YES

We show the termination of the TRS R:

  first(|0|(),X) -> nil()
  first(s(X),cons(Y,Z)) -> cons(Y,n__first(X,activate(Z)))
  from(X) -> cons(X,n__from(n__s(X)))
  first(X1,X2) -> n__first(X1,X2)
  from(X) -> n__from(X)
  s(X) -> n__s(X)
  activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
  activate(n__from(X)) -> from(activate(X))
  activate(n__s(X)) -> s(activate(X))
  activate(X) -> X

-- SCC decomposition.

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

p1: first#(s(X),cons(Y,Z)) -> activate#(Z)
p2: activate#(n__first(X1,X2)) -> first#(activate(X1),activate(X2))
p3: activate#(n__first(X1,X2)) -> activate#(X1)
p4: activate#(n__first(X1,X2)) -> activate#(X2)
p5: activate#(n__from(X)) -> from#(activate(X))
p6: activate#(n__from(X)) -> activate#(X)
p7: activate#(n__s(X)) -> s#(activate(X))
p8: activate#(n__s(X)) -> activate#(X)

and R consists of:

r1: first(|0|(),X) -> nil()
r2: first(s(X),cons(Y,Z)) -> cons(Y,n__first(X,activate(Z)))
r3: from(X) -> cons(X,n__from(n__s(X)))
r4: first(X1,X2) -> n__first(X1,X2)
r5: from(X) -> n__from(X)
r6: s(X) -> n__s(X)
r7: activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
r8: activate(n__from(X)) -> from(activate(X))
r9: activate(n__s(X)) -> s(activate(X))
r10: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p6, p8}


-- Reduction pair.

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

p1: first#(s(X),cons(Y,Z)) -> activate#(Z)
p2: activate#(n__s(X)) -> activate#(X)
p3: activate#(n__from(X)) -> activate#(X)
p4: activate#(n__first(X1,X2)) -> activate#(X2)
p5: activate#(n__first(X1,X2)) -> activate#(X1)
p6: activate#(n__first(X1,X2)) -> first#(activate(X1),activate(X2))

and R consists of:

r1: first(|0|(),X) -> nil()
r2: first(s(X),cons(Y,Z)) -> cons(Y,n__first(X,activate(Z)))
r3: from(X) -> cons(X,n__from(n__s(X)))
r4: first(X1,X2) -> n__first(X1,X2)
r5: from(X) -> n__from(X)
r6: s(X) -> n__s(X)
r7: activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
r8: activate(n__from(X)) -> from(activate(X))
r9: activate(n__s(X)) -> s(activate(X))
r10: activate(X) -> X

The set of usable rules consists of

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

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          first#_A(x1,x2) = ((0,0),(1,0)) x1 + ((0,0),(1,0)) x2 + (1,7)
          s_A(x1) = ((1,0),(0,0)) x1 + (0,5)
          cons_A(x1,x2) = ((1,0),(1,0)) x2
          activate#_A(x1) = ((0,0),(1,0)) x1 + (1,4)
          n__s_A(x1) = ((1,0),(0,0)) x1 + (0,1)
          n__from_A(x1) = ((1,0),(1,1)) x1 + (3,10)
          n__first_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,0)) x2 + (4,6)
          activate_A(x1) = ((1,0),(1,1)) x1 + (0,9)
          first_A(x1,x2) = ((1,0),(1,0)) x1 + ((1,0),(1,0)) x2 + (4,10)
          |0|_A() = (1,1)
          nil_A() = (2,2)
          from_A(x1) = ((1,0),(1,1)) x1 + (3,12)
  
    precedence:
    
      first# = nil > n__from = activate = |0| = from > s = n__s = first > activate# = n__first > cons
    
    partial status:
  
      pi(first#) = []
      pi(s) = []
      pi(cons) = []
      pi(activate#) = []
      pi(n__s) = []
      pi(n__from) = [1]
      pi(n__first) = []
      pi(activate) = [1]
      pi(first) = []
      pi(|0|) = []
      pi(nil) = []
      pi(from) = [1]

The next rules are strictly ordered:

  p6

We remove them from the problem.

-- SCC decomposition.

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

p1: first#(s(X),cons(Y,Z)) -> activate#(Z)
p2: activate#(n__s(X)) -> activate#(X)
p3: activate#(n__from(X)) -> activate#(X)
p4: activate#(n__first(X1,X2)) -> activate#(X2)
p5: activate#(n__first(X1,X2)) -> activate#(X1)

and R consists of:

r1: first(|0|(),X) -> nil()
r2: first(s(X),cons(Y,Z)) -> cons(Y,n__first(X,activate(Z)))
r3: from(X) -> cons(X,n__from(n__s(X)))
r4: first(X1,X2) -> n__first(X1,X2)
r5: from(X) -> n__from(X)
r6: s(X) -> n__s(X)
r7: activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
r8: activate(n__from(X)) -> from(activate(X))
r9: activate(n__s(X)) -> s(activate(X))
r10: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p2, p3, p4, p5}


-- Reduction pair.

