YES

We show the termination of the TRS R:

  a__from(X) -> cons(mark(X),from(s(X)))
  a__length(nil()) -> |0|()
  a__length(cons(X,Y)) -> s(a__length1(Y))
  a__length1(X) -> a__length(X)
  mark(from(X)) -> a__from(mark(X))
  mark(length(X)) -> a__length(X)
  mark(length1(X)) -> a__length1(X)
  mark(cons(X1,X2)) -> cons(mark(X1),X2)
  mark(s(X)) -> s(mark(X))
  mark(nil()) -> nil()
  mark(|0|()) -> |0|()
  a__from(X) -> from(X)
  a__length(X) -> length(X)
  a__length1(X) -> length1(X)

-- SCC decomposition.

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

p1: a__from#(X) -> mark#(X)
p2: a__length#(cons(X,Y)) -> a__length1#(Y)
p3: a__length1#(X) -> a__length#(X)
p4: mark#(from(X)) -> a__from#(mark(X))
p5: mark#(from(X)) -> mark#(X)
p6: mark#(length(X)) -> a__length#(X)
p7: mark#(length1(X)) -> a__length1#(X)
p8: mark#(cons(X1,X2)) -> mark#(X1)
p9: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__from(X) -> cons(mark(X),from(s(X)))
r2: a__length(nil()) -> |0|()
r3: a__length(cons(X,Y)) -> s(a__length1(Y))
r4: a__length1(X) -> a__length(X)
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(length(X)) -> a__length(X)
r7: mark(length1(X)) -> a__length1(X)
r8: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r9: mark(s(X)) -> s(mark(X))
r10: mark(nil()) -> nil()
r11: mark(|0|()) -> |0|()
r12: a__from(X) -> from(X)
r13: a__length(X) -> length(X)
r14: a__length1(X) -> length1(X)

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: a__from#(X) -> mark#(X)
p2: mark#(s(X)) -> mark#(X)
p3: mark#(cons(X1,X2)) -> mark#(X1)
p4: mark#(from(X)) -> mark#(X)
p5: mark#(from(X)) -> a__from#(mark(X))

and R consists of:

r1: a__from(X) -> cons(mark(X),from(s(X)))
r2: a__length(nil()) -> |0|()
r3: a__length(cons(X,Y)) -> s(a__length1(Y))
r4: a__length1(X) -> a__length(X)
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(length(X)) -> a__length(X)
r7: mark(length1(X)) -> a__length1(X)
r8: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r9: mark(s(X)) -> s(mark(X))
r10: mark(nil()) -> nil()
r11: mark(|0|()) -> |0|()
r12: a__from(X) -> from(X)
r13: a__length(X) -> length(X)
r14: a__length1(X) -> length1(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a__from#_A(x1) = ((1,0),(0,0)) x1 + (7,1)
            mark#_A(x1) = ((1,0),(0,0)) x1 + (6,9)
            s_A(x1) = ((1,0),(0,0)) x1 + (0,1)
            cons_A(x1,x2) = ((1,0),(0,0)) x1 + ((0,0),(1,0)) x2 + (0,4)
            from_A(x1) = ((1,0),(1,1)) x1 + (5,8)
            mark_A(x1) = ((1,0),(1,1)) x1 + (2,5)
            a__from_A(x1) = ((1,0),(1,1)) x1 + (5,10)
            a__length_A(x1) = (2,2)
            nil_A() = (2,2)
            |0|_A() = (1,1)
            a__length1_A(x1) = (2,3)
            length_A(x1) = (1,1)
            length1_A(x1) = (1,2)
    
      precedence:
      
        nil = a__length1 > mark = a__from = a__length > length > |0| > length1 > from > a__from# = mark# = s = cons
      
      partial status:
    
        pi(a__from#) = []
        pi(mark#) = []
        pi(s) = []
        pi(cons) = []
        pi(from) = [1]
        pi(mark) = [1]
        pi(a__from) = [1]
        pi(a__length) = []
        pi(nil) = []
        pi(|0|) = []
        pi(a__length1) = []
        pi(length) = []
        pi(length1) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a__from#_A(x1) = (0,1)
            mark#_A(x1) = (0,1)
            s_A(x1) = (1,0)
            cons_A(x1,x2) = (0,0)
            from_A(x1) = x1 + (0,2)
            mark_A(x1) = ((1,0),(1,1)) x1
            a__from_A(x1) = (2,1)
            a__length_A(x1) = (2,2)
            nil_A() = (1,1)
            |0|_A() = (0,3)
            a__length1_A(x1) = (3,0)
            length_A(x1) = (2,1)
            length1_A(x1) = (1,0)
    
      precedence:
      
        |0| > a__length > nil > s = mark = a__from > from > a__from# = mark# > cons = a__length1 = length = length1
      
      partial status:
    
        pi(a__from#) = []
        pi(mark#) = []
        pi(s) = []
        pi(cons) = []
        pi(from) = []
        pi(mark) = []
        pi(a__from) = []
        pi(a__length) = []
        pi(nil) = []
        pi(|0|) = []
        pi(a__length1) = []
        pi(length) = []
        pi(length1) = []
    

