YES

We show the termination of the TRS R:

  a__first(|0|(),X) -> nil()
  a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
  a__from(X) -> cons(mark(X),from(s(X)))
  mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
  mark(from(X)) -> a__from(mark(X))
  mark(|0|()) -> |0|()
  mark(nil()) -> nil()
  mark(s(X)) -> s(mark(X))
  mark(cons(X1,X2)) -> cons(mark(X1),X2)
  a__first(X1,X2) -> first(X1,X2)
  a__from(X) -> from(X)

-- SCC decomposition.

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

p1: a__first#(s(X),cons(Y,Z)) -> mark#(Y)
p2: a__from#(X) -> mark#(X)
p3: mark#(first(X1,X2)) -> a__first#(mark(X1),mark(X2))
p4: mark#(first(X1,X2)) -> mark#(X1)
p5: mark#(first(X1,X2)) -> mark#(X2)
p6: mark#(from(X)) -> a__from#(mark(X))
p7: mark#(from(X)) -> mark#(X)
p8: mark#(s(X)) -> mark#(X)
p9: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__first(|0|(),X) -> nil()
r2: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(nil()) -> nil()
r8: mark(s(X)) -> s(mark(X))
r9: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r10: a__first(X1,X2) -> first(X1,X2)
r11: a__from(X) -> from(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__first#(s(X),cons(Y,Z)) -> mark#(Y)
p2: mark#(cons(X1,X2)) -> mark#(X1)
p3: mark#(s(X)) -> mark#(X)
p4: mark#(from(X)) -> mark#(X)
p5: mark#(from(X)) -> a__from#(mark(X))
p6: a__from#(X) -> mark#(X)
p7: mark#(first(X1,X2)) -> mark#(X2)
p8: mark#(first(X1,X2)) -> mark#(X1)
p9: mark#(first(X1,X2)) -> a__first#(mark(X1),mark(X2))

and R consists of:

r1: a__first(|0|(),X) -> nil()
r2: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(nil()) -> nil()
r8: mark(s(X)) -> s(mark(X))
r9: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r10: a__first(X1,X2) -> first(X1,X2)
r11: a__from(X) -> from(X)

The set of usable rules consists of

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          a__first#_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2
          s_A(x1) = ((0,0),(0,1)) x1 + (11,7)
          cons_A(x1,x2) = ((0,1),(0,1)) x1 + (1,0)
          mark#_A(x1) = ((0,1),(0,0)) x1 + (8,0)
          from_A(x1) = ((0,0),(0,1)) x1 + (9,6)
          a__from#_A(x1) = ((0,1),(0,0)) x1 + (8,0)
          mark_A(x1) = ((0,1),(0,1)) x1 + (4,0)
          first_A(x1,x2) = ((0,0),(0,1)) x1 + ((0,0),(0,1)) x2 + (2,0)
          a__first_A(x1,x2) = ((0,0),(0,1)) x1 + ((0,1),(0,1)) x2 + (3,0)
          |0|_A() = (1,1)
          nil_A() = (0,1)
          a__from_A(x1) = ((0,1),(0,1)) x1 + (9,6)
  
    precedence:
    
      from = a__from > cons = mark = a__first = nil > a__first# = s = mark# = a__from# = first = |0|
    
    partial status:
  
      pi(a__first#) = []
      pi(s) = []
      pi(cons) = []
      pi(mark#) = []
      pi(from) = []
      pi(a__from#) = []
      pi(mark) = []
      pi(first) = []
      pi(a__first) = []
      pi(|0|) = []
      pi(nil) = []
      pi(a__from) = []

The next rules are strictly ordered:

  p4

We remove them from the problem.

-- SCC decomposition.

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

p1: a__first#(s(X),cons(Y,Z)) -> mark#(Y)
p2: mark#(cons(X1,X2)) -> mark#(X1)
p3: mark#(s(X)) -> mark#(X)
p4: mark#(from(X)) -> a__from#(mark(X))
p5: a__from#(X) -> mark#(X)
p6: mark#(first(X1,X2)) -> mark#(X2)
p7: mark#(first(X1,X2)) -> mark#(X1)
p8: mark#(first(X1,X2)) -> a__first#(mark(X1),mark(X2))

and R consists of:

r1: a__first(|0|(),X) -> nil()
r2: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(nil()) -> nil()
r8: mark(s(X)) -> s(mark(X))
r9: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r10: a__first(X1,X2) -> first(X1,X2)
r11: a__from(X) -> from(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__first#(s(X),cons(Y,Z)) -> mark#(Y)
p2: mark#(first(X1,X2)) -> a__first#(mark(X1),mark(X2))
p3: mark#(first(X1,X2)) -> mark#(X1)
p4: mark#(first(X1,X2)) -> mark#(X2)
p5: mark#(from(X)) -> a__from#(mark(X))
p6: a__from#(X) -> mark#(X)
p7: mark#(s(X)) -> mark#(X)
p8: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__first(|0|(),X) -> nil()
r2: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(nil()) -> nil()
r8: mark(s(X)) -> s(mark(X))
r9: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r10: a__first(X1,X2) -> first(X1,X2)
r11: a__from(X) -> from(X)

