YES

We show the termination of the TRS R:

  from(X) -> cons(X,n__from(n__s(X)))
  head(cons(X,XS)) -> X
  |2nd|(cons(X,XS)) -> head(activate(XS))
  take(|0|(),XS) -> nil()
  take(s(N),cons(X,XS)) -> cons(X,n__take(N,activate(XS)))
  sel(|0|(),cons(X,XS)) -> X
  sel(s(N),cons(X,XS)) -> sel(N,activate(XS))
  from(X) -> n__from(X)
  s(X) -> n__s(X)
  take(X1,X2) -> n__take(X1,X2)
  activate(n__from(X)) -> from(activate(X))
  activate(n__s(X)) -> s(activate(X))
  activate(n__take(X1,X2)) -> take(activate(X1),activate(X2))
  activate(X) -> X

-- SCC decomposition.

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

p1: |2nd|#(cons(X,XS)) -> head#(activate(XS))
p2: |2nd|#(cons(X,XS)) -> activate#(XS)
p3: take#(s(N),cons(X,XS)) -> activate#(XS)
p4: sel#(s(N),cons(X,XS)) -> sel#(N,activate(XS))
p5: sel#(s(N),cons(X,XS)) -> activate#(XS)
p6: activate#(n__from(X)) -> from#(activate(X))
p7: activate#(n__from(X)) -> activate#(X)
p8: activate#(n__s(X)) -> s#(activate(X))
p9: activate#(n__s(X)) -> activate#(X)
p10: activate#(n__take(X1,X2)) -> take#(activate(X1),activate(X2))
p11: activate#(n__take(X1,X2)) -> activate#(X1)
p12: activate#(n__take(X1,X2)) -> activate#(X2)

and R consists of:

r1: from(X) -> cons(X,n__from(n__s(X)))
r2: head(cons(X,XS)) -> X
r3: |2nd|(cons(X,XS)) -> head(activate(XS))
r4: take(|0|(),XS) -> nil()
r5: take(s(N),cons(X,XS)) -> cons(X,n__take(N,activate(XS)))
r6: sel(|0|(),cons(X,XS)) -> X
r7: sel(s(N),cons(X,XS)) -> sel(N,activate(XS))
r8: from(X) -> n__from(X)
r9: s(X) -> n__s(X)
r10: take(X1,X2) -> n__take(X1,X2)
r11: activate(n__from(X)) -> from(activate(X))
r12: activate(n__s(X)) -> s(activate(X))
r13: activate(n__take(X1,X2)) -> take(activate(X1),activate(X2))
r14: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p4}
  {p3, p7, p9, p10, p11, p12}


-- Reduction pair.

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

p1: sel#(s(N),cons(X,XS)) -> sel#(N,activate(XS))

and R consists of:

r1: from(X) -> cons(X,n__from(n__s(X)))
r2: head(cons(X,XS)) -> X
r3: |2nd|(cons(X,XS)) -> head(activate(XS))
r4: take(|0|(),XS) -> nil()
r5: take(s(N),cons(X,XS)) -> cons(X,n__take(N,activate(XS)))
r6: sel(|0|(),cons(X,XS)) -> X
r7: sel(s(N),cons(X,XS)) -> sel(N,activate(XS))
r8: from(X) -> n__from(X)
r9: s(X) -> n__s(X)
r10: take(X1,X2) -> n__take(X1,X2)
r11: activate(n__from(X)) -> from(activate(X))
r12: activate(n__s(X)) -> s(activate(X))
r13: activate(n__take(X1,X2)) -> take(activate(X1),activate(X2))
r14: activate(X) -> X

The set of usable rules consists of

  r1, r4, r5, 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:
          sel#_A(x1,x2) = x1 + (0,1)
          s_A(x1) = x1 + (0,2)
          cons_A(x1,x2) = (1,0)
          activate_A(x1) = x1 + (7,5)
          from_A(x1) = ((0,0),(0,1)) x1 + (7,1)
          n__from_A(x1) = ((0,0),(0,1)) x1 + (1,1)
          n__s_A(x1) = x1 + (0,2)
          take_A(x1,x2) = ((0,1),(0,0)) x1 + (3,4)
          |0|_A() = (2,1)
          nil_A() = (1,1)
          n__take_A(x1,x2) = ((0,1),(0,0)) x1 + (2,3)
  
    precedence:
    
      cons > take > activate = from > s = n__s > sel# = n__take > n__from > |0| > nil
    
    partial status:
  
      pi(sel#) = [1]
      pi(s) = [1]
      pi(cons) = []
      pi(activate) = [1]
      pi(from) = []
      pi(n__from) = []
      pi(n__s) = [1]
      pi(take) = []
      pi(|0|) = []
      pi(nil) = []
      pi(n__take) = []

