YES

We show the termination of the TRS R:

  a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
  a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
  a__from(X) -> cons(mark(X),from(s(X)))
  mark(|2nd|(X)) -> a__2nd(mark(X))
  mark(from(X)) -> a__from(mark(X))
  mark(cons(X1,X2)) -> cons(mark(X1),X2)
  mark(s(X)) -> s(mark(X))
  mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
  a__2nd(X) -> |2nd|(X)
  a__from(X) -> from(X)

-- SCC decomposition.

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

p1: a__2nd#(cons1(X,cons(Y,Z))) -> mark#(Y)
p2: a__2nd#(cons(X,X1)) -> a__2nd#(cons1(mark(X),mark(X1)))
p3: a__2nd#(cons(X,X1)) -> mark#(X)
p4: a__2nd#(cons(X,X1)) -> mark#(X1)
p5: a__from#(X) -> mark#(X)
p6: mark#(|2nd|(X)) -> a__2nd#(mark(X))
p7: mark#(|2nd|(X)) -> mark#(X)
p8: mark#(from(X)) -> a__from#(mark(X))
p9: mark#(from(X)) -> mark#(X)
p10: mark#(cons(X1,X2)) -> mark#(X1)
p11: mark#(s(X)) -> mark#(X)
p12: mark#(cons1(X1,X2)) -> mark#(X1)
p13: mark#(cons1(X1,X2)) -> mark#(X2)

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p13}


-- Reduction pair.

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

p1: a__2nd#(cons1(X,cons(Y,Z))) -> mark#(Y)
p2: mark#(cons1(X1,X2)) -> mark#(X2)
p3: mark#(cons1(X1,X2)) -> mark#(X1)
p4: mark#(s(X)) -> mark#(X)
p5: mark#(cons(X1,X2)) -> mark#(X1)
p6: mark#(from(X)) -> mark#(X)
p7: mark#(from(X)) -> a__from#(mark(X))
p8: a__from#(X) -> mark#(X)
p9: mark#(|2nd|(X)) -> mark#(X)
p10: mark#(|2nd|(X)) -> a__2nd#(mark(X))
p11: a__2nd#(cons(X,X1)) -> mark#(X1)
p12: a__2nd#(cons(X,X1)) -> mark#(X)
p13: a__2nd#(cons(X,X1)) -> a__2nd#(cons1(mark(X),mark(X1)))

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(X)

The set of usable rules consists of

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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            a__2nd#_A(x1) = ((1,1),(0,0)) x1 + (2,6)
            cons1_A(x1,x2) = ((0,0),(0,1)) x1 + ((0,0),(0,1)) x2 + (4,1)
            cons_A(x1,x2) = ((0,0),(1,1)) x1 + ((0,1),(0,0)) x2 + (0,6)
            mark#_A(x1) = ((0,1),(0,0)) x1 + (3,6)
            s_A(x1) = ((0,0),(1,1)) x1 + (1,1)
            from_A(x1) = ((1,1),(1,1)) x1 + (9,6)
            a__from#_A(x1) = ((0,1),(0,0)) x1 + (9,6)
            mark_A(x1) = x1
            |2nd|_A(x1) = ((1,1),(1,1)) x1 + (1,2)
            a__2nd_A(x1) = ((1,1),(1,1)) x1 + (1,2)
            a__from_A(x1) = ((1,1),(1,1)) x1 + (9,6)
    
      precedence:
      
        a__2nd# = cons1 = cons = mark# = s = from = a__from# = mark = |2nd| = a__2nd = a__from
      
      partial status:
    
        pi(a__2nd#) = []
        pi(cons1) = []
        pi(cons) = []
        pi(mark#) = []
        pi(s) = []
        pi(from) = []
        pi(a__from#) = []
        pi(mark) = []
        pi(|2nd|) = []
        pi(a__2nd) = []
        pi(a__from) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            a__2nd#_A(x1) = x1
            cons1_A(x1,x2) = (3,1)
            cons_A(x1,x2) = (4,2)
            mark#_A(x1) = (2,1)
            s_A(x1) = (3,1)
            from_A(x1) = (5,1)
            a__from#_A(x1) = (2,1)
            mark_A(x1) = ((1,1),(0,1)) x1 + (1,1)
            |2nd|_A(x1) = (3,2)
            a__2nd_A(x1) = (5,3)
            a__from_A(x1) = (5,1)
    
      precedence:
      
        a__2nd > from = mark = a__from > s > cons1 = cons > mark# = a__from# > a__2nd# > |2nd|
      
      partial status:
    
        pi(a__2nd#) = [1]
        pi(cons1) = []
        pi(cons) = []
        pi(mark#) = []
        pi(s) = []
        pi(from) = []
        pi(a__from#) = []
        pi(mark) = [1]
        pi(|2nd|) = []
        pi(a__2nd) = []
        pi(a__from) = []
    