The set of usable rules consists of

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          a__first#_A(x1,x2) = ((1,1),(0,0)) x1 + ((1,1),(0,1)) x2 + (1,1)
          s_A(x1) = ((1,1),(0,1)) x1 + (6,1)
          cons_A(x1,x2) = x1 + (1,6)
          mark#_A(x1) = ((1,1),(0,1)) x1 + (2,1)
          first_A(x1,x2) = ((1,1),(0,1)) x1 + ((1,1),(0,1)) x2 + (3,6)
          mark_A(x1) = ((1,1),(0,1)) x1 + (4,0)
          from_A(x1) = ((1,1),(0,1)) x1 + (8,8)
          a__from#_A(x1) = ((1,1),(0,1)) x1 + (3,1)
          a__first_A(x1,x2) = ((1,1),(0,1)) x1 + ((1,1),(0,1)) x2 + (4,6)
          |0|_A() = (2,1)
          nil_A() = (1,1)
          a__from_A(x1) = ((1,1),(0,1)) x1 + (15,8)
  
    precedence:
    
      first = from > a__first# = s = mark# = mark = a__from# = |0| = nil > a__first > cons = a__from
    
    partial status:
  
      pi(a__first#) = []
      pi(s) = [1]
      pi(cons) = [1]
      pi(mark#) = [1]
      pi(first) = [1, 2]
      pi(mark) = [1]
      pi(from) = []
      pi(a__from#) = [1]
      pi(a__first) = [1, 2]
      pi(|0|) = []
      pi(nil) = []
      pi(a__from) = []

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__first#(s(X),cons(Y,Z)) -> mark#(Y)
p2: mark#(first(X1,X2)) -> a__first#(mark(X1),mark(X2))
p3: mark#(first(X1,X2)) -> mark#(X1)
p4: mark#(first(X1,X2)) -> mark#(X2)
p5: a__from#(X) -> mark#(X)
p6: mark#(s(X)) -> mark#(X)
p7: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__first(|0|(),X) -> nil()
r2: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(nil()) -> nil()
r8: mark(s(X)) -> s(mark(X))
r9: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r10: a__first(X1,X2) -> first(X1,X2)
r11: a__from(X) -> from(X)

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: a__first#(s(X),cons(Y,Z)) -> mark#(Y)
p2: mark#(cons(X1,X2)) -> mark#(X1)
p3: mark#(s(X)) -> mark#(X)
p4: mark#(first(X1,X2)) -> mark#(X2)
p5: mark#(first(X1,X2)) -> mark#(X1)
p6: mark#(first(X1,X2)) -> a__first#(mark(X1),mark(X2))

and R consists of:

r1: a__first(|0|(),X) -> nil()
r2: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(nil()) -> nil()
r8: mark(s(X)) -> s(mark(X))
r9: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r10: a__first(X1,X2) -> first(X1,X2)
r11: a__from(X) -> from(X)

The set of usable rules consists of

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          a__first#_A(x1,x2) = ((0,1),(1,0)) x1 + ((1,1),(1,1)) x2 + (1,1)
          s_A(x1) = ((0,1),(1,1)) x1 + (2,3)
          cons_A(x1,x2) = x1
          mark#_A(x1) = ((1,1),(1,1)) x1 + (2,1)
          first_A(x1,x2) = ((0,1),(1,1)) x1 + ((0,1),(1,1)) x2
          mark_A(x1) = ((0,1),(1,1)) x1
          a__first_A(x1,x2) = ((0,1),(1,1)) x1 + ((0,1),(1,1)) x2
          |0|_A() = (0,1)
          nil_A() = (0,1)
          a__from_A(x1) = ((0,1),(1,1)) x1 + (0,3)
          from_A(x1) = ((0,1),(1,1)) x1 + (0,3)
  
    precedence:
    
      a__first# = mark# > |0| = nil > s = cons = first = mark = a__first = a__from > from
    
    partial status:
  
      pi(a__first#) = []
      pi(s) = []
      pi(cons) = []
      pi(mark#) = []
      pi(first) = []
      pi(mark) = []
      pi(a__first) = []
      pi(|0|) = []
      pi(nil) = []
      pi(a__from) = []
      pi(from) = []

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__first#(s(X),cons(Y,Z)) -> mark#(Y)
p2: mark#(cons(X1,X2)) -> mark#(X1)
p3: mark#(first(X1,X2)) -> mark#(X2)
p4: mark#(first(X1,X2)) -> mark#(X1)
p5: mark#(first(X1,X2)) -> a__first#(mark(X1),mark(X2))

and R consists of:

r1: a__first(|0|(),X) -> nil()
r2: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(nil()) -> nil()
r8: mark(s(X)) -> s(mark(X))
r9: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r10: a__first(X1,X2) -> first(X1,X2)
r11: a__from(X) -> from(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5}


-- Reduction pair.

