YES

We show the termination of the TRS R:

  and(true(),X) -> activate(X)
  and(false(),Y) -> false()
  if(true(),X,Y) -> activate(X)
  if(false(),X,Y) -> activate(Y)
  add(|0|(),X) -> activate(X)
  add(s(X),Y) -> s(n__add(activate(X),activate(Y)))
  first(|0|(),X) -> nil()
  first(s(X),cons(Y,Z)) -> cons(activate(Y),n__first(activate(X),activate(Z)))
  from(X) -> cons(activate(X),n__from(n__s(activate(X))))
  add(X1,X2) -> n__add(X1,X2)
  first(X1,X2) -> n__first(X1,X2)
  from(X) -> n__from(X)
  s(X) -> n__s(X)
  activate(n__add(X1,X2)) -> add(activate(X1),X2)
  activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
  activate(n__from(X)) -> from(X)
  activate(n__s(X)) -> s(X)
  activate(X) -> X

-- SCC decomposition.

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

p1: and#(true(),X) -> activate#(X)
p2: if#(true(),X,Y) -> activate#(X)
p3: if#(false(),X,Y) -> activate#(Y)
p4: add#(|0|(),X) -> activate#(X)
p5: add#(s(X),Y) -> s#(n__add(activate(X),activate(Y)))
p6: add#(s(X),Y) -> activate#(X)
p7: add#(s(X),Y) -> activate#(Y)
p8: first#(s(X),cons(Y,Z)) -> activate#(Y)
p9: first#(s(X),cons(Y,Z)) -> activate#(X)
p10: first#(s(X),cons(Y,Z)) -> activate#(Z)
p11: from#(X) -> activate#(X)
p12: activate#(n__add(X1,X2)) -> add#(activate(X1),X2)
p13: activate#(n__add(X1,X2)) -> activate#(X1)
p14: activate#(n__first(X1,X2)) -> first#(activate(X1),activate(X2))
p15: activate#(n__first(X1,X2)) -> activate#(X1)
p16: activate#(n__first(X1,X2)) -> activate#(X2)
p17: activate#(n__from(X)) -> from#(X)
p18: activate#(n__s(X)) -> s#(X)

and R consists of:

r1: and(true(),X) -> activate(X)
r2: and(false(),Y) -> false()
r3: if(true(),X,Y) -> activate(X)
r4: if(false(),X,Y) -> activate(Y)
r5: add(|0|(),X) -> activate(X)
r6: add(s(X),Y) -> s(n__add(activate(X),activate(Y)))
r7: first(|0|(),X) -> nil()
r8: first(s(X),cons(Y,Z)) -> cons(activate(Y),n__first(activate(X),activate(Z)))
r9: from(X) -> cons(activate(X),n__from(n__s(activate(X))))
r10: add(X1,X2) -> n__add(X1,X2)
r11: first(X1,X2) -> n__first(X1,X2)
r12: from(X) -> n__from(X)
r13: s(X) -> n__s(X)
r14: activate(n__add(X1,X2)) -> add(activate(X1),X2)
r15: activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
r16: activate(n__from(X)) -> from(X)
r17: activate(n__s(X)) -> s(X)
r18: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p4, p6, p7, p8, p9, p10, p11, p12, p13, p14, p15, p16, p17}


-- Reduction pair.

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

p1: activate#(n__add(X1,X2)) -> add#(activate(X1),X2)
p2: add#(s(X),Y) -> activate#(Y)
p3: activate#(n__from(X)) -> from#(X)
p4: from#(X) -> activate#(X)
p5: activate#(n__first(X1,X2)) -> activate#(X2)
p6: activate#(n__first(X1,X2)) -> activate#(X1)
p7: activate#(n__first(X1,X2)) -> first#(activate(X1),activate(X2))
p8: first#(s(X),cons(Y,Z)) -> activate#(Z)
p9: activate#(n__add(X1,X2)) -> activate#(X1)
p10: first#(s(X),cons(Y,Z)) -> activate#(X)
p11: first#(s(X),cons(Y,Z)) -> activate#(Y)
p12: add#(s(X),Y) -> activate#(X)
p13: add#(|0|(),X) -> activate#(X)

and R consists of:

