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:
  
      max/plus interpretations on natural numbers:
      
        sel#_A(x1,x2) = x1 + 27
        s_A(x1) = max{11, x1 + 7}
        cons_A(x1,x2) = max{12, x1 + 9}
        activate_A(x1) = x1 + 14
        from_A(x1) = x1 + 13
        n__from_A(x1) = x1 + 13
        n__s_A(x1) = x1 + 7
        take_A(x1,x2) = max{x1 + 6, x2 + 9}
        |0|_A = 3
        nil_A = 8
        n__take_A(x1,x2) = max{x1 + 6, x2 + 9}
  
    precedence:
    
      cons = activate > sel# = s = from = n__from = n__s = take = |0| = nil = n__take
    
    partial status:
  
      pi(sel#) = []
      pi(s) = [1]
      pi(cons) = [1]
      pi(activate) = []
      pi(from) = [1]
      pi(n__from) = [1]
      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:
  
      max/plus interpretations on natural numbers:
      
        take#_A(x1,x2) = max{27, x1 + 4, x2 + 22}
        s_A(x1) = max{25, x1 + 13}
        cons_A(x1,x2) = max{x1 + 4, x2 - 13}
        activate#_A(x1) = max{25, x1 + 3}
        n__take_A(x1,x2) = max{24, x1 + 1, x2 + 19}
        activate_A(x1) = max{5, x1}
        n__s_A(x1) = max{25, x1 + 13}
        n__from_A(x1) = max{18, x1 + 6}
        from_A(x1) = max{18, x1 + 6}
        take_A(x1,x2) = max{24, x1 + 1, x2 + 19}
        |0|_A = 26
        nil_A = 25
  
    precedence:
    
      take# = s = cons = activate# = n__take = activate = n__s = from = take > n__from = |0| > nil
    
    partial status:
  
      pi(take#) = [1, 2]
      pi(s) = [1]
      pi(cons) = []
      pi(activate#) = [1]
      pi(n__take) = []
      pi(activate) = [1]
      pi(n__s) = [1]
      pi(n__from) = []
      pi(from) = [1]
      pi(take) = []
      pi(|0|) = []
      pi(nil) = []

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: activate#(n__take(X1,X2)) -> activate#(X2)
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, p4, p5}


-- Reduction pair.

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

p1: activate#(n__take(X1,X2)) -> activate#(X2)
p2: activate#(n__from(X)) -> activate#(X)
p3: activate#(n__s(X)) -> activate#(X)
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 set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        activate#_A(x1) = x1 + 1
        n__take_A(x1,x2) = max{x1, x2}
        n__from_A(x1) = x1
        n__s_A(x1) = x1
  
    precedence:
    
      activate# = n__take = n__from = n__s
    
    partial status:
  
      pi(activate#) = [1]
      pi(n__take) = [1, 2]
      pi(n__from) = [1]
      pi(n__s) = [1]

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: activate#(n__from(X)) -> activate#(X)
p2: activate#(n__s(X)) -> activate#(X)
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 estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

p1: activate#(n__from(X)) -> activate#(X)
p2: activate#(n__take(X1,X2)) -> activate#(X1)
p3: 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

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        activate#_A(x1) = x1 + 1
        n__from_A(x1) = x1
        n__take_A(x1,x2) = max{x1, x2}
        n__s_A(x1) = x1
  
    precedence:
    
      activate# = n__from = n__take = n__s
    
    partial status:
  
      pi(activate#) = [1]
      pi(n__from) = [1]
      pi(n__take) = [1, 2]
      pi(n__s) = [1]

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: activate#(n__from(X)) -> activate#(X)
p2: 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 estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: activate#(n__from(X)) -> activate#(X)
p2: 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

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        activate#_A(x1) = max{3, x1 + 2}
        n__from_A(x1) = x1 + 3
        n__s_A(x1) = max{1, x1}
  
    precedence:
    
      activate# = n__from = n__s
    
    partial status:
  
      pi(activate#) = [1]
      pi(n__from) = [1]
      pi(n__s) = [1]

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: 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 estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: 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

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        activate#_A(x1) = x1 + 2
        n__s_A(x1) = x1 + 2
  
    precedence:
    
      activate# = n__s
    
    partial status:
  
      pi(activate#) = [1]
      pi(n__s) = [1]

The next rules are strictly ordered:

  p1

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