The next rules are strictly ordered:

  p1, p10, p11, p12

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(cons1(X1,X2)) -> mark#(X2)
p2: mark#(cons1(X1,X2)) -> mark#(X1)
p3: mark#(s(X)) -> mark#(X)
p4: mark#(cons(X1,X2)) -> mark#(X1)
p5: mark#(from(X)) -> mark#(X)
p6: mark#(from(X)) -> a__from#(mark(X))
p7: a__from#(X) -> mark#(X)
p8: mark#(|2nd|(X)) -> mark#(X)
p9: a__2nd#(cons(X,X1)) -> a__2nd#(cons1(mark(X),mark(X1)))

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: 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: mark#(cons1(X1,X2)) -> mark#(X2)
p2: mark#(|2nd|(X)) -> mark#(X)
p3: mark#(from(X)) -> a__from#(mark(X))
p4: a__from#(X) -> mark#(X)
p5: mark#(from(X)) -> mark#(X)
p6: mark#(cons(X1,X2)) -> mark#(X1)
p7: mark#(s(X)) -> mark#(X)
p8: mark#(cons1(X1,X2)) -> mark#(X1)

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(X)

The set of usable rules consists of

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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            mark#_A(x1) = ((1,0),(0,0)) x1
            cons1_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (1,0)
            |2nd|_A(x1) = ((1,1),(1,1)) x1 + (1,1)
            from_A(x1) = ((1,1),(1,0)) x1 + (3,6)
            a__from#_A(x1) = ((1,0),(0,0)) x1
            mark_A(x1) = x1
            cons_A(x1,x2) = ((1,1),(0,0)) x1 + ((0,0),(1,0)) x2 + (2,1)
            s_A(x1) = ((1,0),(0,0)) x1 + (1,1)
            a__2nd_A(x1) = ((1,1),(1,1)) x1 + (1,1)
            a__from_A(x1) = ((1,1),(1,0)) x1 + (3,6)
    
      precedence:
      
        mark# = cons1 = |2nd| = from = a__from# = mark = cons = s = a__2nd = a__from
      
      partial status:
    
        pi(mark#) = []
        pi(cons1) = []
        pi(|2nd|) = []
        pi(from) = []
        pi(a__from#) = []
        pi(mark) = []
        pi(cons) = []
        pi(s) = []
        pi(a__2nd) = []
        pi(a__from) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            mark#_A(x1) = ((0,1),(0,0)) x1 + (2,0)
            cons1_A(x1,x2) = x1 + x2 + (3,11)
            |2nd|_A(x1) = ((0,0),(0,1)) x1 + (1,1)
            from_A(x1) = ((0,1),(0,1)) x1 + (6,0)
            a__from#_A(x1) = ((0,1),(0,0)) x1 + (2,0)
            mark_A(x1) = ((0,1),(0,1)) x1 + (7,0)
            cons_A(x1,x2) = (3,0)
            s_A(x1) = ((0,0),(0,1)) x1 + (7,0)
            a__2nd_A(x1) = x1 + (1,1)
            a__from_A(x1) = ((0,1),(0,1)) x1 + (6,0)
    
      precedence:
      
        mark > s = a__from > mark# = cons1 = |2nd| = from = a__from# = cons = a__2nd
      
      partial status:
    
        pi(mark#) = []
        pi(cons1) = [1]
        pi(|2nd|) = []
        pi(from) = []
        pi(a__from#) = []
        pi(mark) = []
        pi(cons) = []
        pi(s) = []
        pi(a__2nd) = [1]
        pi(a__from) = []
    

The next rules are strictly ordered:

  p1, p6, p8

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(|2nd|(X)) -> mark#(X)
p2: mark#(from(X)) -> a__from#(mark(X))
p3: a__from#(X) -> mark#(X)
p4: mark#(from(X)) -> mark#(X)
p5: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: 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: mark#(|2nd|(X)) -> mark#(X)
p2: mark#(s(X)) -> mark#(X)
p3: mark#(from(X)) -> mark#(X)
p4: mark#(from(X)) -> a__from#(mark(X))
p5: a__from#(X) -> mark#(X)

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(X)

The set of usable rules consists of

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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            mark#_A(x1) = ((0,1),(0,1)) x1
            |2nd|_A(x1) = ((0,1),(0,1)) x1 + (1,2)
            s_A(x1) = x1 + (1,0)
            from_A(x1) = ((1,1),(1,1)) x1 + (2,5)
            a__from#_A(x1) = ((1,1),(0,1)) x1 + (1,0)
            mark_A(x1) = x1
            a__2nd_A(x1) = ((0,1),(0,1)) x1 + (1,2)
            cons1_A(x1,x2) = ((1,0),(1,0)) x2 + (0,1)
            cons_A(x1,x2) = ((1,1),(0,0)) x1 + ((0,0),(1,0)) x2 + (1,2)
            a__from_A(x1) = ((1,1),(1,1)) x1 + (2,5)
    
      precedence:
      
        mark# = |2nd| = s = from = a__from# = mark = a__2nd = cons1 = cons = a__from
      
      partial status:
    