The next rules are strictly ordered:

  p1

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: take#(s(N),cons(X,XS)) -> activate#(XS)
p2: activate#(n__take(X1,X2)) -> activate#(X2)
p3: activate#(n__take(X1,X2)) -> activate#(X1)
p4: activate#(n__take(X1,X2)) -> take#(activate(X1),activate(X2))
p5: activate#(n__s(X)) -> activate#(X)
p6: activate#(n__from(X)) -> activate#(X)

and R consists of:

r1: from(X) -> cons(X,n__from(n__s(X)))
r2: head(cons(X,XS)) -> X
r3: |2nd|(cons(X,XS)) -> head(activate(XS))
r4: take(|0|(),XS) -> nil()
r5: take(s(N),cons(X,XS)) -> cons(X,n__take(N,activate(XS)))
r6: sel(|0|(),cons(X,XS)) -> X
r7: sel(s(N),cons(X,XS)) -> sel(N,activate(XS))
r8: from(X) -> n__from(X)
r9: s(X) -> n__s(X)
r10: take(X1,X2) -> n__take(X1,X2)
r11: activate(n__from(X)) -> from(activate(X))
r12: activate(n__s(X)) -> s(activate(X))
r13: activate(n__take(X1,X2)) -> take(activate(X1),activate(X2))
r14: activate(X) -> X

The set of usable rules consists of

  r1, r4, r5, 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:
          take#_A(x1,x2) = ((1,0),(0,0)) x2 + (2,3)
          s_A(x1) = x1 + (0,1)
          cons_A(x1,x2) = ((1,0),(0,0)) x2 + (0,2)
          activate#_A(x1) = ((1,0),(0,0)) x1 + (1,3)
          n__take_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (2,3)
          activate_A(x1) = x1 + (0,4)
          n__s_A(x1) = x1 + (0,1)
          n__from_A(x1) = ((1,0),(0,0)) x1 + (2,3)
          from_A(x1) = ((1,0),(0,0)) x1 + (2,4)
          take_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (2,3)
          |0|_A() = (2,0)
          nil_A() = (1,0)
  
    precedence:
    
      take# = cons = activate# = n__take = activate = n__from = from = take = |0| = nil > s > n__s
    
    partial status:
  
      pi(take#) = []
      pi(s) = []
      pi(cons) = []
      pi(activate#) = []
      pi(n__take) = []
      pi(activate) = []
      pi(n__s) = []
      pi(n__from) = []
      pi(from) = []
      pi(take) = []
      pi(|0|) = []
      pi(nil) = []

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: take#(s(N),cons(X,XS)) -> activate#(XS)
p2: activate#(n__take(X1,X2)) -> activate#(X1)
p3: activate#(n__take(X1,X2)) -> take#(activate(X1),activate(X2))
p4: activate#(n__s(X)) -> activate#(X)
p5: activate#(n__from(X)) -> activate#(X)

and R consists of:

r1: from(X) -> cons(X,n__from(n__s(X)))
r2: head(cons(X,XS)) -> X
r3: |2nd|(cons(X,XS)) -> head(activate(XS))
r4: take(|0|(),XS) -> nil()
r5: take(s(N),cons(X,XS)) -> cons(X,n__take(N,activate(XS)))
r6: sel(|0|(),cons(X,XS)) -> X
r7: sel(s(N),cons(X,XS)) -> sel(N,activate(XS))
r8: from(X) -> n__from(X)
r9: s(X) -> n__s(X)
r10: take(X1,X2) -> n__take(X1,X2)
r11: activate(n__from(X)) -> from(activate(X))
r12: activate(n__s(X)) -> s(activate(X))
r13: activate(n__take(X1,X2)) -> take(activate(X1),activate(X2))
r14: activate(X) -> 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: take#(s(N),cons(X,XS)) -> activate#(XS)
p2: activate#(n__from(X)) -> activate#(X)
p3: activate#(n__s(X)) -> activate#(X)
p4: activate#(n__take(X1,X2)) -> take#(activate(X1),activate(X2))
p5: activate#(n__take(X1,X2)) -> activate#(X1)

and R consists of:

r1: from(X) -> cons(X,n__from(n__s(X)))
r2: head(cons(X,XS)) -> X
r3: |2nd|(cons(X,XS)) -> head(activate(XS))
r4: take(|0|(),XS) -> nil()
r5: take(s(N),cons(X,XS)) -> cons(X,n__take(N,activate(XS)))
r6: sel(|0|(),cons(X,XS)) -> X
r7: sel(s(N),cons(X,XS)) -> sel(N,activate(XS))
r8: from(X) -> n__from(X)
r9: s(X) -> n__s(X)
r10: take(X1,X2) -> n__take(X1,X2)
r11: activate(n__from(X)) -> from(activate(X))
r12: activate(n__s(X)) -> s(activate(X))
r13: activate(n__take(X1,X2)) -> take(activate(X1),activate(X2))
r14: activate(X) -> X