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__from#(X) -> mark#(X)
p2: mark#(s(X)) -> mark#(X)
p3: mark#(cons(X1,X2)) -> mark#(X1)
p4: mark#(from(X)) -> a__from#(mark(X))

and R consists of:

r1: a__from(X) -> cons(mark(X),from(s(X)))
r2: a__length(nil()) -> |0|()
r3: a__length(cons(X,Y)) -> s(a__length1(Y))
r4: a__length1(X) -> a__length(X)
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(length(X)) -> a__length(X)
r7: mark(length1(X)) -> a__length1(X)
r8: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r9: mark(s(X)) -> s(mark(X))
r10: mark(nil()) -> nil()
r11: mark(|0|()) -> |0|()
r12: a__from(X) -> from(X)
r13: a__length(X) -> length(X)
r14: a__length1(X) -> length1(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: a__from#(X) -> mark#(X)
p2: mark#(from(X)) -> a__from#(mark(X))
p3: mark#(cons(X1,X2)) -> mark#(X1)
p4: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__from(X) -> cons(mark(X),from(s(X)))
r2: a__length(nil()) -> |0|()
r3: a__length(cons(X,Y)) -> s(a__length1(Y))
r4: a__length1(X) -> a__length(X)
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(length(X)) -> a__length(X)
r7: mark(length1(X)) -> a__length1(X)
r8: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r9: mark(s(X)) -> s(mark(X))
r10: mark(nil()) -> nil()
r11: mark(|0|()) -> |0|()
r12: a__from(X) -> from(X)
r13: a__length(X) -> length(X)
r14: a__length1(X) -> length1(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a__from#_A(x1) = ((1,0),(0,0)) x1 + (4,2)
            mark#_A(x1) = ((1,0),(0,0)) x1 + (0,3)
            from_A(x1) = ((1,0),(0,0)) x1 + (4,6)
            mark_A(x1) = x1 + (0,2)
            cons_A(x1,x2) = ((1,0),(0,0)) x1 + (3,6)
            s_A(x1) = ((1,0),(0,0)) x1 + (0,3)
            a__from_A(x1) = ((1,0),(0,0)) x1 + (4,7)
            a__length_A(x1) = (2,4)
            nil_A() = (1,2)
            |0|_A() = (1,1)
            a__length1_A(x1) = ((0,0),(1,0)) x1 + (2,5)
            length_A(x1) = (2,3)
            length1_A(x1) = ((0,0),(1,0)) x1 + (2,4)
    
      precedence:
      
        a__from# > mark# > mark = nil = a__length1 > a__from > from = cons = length1 > a__length > length > s = |0|
      
      partial status:
    
        pi(a__from#) = []
        pi(mark#) = []
        pi(from) = []
        pi(mark) = [1]
        pi(cons) = []
        pi(s) = []
        pi(a__from) = []
        pi(a__length) = []
        pi(nil) = []
        pi(|0|) = []
        pi(a__length1) = []
        pi(length) = []
        pi(length1) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a__from#_A(x1) = (1,1)
            mark#_A(x1) = (2,2)
            from_A(x1) = (6,2)
            mark_A(x1) = ((1,0),(0,0)) x1 + (8,0)
            cons_A(x1,x2) = (6,0)
            s_A(x1) = (3,4)
            a__from_A(x1) = (7,3)
            a__length_A(x1) = (4,5)
            nil_A() = (6,7)
            |0|_A() = (5,6)
            a__length1_A(x1) = (5,6)
            length_A(x1) = (1,1)
            length1_A(x1) = (4,7)
    
      precedence:
      
        |0| > mark = a__from = a__length > nil > length > from > cons = a__length1 = length1 > mark# = s > a__from#
      
      partial status:
    
        pi(a__from#) = []
        pi(mark#) = []
        pi(from) = []
        pi(mark) = []
        pi(cons) = []
        pi(s) = []
        pi(a__from) = []
        pi(a__length) = []
        pi(nil) = []
        pi(|0|) = []
        pi(a__length1) = []
        pi(length) = []
        pi(length1) = []
    

