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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          a__from#_A(x1) = ((1,1),(1,1)) x1 + (4,9)
          mark#_A(x1) = ((0,1),(0,1)) x1 + (3,9)
          s_A(x1) = ((0,0),(0,1)) x1 + (1,0)
          cons_A(x1,x2) = ((0,0),(0,1)) x1 + (2,1)
          from_A(x1) = ((0,0),(0,1)) x1 + (1,8)
          mark_A(x1) = ((0,0),(0,1)) x1 + (4,2)
          a__from_A(x1) = ((0,0),(0,1)) x1 + (3,8)
          a__length_A(x1) = (2,2)
          nil_A() = (2,1)
          |0|_A() = (1,1)
          a__length1_A(x1) = (3,2)
          length_A(x1) = (1,1)
          length1_A(x1) = (1,1)
  
    precedence:
    
      |0| > cons = mark = a__from = a__length = nil = a__length1 > length > a__from# = mark# = s = from > 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:

  p3

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#(from(X)) -> mark#(X)
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#(from(X)) -> mark#(X)
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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          a__from#_A(x1) = ((0,1),(0,0)) x1 + (2,4)
          mark#_A(x1) = ((0,1),(0,0)) x1 + (1,4)
          from_A(x1) = ((0,0),(0,1)) x1 + (0,3)
          mark_A(x1) = x1 + (0,1)
          s_A(x1) = ((0,0),(0,1)) x1 + (1,0)
          a__from_A(x1) = ((0,0),(0,1)) x1 + (0,3)
          cons_A(x1,x2) = ((1,0),(1,1)) x2
          a__length_A(x1) = ((0,0),(1,1)) x1 + (2,2)
          nil_A() = (2,1)
          |0|_A() = (1,2)
          a__length1_A(x1) = ((1,1),(1,1)) x1 + (3,2)
          length_A(x1) = ((0,0),(1,1)) x1 + (2,1)
          length1_A(x1) = ((1,1),(1,1)) x1 + (3,2)
  
    precedence:
    
      nil > a__from# = mark# = from = mark = s = a__from = cons = a__length = |0| = a__length1 = length = length1
    
    partial status:
  
      pi(a__from#) = []
      pi(mark#) = []
      pi(from) = []
      pi(mark) = [1]
      pi(s) = []
      pi(a__from) = []
      pi(cons) = []
      pi(a__length) = []
      pi(nil) = []
      pi(|0|) = []
      pi(a__length1) = [1]
      pi(length) = []
      pi(length1) = [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: a__from#(X) -> mark#(X)
p2: mark#(from(X)) -> a__from#(mark(X))
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:

  {p1, 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#(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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          a__from#_A(x1) = ((1,0),(0,0)) x1 + (3,4)
          mark#_A(x1) = ((1,0),(0,0)) x1 + (2,4)
          s_A(x1) = ((1,0),(0,0)) x1 + (1,3)
          from_A(x1) = ((1,0),(1,0)) x1 + (6,0)
          mark_A(x1) = x1 + (4,14)
          a__from_A(x1) = ((1,0),(1,0)) x1 + (6,10)
          cons_A(x1,x2) = ((0,1),(1,0)) x2 + (0,3)
          a__length_A(x1) = ((1,1),(0,1)) x1 + (2,0)
          nil_A() = (0,0)
          |0|_A() = (1,0)
          a__length1_A(x1) = ((1,1),(0,1)) x1 + (3,0)
          length_A(x1) = ((1,1),(0,1)) x1
          length1_A(x1) = ((1,1),(0,1)) x1
  
    precedence:
    
      nil > from = mark = a__from = a__length1 > cons > s = a__length = |0| = length > length1 > a__from# = mark#
    
    partial status:
  
      pi(a__from#) = []
      pi(mark#) = []
      pi(s) = []
      pi(from) = []
      pi(mark) = [1]
      pi(a__from) = []
      pi(cons) = []
      pi(a__length) = []
      pi(nil) = []
      pi(|0|) = []
      pi(a__length1) = []
      pi(length) = []
      pi(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__from#(X) -> mark#(X)
p2: 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}


-- 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))

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:

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

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)

-- 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:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          a__length#_A(x1) = ((0,1),(0,0)) x1 + (1,3)
          cons_A(x1,x2) = ((1,1),(1,1)) x1 + ((0,0),(0,1)) x2 + (1,2)
          a__length1#_A(x1) = ((0,1),(0,0)) x1 + (2,3)
  
    precedence:
    
      a__length# > cons = a__length1#
    
    partial status:
  
      pi(a__length#) = []
      pi(cons) = [1]
      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)