The set of usable rules consists of

  r1, r4, r5, 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:
          take#_A(x1,x2) = ((0,1),(0,0)) x1 + ((0,1),(0,0)) x2 + (5,2)
          s_A(x1) = ((0,1),(0,1)) x1 + (2,0)
          cons_A(x1,x2) = ((0,0),(0,1)) x2 + (3,0)
          activate#_A(x1) = ((0,1),(0,0)) x1 + (4,2)
          n__from_A(x1) = ((1,1),(0,1)) x1 + (5,2)
          n__s_A(x1) = ((0,0),(0,1)) x1 + (2,0)
          n__take_A(x1,x2) = x1 + ((0,1),(0,1)) x2 + (1,2)
          activate_A(x1) = ((1,1),(0,1)) x1
          from_A(x1) = ((1,1),(0,1)) x1 + (6,2)
          take_A(x1,x2) = x1 + ((0,1),(0,1)) x2 + (2,2)
          |0|_A() = (1,1)
          nil_A() = (2,2)
  
    precedence:
    
      s = activate = take > |0| > nil > n__from = from > cons > take# = activate# = n__s = n__take
    
    partial status:
  
      pi(take#) = []
      pi(s) = []
      pi(cons) = []
      pi(activate#) = []
      pi(n__from) = []
      pi(n__s) = []
      pi(n__take) = [1]
      pi(activate) = [1]
      pi(from) = []
      pi(take) = []
      pi(|0|) = []
      pi(nil) = []

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: take#(s(N),cons(X,XS)) -> activate#(XS)
p2: activate#(n__s(X)) -> activate#(X)
p3: activate#(n__take(X1,X2)) -> take#(activate(X1),activate(X2))
p4: activate#(n__take(X1,X2)) -> activate#(X1)

and R consists of:

r1: from(X) -> cons(X,n__from(n__s(X)))
r2: head(cons(X,XS)) -> X
r3: |2nd|(cons(X,XS)) -> head(activate(XS))
r4: take(|0|(),XS) -> nil()
r5: take(s(N),cons(X,XS)) -> cons(X,n__take(N,activate(XS)))
r6: sel(|0|(),cons(X,XS)) -> X
r7: sel(s(N),cons(X,XS)) -> sel(N,activate(XS))
r8: from(X) -> n__from(X)
r9: s(X) -> n__s(X)
r10: take(X1,X2) -> n__take(X1,X2)
r11: activate(n__from(X)) -> from(activate(X))
r12: activate(n__s(X)) -> s(activate(X))
r13: activate(n__take(X1,X2)) -> take(activate(X1),activate(X2))
r14: activate(X) -> 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: take#(s(N),cons(X,XS)) -> activate#(XS)
p2: activate#(n__take(X1,X2)) -> activate#(X1)
p3: activate#(n__take(X1,X2)) -> take#(activate(X1),activate(X2))
p4: activate#(n__s(X)) -> activate#(X)

and R consists of:

r1: from(X) -> cons(X,n__from(n__s(X)))
r2: head(cons(X,XS)) -> X
r3: |2nd|(cons(X,XS)) -> head(activate(XS))
r4: take(|0|(),XS) -> nil()
r5: take(s(N),cons(X,XS)) -> cons(X,n__take(N,activate(XS)))
r6: sel(|0|(),cons(X,XS)) -> X
r7: sel(s(N),cons(X,XS)) -> sel(N,activate(XS))
r8: from(X) -> n__from(X)
r9: s(X) -> n__s(X)
r10: take(X1,X2) -> n__take(X1,X2)
r11: activate(n__from(X)) -> from(activate(X))
r12: activate(n__s(X)) -> s(activate(X))
r13: activate(n__take(X1,X2)) -> take(activate(X1),activate(X2))
r14: activate(X) -> X

