YES

We show the termination of the TRS R:

  |2nd|(cons(X,n__cons(Y,Z))) -> activate(Y)
  from(X) -> cons(X,n__from(n__s(X)))
  cons(X1,X2) -> n__cons(X1,X2)
  from(X) -> n__from(X)
  s(X) -> n__s(X)
  activate(n__cons(X1,X2)) -> cons(activate(X1),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: |2nd|#(cons(X,n__cons(Y,Z))) -> activate#(Y)
p2: from#(X) -> cons#(X,n__from(n__s(X)))
p3: activate#(n__cons(X1,X2)) -> cons#(activate(X1),X2)
p4: activate#(n__cons(X1,X2)) -> activate#(X1)
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: |2nd|(cons(X,n__cons(Y,Z))) -> activate(Y)
r2: from(X) -> cons(X,n__from(n__s(X)))
r3: cons(X1,X2) -> n__cons(X1,X2)
r4: from(X) -> n__from(X)
r5: s(X) -> n__s(X)
r6: activate(n__cons(X1,X2)) -> cons(activate(X1),X2)
r7: activate(n__from(X)) -> from(activate(X))
r8: activate(n__s(X)) -> s(activate(X))
r9: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p4, p6, p8}


-- 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__cons(X1,X2)) -> activate#(X1)

and R consists of:

r1: |2nd|(cons(X,n__cons(Y,Z))) -> activate(Y)
r2: from(X) -> cons(X,n__from(n__s(X)))
r3: cons(X1,X2) -> n__cons(X1,X2)
r4: from(X) -> n__from(X)
r5: s(X) -> n__s(X)
r6: activate(n__cons(X1,X2)) -> cons(activate(X1),X2)
r7: activate(n__from(X)) -> from(activate(X))
r8: activate(n__s(X)) -> s(activate(X))
r9: 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__cons_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (2,3)
  
    precedence:
    
      n__s = n__from > activate# = n__cons
    
    partial status:
  
      pi(activate#) = [1]
      pi(n__s) = [1]
      pi(n__from) = [1]
      pi(n__cons) = [2]

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__cons(X1,X2)) -> activate#(X1)

and R consists of:

r1: |2nd|(cons(X,n__cons(Y,Z))) -> activate(Y)
r2: from(X) -> cons(X,n__from(n__s(X)))
r3: cons(X1,X2) -> n__cons(X1,X2)
r4: from(X) -> n__from(X)
r5: s(X) -> n__s(X)
r6: activate(n__cons(X1,X2)) -> cons(activate(X1),X2)
r7: activate(n__from(X)) -> from(activate(X))
r8: activate(n__s(X)) -> s(activate(X))
r9: 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__cons(X1,X2)) -> activate#(X1)

and R consists of:

r1: |2nd|(cons(X,n__cons(Y,Z))) -> activate(Y)
r2: from(X) -> cons(X,n__from(n__s(X)))
r3: cons(X1,X2) -> n__cons(X1,X2)
r4: from(X) -> n__from(X)
r5: s(X) -> n__s(X)
r6: activate(n__cons(X1,X2)) -> cons(activate(X1),X2)
r7: activate(n__from(X)) -> from(activate(X))
r8: activate(n__s(X)) -> s(activate(X))
r9: 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__cons_A(x1,x2) = ((1,0),(0,0)) x1 + x2 + (2,1)
  
    precedence:
    
      activate# = n__s = n__cons
    
    partial status:
  
      pi(activate#) = []
      pi(n__s) = []
      pi(n__cons) = [2]

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__cons(X1,X2)) -> activate#(X1)

and R consists of:

r1: |2nd|(cons(X,n__cons(Y,Z))) -> activate(Y)
r2: from(X) -> cons(X,n__from(n__s(X)))
r3: cons(X1,X2) -> n__cons(X1,X2)
r4: from(X) -> n__from(X)
r5: s(X) -> n__s(X)
r6: activate(n__cons(X1,X2)) -> cons(activate(X1),X2)
r7: activate(n__from(X)) -> from(activate(X))
r8: activate(n__s(X)) -> s(activate(X))
r9: 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__cons(X1,X2)) -> activate#(X1)

and R consists of:

r1: |2nd|(cons(X,n__cons(Y,Z))) -> activate(Y)
r2: from(X) -> cons(X,n__from(n__s(X)))
r3: cons(X1,X2) -> n__cons(X1,X2)
r4: from(X) -> n__from(X)
r5: s(X) -> n__s(X)
r6: activate(n__cons(X1,X2)) -> cons(activate(X1),X2)
r7: activate(n__from(X)) -> from(activate(X))
r8: activate(n__s(X)) -> s(activate(X))
r9: activate(X) -> X

The set of usable rules consists of

  (no rules)

Take the monotone 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__cons_A(x1,x2) = ((1,0),(1,1)) x1 + x2 + (1,1)
  
    precedence:
    
      activate# > n__cons
    
    partial status:
  
      pi(activate#) = [1]
      pi(n__cons) = [1, 2]

The next rules are strictly ordered:

  p1

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