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

p1: a__first#(s(X),cons(Y,Z)) -> mark#(Y)
p2: mark#(first(X1,X2)) -> a__first#(mark(X1),mark(X2))
p3: mark#(first(X1,X2)) -> mark#(X1)
p4: mark#(first(X1,X2)) -> mark#(X2)
p5: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__first(|0|(),X) -> nil()
r2: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(nil()) -> nil()
r8: mark(s(X)) -> s(mark(X))
r9: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r10: a__first(X1,X2) -> first(X1,X2)
r11: a__from(X) -> from(X)

The set of usable rules consists of

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          a__first#_A(x1,x2) = ((0,1),(0,0)) x1 + ((1,0),(1,0)) x2 + (1,3)
          s_A(x1) = ((0,1),(1,0)) x1 + (1,2)
          cons_A(x1,x2) = ((0,1),(0,1)) x1
          mark#_A(x1) = ((0,1),(0,1)) x1 + (2,3)
          first_A(x1,x2) = ((0,0),(0,1)) x1 + ((0,1),(1,1)) x2
          mark_A(x1) = x1
          a__first_A(x1,x2) = ((0,0),(0,1)) x1 + ((0,1),(1,1)) x2
          |0|_A() = (0,1)
          nil_A() = (0,0)
          a__from_A(x1) = ((1,1),(1,1)) x1 + (1,2)
          from_A(x1) = ((1,1),(1,1)) x1 + (1,2)
  
    precedence:
    
      first = a__first = nil > mark# > a__first# = cons = a__from = from > mark > s = |0|
    
    partial status:
  
      pi(a__first#) = []
      pi(s) = []
      pi(cons) = []
      pi(mark#) = []
      pi(first) = []
      pi(mark) = [1]
      pi(a__first) = []
      pi(|0|) = []
      pi(nil) = []
      pi(a__from) = []
      pi(from) = []

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: a__first#(s(X),cons(Y,Z)) -> mark#(Y)
p2: mark#(first(X1,X2)) -> mark#(X1)
p3: mark#(first(X1,X2)) -> mark#(X2)
p4: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__first(|0|(),X) -> nil()
r2: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(nil()) -> nil()
r8: mark(s(X)) -> s(mark(X))
r9: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r10: a__first(X1,X2) -> first(X1,X2)
r11: a__from(X) -> from(X)

The estimated dependency graph contains the following SCCs:

  {p2, p3, p4}


-- Reduction pair.

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

p1: mark#(first(X1,X2)) -> mark#(X1)
p2: mark#(cons(X1,X2)) -> mark#(X1)
p3: mark#(first(X1,X2)) -> mark#(X2)

and R consists of:

r1: a__first(|0|(),X) -> nil()
r2: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(nil()) -> nil()
r8: mark(s(X)) -> s(mark(X))
r9: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r10: a__first(X1,X2) -> first(X1,X2)
r11: a__from(X) -> from(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: standard order
        interpretations:
          mark#_A(x1) = ((1,1),(0,1)) x1 + (1,1)
          first_A(x1,x2) = x1 + ((1,1),(0,1)) x2 + (2,1)
          cons_A(x1,x2) = ((1,1),(0,1)) x1 + ((1,1),(1,0)) x2 + (2,1)
  
    precedence:
    
      mark# = first = cons
    
    partial status:
  
      pi(mark#) = [1]
      pi(first) = [2]
      pi(cons) = [1]

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: mark#(first(X1,X2)) -> mark#(X1)
p2: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__first(|0|(),X) -> nil()
r2: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(nil()) -> nil()
r8: mark(s(X)) -> s(mark(X))
r9: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r10: a__first(X1,X2) -> first(X1,X2)
r11: a__from(X) -> from(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: mark#(first(X1,X2)) -> mark#(X1)
p2: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__first(|0|(),X) -> nil()
r2: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(nil()) -> nil()
r8: mark(s(X)) -> s(mark(X))
r9: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r10: a__first(X1,X2) -> first(X1,X2)
r11: a__from(X) -> from(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: standard order
        interpretations:
          mark#_A(x1) = ((0,1),(0,0)) x1 + (2,2)
          first_A(x1,x2) = ((0,0),(0,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
          cons_A(x1,x2) = ((0,0),(0,1)) x1 + ((1,1),(1,1)) x2 + (1,1)
  
    precedence:
    
      mark# = first > cons
    
    partial status:
  
      pi(mark#) = []
      pi(first) = [2]
      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: mark#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__first(|0|(),X) -> nil()
r2: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(nil()) -> nil()
r8: mark(s(X)) -> s(mark(X))
r9: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r10: a__first(X1,X2) -> first(X1,X2)
r11: a__from(X) -> from(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#(cons(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__first(|0|(),X) -> nil()
r2: a__first(s(X),cons(Y,Z)) -> cons(mark(Y),first(X,Z))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(|0|()) -> |0|()
r7: mark(nil()) -> nil()
r8: mark(s(X)) -> s(mark(X))
r9: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r10: a__first(X1,X2) -> first(X1,X2)
r11: a__from(X) -> from(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

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

The next rules are strictly ordered:

  p1

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