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#(from(X)) -> a__from#(mark(X))
p2: mark#(cons(X1,X2)) -> mark#(X1)
p3: mark#(s(X)) -> mark#(X)

and R consists of:

r1: a__from(X) -> cons(mark(X),from(s(X)))
r2: a__length(nil()) -> |0|()
r3: a__length(cons(X,Y)) -> s(a__length1(Y))
r4: a__length1(X) -> a__length(X)
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(length(X)) -> a__length(X)
r7: mark(length1(X)) -> a__length1(X)
r8: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r9: mark(s(X)) -> s(mark(X))
r10: mark(nil()) -> nil()
r11: mark(|0|()) -> |0|()
r12: a__from(X) -> from(X)
r13: a__length(X) -> length(X)
r14: a__length1(X) -> length1(X)

The estimated dependency graph contains the following SCCs:

  {p2, p3}


-- Reduction pair.

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

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

and R consists of:

r1: a__from(X) -> cons(mark(X),from(s(X)))
r2: a__length(nil()) -> |0|()
r3: a__length(cons(X,Y)) -> s(a__length1(Y))
r4: a__length1(X) -> a__length(X)
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(length(X)) -> a__length(X)
r7: mark(length1(X)) -> a__length1(X)
r8: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r9: mark(s(X)) -> s(mark(X))
r10: mark(nil()) -> nil()
r11: mark(|0|()) -> |0|()
r12: a__from(X) -> from(X)
r13: a__length(X) -> length(X)
r14: a__length1(X) -> length1(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
            cons_A(x1,x2) = ((1,0),(1,1)) x1 + ((1,0),(1,1)) x2 + (1,1)
            s_A(x1) = ((1,0),(1,1)) x1 + (1,1)
    
      precedence:
      
        mark# = cons = s
      
      partial status:
    
        pi(mark#) = [1]
        pi(cons) = [1, 2]
        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
            cons_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (1,1)
            s_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        s > mark# = cons
      
      partial status:
    
        pi(mark#) = []
        pi(cons) = [2]
        pi(s) = []
    

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: a__length#(cons(X,Y)) -> a__length1#(Y)
p2: a__length1#(X) -> a__length#(X)

and R consists of:

r1: a__from(X) -> cons(mark(X),from(s(X)))
r2: a__length(nil()) -> |0|()
r3: a__length(cons(X,Y)) -> s(a__length1(Y))
r4: a__length1(X) -> a__length(X)
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(length(X)) -> a__length(X)
r7: mark(length1(X)) -> a__length1(X)
r8: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r9: mark(s(X)) -> s(mark(X))
r10: mark(nil()) -> nil()
r11: mark(|0|()) -> |0|()
r12: a__from(X) -> from(X)
r13: a__length(X) -> length(X)
r14: a__length1(X) -> length1(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:
            a__length#_A(x1) = x1
            cons_A(x1,x2) = x1 + x2 + (2,2)
            a__length1#_A(x1) = ((1,0),(0,0)) x1 + (1,1)
    
      precedence:
      
        cons = a__length1# > a__length#
      
      partial status:
    
        pi(a__length#) = [1]
        pi(cons) = [1, 2]
        pi(a__length1#) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: lexicographic order
          interpretations:
            a__length#_A(x1) = ((1,0),(1,1)) x1 + (1,1)
            cons_A(x1,x2) = x1 + ((0,0),(1,0)) x2 + (2,3)
            a__length1#_A(x1) = (2,2)
    
      precedence:
      
        a__length# = cons = a__length1#
      
      partial status:
    
        pi(a__length#) = [1]
        pi(cons) = []
        pi(a__length1#) = []
    

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__length#(cons(X,Y)) -> a__length1#(Y)

and R consists of:

r1: a__from(X) -> cons(mark(X),from(s(X)))
r2: a__length(nil()) -> |0|()
r3: a__length(cons(X,Y)) -> s(a__length1(Y))
r4: a__length1(X) -> a__length(X)
r5: mark(from(X)) -> a__from(mark(X))
r6: mark(length(X)) -> a__length(X)
r7: mark(length1(X)) -> a__length1(X)
r8: mark(cons(X1,X2)) -> cons(mark(X1),X2)
r9: mark(s(X)) -> s(mark(X))
r10: mark(nil()) -> nil()
r11: mark(|0|()) -> |0|()
r12: a__from(X) -> from(X)
r13: a__length(X) -> length(X)
r14: a__length1(X) -> length1(X)

The estimated dependency graph contains the following SCCs:

  (no SCCs)