        pi(mark#) = []
        pi(|2nd|) = []
        pi(s) = []
        pi(from) = []
        pi(a__from#) = []
        pi(mark) = []
        pi(a__2nd) = []
        pi(cons1) = []
        pi(cons) = []
        pi(a__from) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            mark#_A(x1) = (2,2)
            |2nd|_A(x1) = (2,2)
            s_A(x1) = ((1,0),(0,0)) x1 + (0,2)
            from_A(x1) = (2,2)
            a__from#_A(x1) = (2,2)
            mark_A(x1) = (2,2)
            a__2nd_A(x1) = (2,2)
            cons1_A(x1,x2) = (1,2)
            cons_A(x1,x2) = (0,1)
            a__from_A(x1) = (2,2)
    
      precedence:
      
        mark > a__2nd > |2nd| = s = a__from > cons > from > mark# = a__from# = cons1
      
      partial status:
    
        pi(mark#) = []
        pi(|2nd|) = []
        pi(s) = []
        pi(from) = []
        pi(a__from#) = []
        pi(mark) = []
        pi(a__2nd) = []
        pi(cons1) = []
        pi(cons) = []
        pi(a__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: mark#(|2nd|(X)) -> mark#(X)
p2: mark#(s(X)) -> mark#(X)
p3: mark#(from(X)) -> a__from#(mark(X))
p4: a__from#(X) -> mark#(X)

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(X)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- Reduction pair.

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

p1: mark#(|2nd|(X)) -> mark#(X)
p2: mark#(from(X)) -> a__from#(mark(X))
p3: a__from#(X) -> mark#(X)
p4: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(X)

The set of usable rules consists of

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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            mark#_A(x1) = ((0,1),(0,0)) x1 + (1,6)
            |2nd|_A(x1) = ((0,1),(0,1)) x1 + (2,6)
            from_A(x1) = ((1,1),(1,1)) x1 + (3,5)
            a__from#_A(x1) = ((1,1),(0,0)) x1 + (2,6)
            mark_A(x1) = x1
            s_A(x1) = ((0,0),(0,1)) x1 + (2,0)
            a__2nd_A(x1) = ((0,1),(0,1)) x1 + (2,6)
            cons1_A(x1,x2) = ((1,0),(1,0)) x2 + (3,0)
            cons_A(x1,x2) = ((1,1),(0,0)) x1 + ((0,0),(1,0)) x2 + (1,0)
            a__from_A(x1) = ((1,1),(1,1)) x1 + (3,5)
    
      precedence:
      
        from = a__from > mark = a__2nd = cons > mark# = |2nd| = s > a__from# > cons1
      
      partial status:
    
        pi(mark#) = []
        pi(|2nd|) = []
        pi(from) = []
        pi(a__from#) = []
        pi(mark) = [1]
        pi(s) = []
        pi(a__2nd) = []
        pi(cons1) = []
        pi(cons) = []
        pi(a__from) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            mark#_A(x1) = (2,2)
            |2nd|_A(x1) = (1,1)
            from_A(x1) = (4,2)
            a__from#_A(x1) = (2,2)
            mark_A(x1) = x1 + (5,4)
            s_A(x1) = (3,3)
            a__2nd_A(x1) = (5,4)
            cons1_A(x1,x2) = (0,3)
            cons_A(x1,x2) = (0,1)
            a__from_A(x1) = (4,2)
    
      precedence:
      
        mark = a__2nd > s = a__from > from > a__from# > cons > mark# = |2nd| > cons1
      
      partial status:
    
        pi(mark#) = []
        pi(|2nd|) = []
        pi(from) = []
        pi(a__from#) = []
        pi(mark) = []
        pi(s) = []
        pi(a__2nd) = []
        pi(cons1) = []
        pi(cons) = []
        pi(a__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: mark#(|2nd|(X)) -> mark#(X)
p2: a__from#(X) -> mark#(X)
p3: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(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#(|2nd|(X)) -> mark#(X)
p2: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__2nd(cons1(X,cons(Y,Z))) -> mark(Y)
r2: a__2nd(cons(X,X1)) -> a__2nd(cons1(mark(X),mark(X1)))
r3: a__from(X) -> cons(mark(X),from(s(X)))
r4: mark(|2nd|(X)) -> a__2nd(mark(X))
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r7: mark(s(X)) -> s(mark(X))
r8: mark(cons1(X1,X2)) -> cons1(mark(X1),mark(X2))
r9: a__2nd(X) -> |2nd|(X)
r10: a__from(X) -> from(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            mark#_A(x1) = x1 + (1,2)
            |2nd|_A(x1) = x1 + (2,1)
            s_A(x1) = ((1,0),(1,1)) x1 + (2,1)
    
      precedence:
      
        s > mark# = |2nd|
      
      partial status:
    
        pi(mark#) = [1]
        pi(|2nd|) = [1]
        pi(s) = [1]
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            mark#_A(x1) = ((1,0),(1,0)) x1 + (0,2)
            |2nd|_A(x1) = ((1,0),(0,0)) x1 + (1,1)
            s_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        mark# = |2nd| = s
      
      partial status:
    
        pi(mark#) = []
        pi(|2nd|) = []
        pi(s) = []
    

The next rules are strictly ordered:

  p1, p2

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