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:
  
      max/plus interpretations on natural numbers:
      
        first#_A(x1,x2) = max{x1 - 4, x2 + 4}
        s_A(x1) = max{12, x1}
        cons_A(x1,x2) = max{25, x1 + 13, x2}
        activate#_A(x1) = max{7, x1 + 4}
        n__s_A(x1) = max{12, x1}
        n__from_A(x1) = max{25, x1 + 13}
        n__first_A(x1,x2) = max{2, x1, x2}
        activate_A(x1) = max{3, x1}
        first_A(x1,x2) = max{2, x1, x2}
        |0|_A = 2
        nil_A = 1
        from_A(x1) = max{25, x1 + 13}
  
    precedence:
    
      first# = s = cons = activate# = n__s = n__from = n__first = activate = first = |0| = nil = from
    
    partial status:
  
      pi(first#) = []
      pi(s) = [1]
      pi(cons) = []
      pi(activate#) = []
      pi(n__s) = [1]
      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:

  p3

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__first(X1,X2)) -> activate#(X2)
p4: activate#(n__first(X1,X2)) -> activate#(X1)
p5: 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, p3, p4, p5}


-- 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))
p3: activate#(n__first(X1,X2)) -> activate#(X1)
p4: activate#(n__first(X1,X2)) -> activate#(X2)
p5: 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 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:
  
      max/plus interpretations on natural numbers:
      
        first#_A(x1,x2) = max{x1 - 1, x2 + 27}
        s_A(x1) = max{42, x1 + 6}
        cons_A(x1,x2) = max{40, x1 - 25, x2 - 22}
        activate#_A(x1) = x1 + 4
        n__first_A(x1,x2) = max{38, x1 + 33, x2 + 29}
        activate_A(x1) = max{5, x1}
        n__s_A(x1) = max{42, x1 + 6}
        first_A(x1,x2) = max{38, x1 + 33, x2 + 29}
        |0|_A = 1
        nil_A = 0
        from_A(x1) = max{41, x1 + 20}
        n__from_A(x1) = max{41, x1 + 20}
  
    precedence:
    
      first# = activate > s = activate# > cons = n__first = n__s = first = |0| = nil = from = n__from
    
    partial status:
  
      pi(first#) = []
      pi(s) = []
      pi(cons) = []
      pi(activate#) = [1]
      pi(n__first) = [1, 2]
      pi(activate) = []
      pi(n__s) = [1]
      pi(first) = [1, 2]
      pi(|0|) = []
      pi(nil) = []
      pi(from) = [1]
      pi(n__from) = [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))
p2: activate#(n__first(X1,X2)) -> activate#(X1)
p3: activate#(n__first(X1,X2)) -> activate#(X2)
p4: 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:

  {p2, p3, p4}


-- Reduction pair.

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

p1: activate#(n__first(X1,X2)) -> activate#(X1)
p2: activate#(n__s(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:
  
      max/plus interpretations on natural numbers:
      
        activate#_A(x1) = x1 + 2
        n__first_A(x1,x2) = max{x1, x2}
        n__s_A(x1) = x1 + 1
  
    precedence:
    
      activate# = n__first = n__s
    
    partial status:
  
      pi(activate#) = []
      pi(n__first) = [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__first(X1,X2)) -> activate#(X1)
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__first(X1,X2)) -> activate#(X1)
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:
  
      max/plus interpretations on natural numbers:
      
        activate#_A(x1) = x1 + 3
        n__first_A(x1,x2) = max{x1 + 1, x2}
  
    precedence:
    
      activate# = n__first
    
    partial status:
  
      pi(activate#) = []
      pi(n__first) = [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__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:
  
      max/plus interpretations on natural numbers:
      
        activate#_A(x1) = x1 + 4
        n__first_A(x1,x2) = max{x1 - 1, x2 + 1}
  
    precedence:
    
      activate# = n__first
    
    partial status:
  
      pi(activate#) = []
      pi(n__first) = [2]

The next rules are strictly ordered:

  p1

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