((1,0),(1,0)) x1 + (10,0)
          |0|_A() = (6,3)
          a__p#_A(x1) = ((1,0),(1,0)) x1 + (2,1)
          mark#_A(x1) = ((1,0),(1,0)) x1 + (11,10)
          f_A(x1) = x1 + (9,19)
          mark_A(x1) = ((1,0),(1,1)) x1 + (2,2)
          p_A(x1) = ((1,0),(0,0)) x1 + (0,4)
          a__f_A(x1) = x1 + (9,23)
          cons_A(x1,x2) = (1,3)
          a__p_A(x1) = ((1,0),(0,0)) x1 + (0,5)
  
    precedence:
    
      |0| = f = mark = a__f = a__p > cons > a__f# = a__p# = mark# > p > s
    
    partial status:
  
      pi(a__f#) = []
      pi(s) = []
      pi(|0|) = []
      pi(a__p#) = []
      pi(mark#) = []
      pi(f) = []
      pi(mark) = []
      pi(p) = []
      pi(a__f) = []
      pi(cons) = []
      pi(a__p) = []

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: a__f#(s(|0|())) -> a__p#(s(|0|()))
p2: a__p#(s(X)) -> mark#(X)
p3: mark#(f(X)) -> mark#(X)
p4: mark#(p(X)) -> a__p#(mark(X))
p5: mark#(p(X)) -> mark#(X)

and R consists of:

r1: a__f(|0|()) -> cons(|0|(),f(s(|0|())))
r2: a__f(s(|0|())) -> a__f(a__p(s(|0|())))
r3: a__p(s(X)) -> mark(X)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(p(X)) -> a__p(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r8: mark(s(X)) -> s(mark(X))
r9: a__f(X) -> f(X)
r10: a__p(X) -> p(X)

The estimated dependency graph contains the following SCCs:

  {p2, p3, p4, p5}


-- Reduction pair.

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

p1: a__p#(s(X)) -> mark#(X)
p2: mark#(p(X)) -> mark#(X)
p3: mark#(p(X)) -> a__p#(mark(X))
p4: mark#(f(X)) -> mark#(X)

and R consists of:

r1: a__f(|0|()) -> cons(|0|(),f(s(|0|())))
r2: a__f(s(|0|())) -> a__f(a__p(s(|0|())))
r3: a__p(s(X)) -> mark(X)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(p(X)) -> a__p(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r8: mark(s(X)) -> s(mark(X))
r9: a__f(X) -> f(X)
r10: a__p(X) -> p(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          a__p#_A(x1) = ((1,0),(1,1)) x1 + (1,1)
          s_A(x1) = x1 + (5,7)
          mark#_A(x1) = ((1,0),(1,1)) x1 + (3,6)
          p_A(x1) = x1 + (0,5)
          mark_A(x1) = x1 + (1,8)
          f_A(x1) = ((1,0),(0,0)) x1 + (4,3)
          a__f_A(x1) = ((1,0),(0,0)) x1 + (4,10)
          |0|_A() = (2,11)
          cons_A(x1,x2) = (1,9)
          a__p_A(x1) = x1 + (0,5)
  
    precedence:
    
      s = mark = a__f = |0| = a__p > mark# = f > p > cons > a__p#
    
    partial status:
  
      pi(a__p#) = [1]
      pi(s) = []
      pi(mark#) = [1]
      pi(p) = []
      pi(mark) = []
      pi(f) = []
      pi(a__f) = []
      pi(|0|) = []
      pi(cons) = []
      pi(a__p) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(p(X)) -> mark#(X)
p2: mark#(p(X)) -> a__p#(mark(X))
p3: mark#(f(X)) -> mark#(X)

and R consists of:

r1: a__f(|0|()) -> cons(|0|(),f(s(|0|())))
r2: a__f(s(|0|())) -> a__f(a__p(s(|0|())))
r3: a__p(s(X)) -> mark(X)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(p(X)) -> a__p(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r8: mark(s(X)) -> s(mark(X))
r9: a__f(X) -> f(X)
r10: a__p(X) -> p(X)

The estimated dependency graph contains the following SCCs:

  {p1, p3}


-- Reduction pair.

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

p1: mark#(p(X)) -> mark#(X)
p2: mark#(f(X)) -> mark#(X)

and R consists of:

r1: a__f(|0|()) -> cons(|0|(),f(s(|0|())))
r2: a__f(s(|0|())) -> a__f(a__p(s(|0|())))
r3: a__p(s(X)) -> mark(X)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(p(X)) -> a__p(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r8: mark(s(X)) -> s(mark(X))
r9: a__f(X) -> f(X)
r10: a__p(X) -> p(X)

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:
          mark#_A(x1) = ((1,0),(0,0)) x1 + (1,1)
          p_A(x1) = x1 + (2,2)
          f_A(x1) = ((1,0),(0,0)) x1 + (2,2)
  
    precedence:
    
      mark# = p = f
    
    partial status:
  
      pi(mark#) = []
      pi(p) = [1]
      pi(f) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(f(X)) -> mark#(X)

and R consists of:

r1: a__f(|0|()) -> cons(|0|(),f(s(|0|())))
r2: a__f(s(|0|())) -> a__f(a__p(s(|0|())))
r3: a__p(s(X)) -> mark(X)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(p(X)) -> a__p(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r8: mark(s(X)) -> s(mark(X))
r9: a__f(X) -> f(X)
r10: a__p(X) -> p(X)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: mark#(f(X)) -> mark#(X)

and R consists of:

r1: a__f(|0|()) -> cons(|0|(),f(s(|0|())))
r2: a__f(s(|0|())) -> a__f(a__p(s(|0|())))
r3: a__p(s(X)) -> mark(X)
r4: mark(f(X)) -> a__f(mark(X))
r5: mark(p(X)) -> a__p(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r8: mark(s(X)) -> s(mark(X))
r9: a__f(X) -> f(X)
r10: a__p(X) -> p(X)

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:
          mark#_A(x1) = ((1,0),(1,0)) x1 + (2,2)
          f_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      mark# = f
    
    partial status:
  
      pi(mark#) = []
      pi(f) = []

The next rules are strictly ordered:

  p1

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