The set of usable rules consists of

  r1, r4, r5, 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:
          take#_A(x1,x2) = ((1,0),(0,0)) x1 + ((1,0),(0,0)) x2 + (3,3)
          s_A(x1) = ((1,0),(1,1)) x1 + (1,3)
          cons_A(x1,x2) = ((1,0),(1,0)) x2
          activate#_A(x1) = ((1,0),(0,0)) x1 + (3,3)
          n__take_A(x1,x2) = x1 + x2 + (4,4)
          activate_A(x1) = ((1,0),(1,1)) x1
          n__s_A(x1) = ((1,0),(1,1)) x1 + (1,3)
          from_A(x1) = ((0,0),(1,0)) x1 + (2,2)
          n__from_A(x1) = ((0,0),(1,0)) x1 + (2,0)
          take_A(x1,x2) = x1 + x2 + (4,5)
          |0|_A() = (1,0)
          nil_A() = (2,1)
  
    precedence:
    
      |0| > cons = n__take = activate = from = take = nil > take# = activate# > s = n__s > n__from
    
    partial status:
  
      pi(take#) = []
      pi(s) = [1]
      pi(cons) = []
      pi(activate#) = []
      pi(n__take) = []
      pi(activate) = [1]
      pi(n__s) = [1]
      pi(from) = []
      pi(n__from) = []
      pi(take) = []
      pi(|0|) = []
      pi(nil) = []

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: take#(s(N),cons(X,XS)) -> activate#(XS)
p2: activate#(n__take(X1,X2)) -> activate#(X1)
p3: activate#(n__take(X1,X2)) -> take#(activate(X1),activate(X2))

and R consists of:

r1: from(X) -> cons(X,n__from(n__s(X)))
r2: head(cons(X,XS)) -> X
r3: |2nd|(cons(X,XS)) -> head(activate(XS))
r4: take(|0|(),XS) -> nil()
r5: take(s(N),cons(X,XS)) -> cons(X,n__take(N,activate(XS)))
r6: sel(|0|(),cons(X,XS)) -> X
r7: sel(s(N),cons(X,XS)) -> sel(N,activate(XS))
r8: from(X) -> n__from(X)
r9: s(X) -> n__s(X)
r10: take(X1,X2) -> n__take(X1,X2)
r11: activate(n__from(X)) -> from(activate(X))
r12: activate(n__s(X)) -> s(activate(X))
r13: activate(n__take(X1,X2)) -> take(activate(X1),activate(X2))
r14: activate(X) -> 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: take#(s(N),cons(X,XS)) -> activate#(XS)
p2: activate#(n__take(X1,X2)) -> take#(activate(X1),activate(X2))
p3: activate#(n__take(X1,X2)) -> activate#(X1)

and R consists of:

r1: from(X) -> cons(X,n__from(n__s(X)))
r2: head(cons(X,XS)) -> X
r3: |2nd|(cons(X,XS)) -> head(activate(XS))
r4: take(|0|(),XS) -> nil()
r5: take(s(N),cons(X,XS)) -> cons(X,n__take(N,activate(XS)))
r6: sel(|0|(),cons(X,XS)) -> X
r7: sel(s(N),cons(X,XS)) -> sel(N,activate(XS))
r8: from(X) -> n__from(X)
r9: s(X) -> n__s(X)
r10: take(X1,X2) -> n__take(X1,X2)
r11: activate(n__from(X)) -> from(activate(X))
r12: activate(n__s(X)) -> s(activate(X))
r13: activate(n__take(X1,X2)) -> take(activate(X1),activate(X2))
r14: activate(X) -> X

The set of usable rules consists of

  r1, r4, r5, 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:
          take#_A(x1,x2) = ((0,1),(0,1)) x2 + (3,5)
          s_A(x1) = x1 + (6,2)
          cons_A(x1,x2) = ((0,0),(0,1)) x2 + (1,0)
          activate#_A(x1) = ((0,1),(0,1)) x1 + (2,5)
          n__take_A(x1,x2) = ((0,0),(0,1)) x1 + ((0,1),(0,1)) x2 + (4,4)
          activate_A(x1) = ((1,1),(0,1)) x1 + (5,0)
          from_A(x1) = (7,1)
          n__from_A(x1) = (7,1)
          n__s_A(x1) = x1 + (6,2)
          take_A(x1,x2) = ((0,0),(0,1)) x1 + ((0,1),(0,1)) x2 + (4,4)
          |0|_A() = (1,1)
          nil_A() = (0,0)
  
    precedence:
    
      take# = activate# > n__take = activate = take = |0| > s = cons = from = n__from = n__s = nil
    
    partial status:
  
      pi(take#) = []
      pi(s) = []
      pi(cons) = []
      pi(activate#) = []
      pi(n__take) = []
      pi(activate) = [1]
      pi(from) = []
      pi(n__from) = []
      pi(n__s) = []
      pi(take) = []
      pi(|0|) = []
      pi(nil) = []