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

p1: activate#(n__s(X)) -> activate#(X)
p2: activate#(n__first(X1,X2)) -> activate#(X1)
p3: activate#(n__first(X1,X2)) -> activate#(X2)
p4: activate#(n__from(X)) -> activate#(X)

and R consists of:

r1: first(|0|(),X) -> nil()
r2: first(s(X),cons(Y,Z)) -> cons(Y,n__first(X,activate(Z)))
r3: from(X) -> cons(X,n__from(n__s(X)))
r4: first(X1,X2) -> n__first(X1,X2)
r5: from(X) -> n__from(X)
r6: s(X) -> n__s(X)
r7: activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
r8: activate(n__from(X)) -> from(activate(X))
r9: activate(n__s(X)) -> s(activate(X))
r10: activate(X) -> 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: lexicographic order
        interpretations:
          activate#_A(x1) = ((1,0),(0,0)) x1 + (2,2)
          n__s_A(x1) = ((1,0),(0,0)) x1 + (1,3)
          n__first_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (3,1)
          n__from_A(x1) = ((1,0),(0,0)) x1 + (3,3)
  
    precedence:
    
      activate# = n__s = n__first = n__from
    
    partial status:
  
      pi(activate#) = []
      pi(n__s) = []
      pi(n__first) = [2]
      pi(n__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: activate#(n__s(X)) -> activate#(X)
p2: activate#(n__first(X1,X2)) -> activate#(X2)
p3: activate#(n__from(X)) -> activate#(X)

and R consists of:

r1: first(|0|(),X) -> nil()
r2: first(s(X),cons(Y,Z)) -> cons(Y,n__first(X,activate(Z)))
r3: from(X) -> cons(X,n__from(n__s(X)))
r4: first(X1,X2) -> n__first(X1,X2)
r5: from(X) -> n__from(X)
r6: s(X) -> n__s(X)
r7: activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
r8: activate(n__from(X)) -> from(activate(X))
r9: activate(n__s(X)) -> s(activate(X))
r10: 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__s(X)) -> activate#(X)
p2: activate#(n__from(X)) -> activate#(X)
p3: activate#(n__first(X1,X2)) -> activate#(X2)

and R consists of:

r1: first(|0|(),X) -> nil()
r2: first(s(X),cons(Y,Z)) -> cons(Y,n__first(X,activate(Z)))
r3: from(X) -> cons(X,n__from(n__s(X)))
r4: first(X1,X2) -> n__first(X1,X2)
r5: from(X) -> n__from(X)
r6: s(X) -> n__s(X)
r7: activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
r8: activate(n__from(X)) -> from(activate(X))
r9: activate(n__s(X)) -> s(activate(X))
r10: activate(X) -> 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: lexicographic order
        interpretations:
          activate#_A(x1) = x1 + (1,2)
          n__s_A(x1) = ((1,0),(1,1)) x1 + (2,3)
          n__from_A(x1) = x1 + (2,1)
          n__first_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (2,3)
  
    precedence:
    
      n__s > n__from > activate# = n__first
    
    partial status:
  
      pi(activate#) = [1]
      pi(n__s) = []
      pi(n__from) = [1]
      pi(n__first) = [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__s(X)) -> activate#(X)
p2: activate#(n__first(X1,X2)) -> activate#(X2)

and R consists of:

r1: first(|0|(),X) -> nil()
r2: first(s(X),cons(Y,Z)) -> cons(Y,n__first(X,activate(Z)))
r3: from(X) -> cons(X,n__from(n__s(X)))
r4: first(X1,X2) -> n__first(X1,X2)
r5: from(X) -> n__from(X)
r6: s(X) -> n__s(X)
r7: activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
r8: activate(n__from(X)) -> from(activate(X))
r9: activate(n__s(X)) -> s(activate(X))
r10: 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__s(X)) -> activate#(X)
p2: activate#(n__first(X1,X2)) -> activate#(X2)

and R consists of:

r1: first(|0|(),X) -> nil()
r2: first(s(X),cons(Y,Z)) -> cons(Y,n__first(X,activate(Z)))
r3: from(X) -> cons(X,n__from(n__s(X)))
r4: first(X1,X2) -> n__first(X1,X2)
r5: from(X) -> n__from(X)
r6: s(X) -> n__s(X)
r7: activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
r8: activate(n__from(X)) -> from(activate(X))
r9: activate(n__s(X)) -> s(activate(X))
r10: activate(X) -> 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: lexicographic order
        interpretations:
          activate#_A(x1) = ((1,0),(1,0)) x1 + (1,2)
          n__s_A(x1) = ((1,0),(0,0)) x1 + (2,1)
          n__first_A(x1,x2) = x1 + ((1,0),(0,0)) x2 + (2,1)
  
    precedence:
    
      activate# = n__s = n__first
    
    partial status:
  
      pi(activate#) = []
      pi(n__s) = []
      pi(n__first) = [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__first(X1,X2)) -> activate#(X2)

and R consists of:

r1: first(|0|(),X) -> nil()
r2: first(s(X),cons(Y,Z)) -> cons(Y,n__first(X,activate(Z)))
r3: from(X) -> cons(X,n__from(n__s(X)))
r4: first(X1,X2) -> n__first(X1,X2)
r5: from(X) -> n__from(X)
r6: s(X) -> n__s(X)
r7: activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
r8: activate(n__from(X)) -> from(activate(X))
r9: activate(n__s(X)) -> s(activate(X))
r10: 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__first(X1,X2)) -> activate#(X2)

and R consists of:

r1: first(|0|(),X) -> nil()
r2: first(s(X),cons(Y,Z)) -> cons(Y,n__first(X,activate(Z)))
r3: from(X) -> cons(X,n__from(n__s(X)))
r4: first(X1,X2) -> n__first(X1,X2)
r5: from(X) -> n__from(X)
r6: s(X) -> n__s(X)
r7: activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
r8: activate(n__from(X)) -> from(activate(X))
r9: activate(n__s(X)) -> s(activate(X))
r10: activate(X) -> 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: lexicographic order
        interpretations:
          activate#_A(x1) = ((1,0),(1,1)) x1
          n__first_A(x1,x2) = x1 + ((1,0),(1,1)) x2 + (1,1)
  
    precedence:
    
      activate# = n__first
    
    partial status:
  
      pi(activate#) = [1]
      pi(n__first) = [2]

The next rules are strictly ordered:

  p1

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