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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          first#_A(x1,x2) = max{x1 + 4, x2 + 5}
          s_A(x1) = x1
          cons_A(x1,x2) = max{x1 + 2, x2}
          activate#_A(x1) = x1 + 5
          n__s_A(x1) = x1
          n__from_A(x1) = max{6, x1 + 4}
          n__first_A(x1,x2) = max{14, x1 + 11, x2 + 8}
          activate_A(x1) = max{2, x1}
          first_A(x1,x2) = max{14, x1 + 11, x2 + 8}
          |0|_A = 1
          nil_A = 0
          from_A(x1) = max{6, x1 + 4}
    
      precedence:
      
        first# = activate# = activate = first > |0| = from > cons = n__from > s > n__s = n__first > nil
      
      partial status:
    
        pi(first#) = [2]
        pi(s) = [1]
        pi(cons) = [2]
        pi(activate#) = [1]
        pi(n__s) = [1]
        pi(n__from) = [1]
        pi(n__first) = [1, 2]
        pi(activate) = [1]
        pi(first) = [2]
        pi(|0|) = []
        pi(nil) = []
        pi(from) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          first#_A(x1,x2) = 50
          s_A(x1) = 15
          cons_A(x1,x2) = 52
          activate#_A(x1) = 50
          n__s_A(x1) = x1 + 15
          n__from_A(x1) = max{42, x1 + 29}
          n__first_A(x1,x2) = max{34, x1 + 7}
          activate_A(x1) = max{44, x1 + 16}
          first_A(x1,x2) = max{51, x2 + 33}
          |0|_A = 26
          nil_A = 32
          from_A(x1) = max{45, x1 + 14}
    
      precedence:
      
        first# = s = cons = activate# = n__s = n__from = n__first = activate = first = |0| = nil = from
      
      partial status:
    
        pi(first#) = []
        pi(s) = []
        pi(cons) = []
        pi(activate#) = []
        pi(n__s) = [1]
        pi(n__from) = [1]
        pi(n__first) = []
        pi(activate) = [1]
        pi(first) = [2]
        pi(|0|) = []
        pi(nil) = []
        pi(from) = []
    

The next rules are strictly ordered:

  p2, p3, p4, p5

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__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 estimated dependency graph contains the following SCCs:

  {p1, p2}


-- 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__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:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          first#_A(x1,x2) = max{0, x1 - 12, x2 - 7}
          s_A(x1) = max{19, x1 + 11}
          cons_A(x1,x2) = x2 + 7
          activate#_A(x1) = max{0, x1 - 2}
          n__first_A(x1,x2) = max{x1, x2 - 4}
          activate_A(x1) = max{7, x1}
          first_A(x1,x2) = max{x1, x2 - 4}
          |0|_A = 1
          nil_A = 0
          from_A(x1) = 7
          n__from_A(x1) = 0
          n__s_A(x1) = max{19, x1 + 11}
    
      precedence:
      
        activate = first > s > nil > from > |0| > n__from > n__first = n__s > cons > activate# > first#
      
      partial status:
    
        pi(first#) = []
        pi(s) = []
        pi(cons) = [2]
        pi(activate#) = []
        pi(n__first) = []
        pi(activate) = [1]
        pi(first) = []
        pi(|0|) = []
        pi(nil) = []
        pi(from) = []
        pi(n__from) = []
        pi(n__s) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          first#_A(x1,x2) = 0
          s_A(x1) = 87
          cons_A(x1,x2) = x2 + 9
          activate#_A(x1) = 1
          n__first_A(x1,x2) = 33
          activate_A(x1) = max{34, x1 - 24}
          first_A(x1,x2) = 8
          |0|_A = 30
          nil_A = 29
          from_A(x1) = 45
          n__from_A(x1) = 35
          n__s_A(x1) = x1 + 112
    
      precedence:
      
        activate# > from > activate > first > cons = nil > n__from > s = n__s > n__first > |0| > first#
      
      partial status:
    
        pi(first#) = []
        pi(s) = []
        pi(cons) = [2]
        pi(activate#) = []
        pi(n__first) = []
        pi(activate) = []
        pi(first) = []
        pi(|0|) = []
        pi(nil) = []
        pi(from) = []
        pi(n__from) = []
        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__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 estimated dependency graph contains the following SCCs:

  (no SCCs)