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: take#(s(N),cons(X,XS)) -> activate#(XS)
p2: activate#(n__take(X1,X2)) -> take#(activate(X1),activate(X2))

and R consists of:

r1: from(X) -> cons(X,n__from(n__s(X)))
r2: head(cons(X,XS)) -> X
r3: |2nd|(cons(X,XS)) -> head(activate(XS))
r4: take(|0|(),XS) -> nil()
r5: take(s(N),cons(X,XS)) -> cons(X,n__take(N,activate(XS)))
r6: sel(|0|(),cons(X,XS)) -> X
r7: sel(s(N),cons(X,XS)) -> sel(N,activate(XS))
r8: from(X) -> n__from(X)
r9: s(X) -> n__s(X)
r10: take(X1,X2) -> n__take(X1,X2)
r11: activate(n__from(X)) -> from(activate(X))
r12: activate(n__s(X)) -> s(activate(X))
r13: activate(n__take(X1,X2)) -> take(activate(X1),activate(X2))
r14: activate(X) -> 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: take#(s(N),cons(X,XS)) -> activate#(XS)
p2: activate#(n__take(X1,X2)) -> take#(activate(X1),activate(X2))

and R consists of:

r1: from(X) -> cons(X,n__from(n__s(X)))
r2: head(cons(X,XS)) -> X
r3: |2nd|(cons(X,XS)) -> head(activate(XS))
r4: take(|0|(),XS) -> nil()
r5: take(s(N),cons(X,XS)) -> cons(X,n__take(N,activate(XS)))
r6: sel(|0|(),cons(X,XS)) -> X
r7: sel(s(N),cons(X,XS)) -> sel(N,activate(XS))
r8: from(X) -> n__from(X)
r9: s(X) -> n__s(X)
r10: take(X1,X2) -> n__take(X1,X2)
r11: activate(n__from(X)) -> from(activate(X))
r12: activate(n__s(X)) -> s(activate(X))
r13: activate(n__take(X1,X2)) -> take(activate(X1),activate(X2))
r14: activate(X) -> X

The set of usable rules consists of

  r1, r4, r5, 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:
          take#_A(x1,x2) = ((1,0),(0,0)) x2 + (1,2)
          s_A(x1) = ((1,0),(1,1)) x1 + (1,3)
          cons_A(x1,x2) = ((1,0),(0,0)) x2 + (0,2)
          activate#_A(x1) = ((1,0),(0,0)) x1 + (0,2)
          n__take_A(x1,x2) = ((1,0),(1,0)) x2 + (2,1)
          activate_A(x1) = ((1,0),(1,1)) x1 + (0,3)
          from_A(x1) = (2,3)
          n__from_A(x1) = (2,3)
          n__s_A(x1) = ((1,0),(1,1)) x1 + (1,3)
          take_A(x1,x2) = ((1,0),(1,0)) x2 + (2,3)
          |0|_A() = (1,0)
          nil_A() = (0,0)
  
    precedence:
    
      s = cons = activate# = activate = from = n__from = take > take# = n__take = n__s = |0| = nil
    
    partial status:
  
      pi(take#) = []
      pi(s) = []
      pi(cons) = []
      pi(activate#) = []
      pi(n__take) = []
      pi(activate) = [1]
      pi(from) = []
      pi(n__from) = []
      pi(n__s) = [1]
      pi(take) = []
      pi(|0|) = []
      pi(nil) = []

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: take#(s(N),cons(X,XS)) -> activate#(XS)

and R consists of:

r1: from(X) -> cons(X,n__from(n__s(X)))
r2: head(cons(X,XS)) -> X
r3: |2nd|(cons(X,XS)) -> head(activate(XS))
r4: take(|0|(),XS) -> nil()
r5: take(s(N),cons(X,XS)) -> cons(X,n__take(N,activate(XS)))
r6: sel(|0|(),cons(X,XS)) -> X
r7: sel(s(N),cons(X,XS)) -> sel(N,activate(XS))
r8: from(X) -> n__from(X)
r9: s(X) -> n__s(X)
r10: take(X1,X2) -> n__take(X1,X2)
r11: activate(n__from(X)) -> from(activate(X))
r12: activate(n__s(X)) -> s(activate(X))
r13: activate(n__take(X1,X2)) -> take(activate(X1),activate(X2))
r14: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  (no SCCs)