r1: and(true(),X) -> activate(X)
r2: and(false(),Y) -> false()
r3: if(true(),X,Y) -> activate(X)
r4: if(false(),X,Y) -> activate(Y)
r5: add(|0|(),X) -> activate(X)
r6: add(s(X),Y) -> s(n__add(activate(X),activate(Y)))
r7: first(|0|(),X) -> nil()
r8: first(s(X),cons(Y,Z)) -> cons(activate(Y),n__first(activate(X),activate(Z)))
r9: from(X) -> cons(activate(X),n__from(n__s(activate(X))))
r10: add(X1,X2) -> n__add(X1,X2)
r11: first(X1,X2) -> n__first(X1,X2)
r12: from(X) -> n__from(X)
r13: s(X) -> n__s(X)
r14: activate(n__add(X1,X2)) -> add(activate(X1),X2)
r15: activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
r16: activate(n__from(X)) -> from(X)
r17: activate(n__s(X)) -> s(X)
r18: activate(X) -> X

The set of usable rules consists of

  r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          activate#_A(x1) = max{9, x1 + 4}
          n__add_A(x1,x2) = max{x1 + 19, x2 + 15}
          add#_A(x1,x2) = max{x1 + 13, x2 + 10}
          activate_A(x1) = x1
          s_A(x1) = max{0, x1 - 3}
          n__from_A(x1) = x1 + 9
          from#_A(x1) = max{9, x1 + 4}
          n__first_A(x1,x2) = max{x1 + 14, x2 + 10}
          first#_A(x1,x2) = max{x1 + 8, x2 + 8}
          cons_A(x1,x2) = max{4, x1 - 4, x2 - 4}
          |0|_A = 0
          add_A(x1,x2) = max{x1 + 19, x2 + 15}
          first_A(x1,x2) = max{x1 + 14, x2 + 10}
          nil_A = 1
          from_A(x1) = x1 + 9
          n__s_A(x1) = max{0, x1 - 3}
    
      precedence:
      
        activate = n__from = from# = |0| = nil = from > s = add = n__s > first > n__add = add# = first# = cons > activate# = n__first
      
      partial status:
    
        pi(activate#) = [1]
        pi(n__add) = [1, 2]
        pi(add#) = [1, 2]
        pi(activate) = [1]
        pi(s) = []
        pi(n__from) = []
        pi(from#) = []
        pi(n__first) = [2]
        pi(first#) = []
        pi(cons) = []
        pi(|0|) = []
        pi(add) = [1, 2]
        pi(first) = [1, 2]
        pi(nil) = []
        pi(from) = [1]
        pi(n__s) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          activate#_A(x1) = 4
          n__add_A(x1,x2) = max{x1 + 116, x2 + 115}
          add#_A(x1,x2) = max{124, x1 + 77, x2 + 116}
          activate_A(x1) = x1 + 50
          s_A(x1) = 164
          n__from_A(x1) = 27
          from#_A(x1) = 3
          n__first_A(x1,x2) = max{4, x2 - 4}
          first#_A(x1,x2) = 4
          cons_A(x1,x2) = 48
          |0|_A = 41
          add_A(x1,x2) = max{166, x1 + 116, x2 + 165}
          first_A(x1,x2) = max{47, x2 + 43}
          nil_A = 42
          from_A(x1) = max{112, x1 + 49}
          n__s_A(x1) = 113
    
      precedence:
      
        from# = n__first > add# = first# > activate# = activate = s = cons = |0| = add > n__add = n__from = first = nil = from = n__s
      
      partial status:
    
        pi(activate#) = []
        pi(n__add) = [2]
        pi(add#) = [1, 2]
        pi(activate) = [1]
        pi(s) = []
        pi(n__from) = []
        pi(from#) = []
        pi(n__first) = []
        pi(first#) = []
        pi(cons) = []
        pi(|0|) = []
        pi(add) = [1, 2]
        pi(first) = []
        pi(nil) = []
        pi(from) = []
        pi(n__s) = []
    

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p7, p9, p10, p11, p12, p13

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: first#(s(X),cons(Y,Z)) -> activate#(Z)

and R consists of:

r1: and(true(),X) -> activate(X)
r2: and(false(),Y) -> false()
r3: if(true(),X,Y) -> activate(X)
r4: if(false(),X,Y) -> activate(Y)
r5: add(|0|(),X) -> activate(X)
r6: add(s(X),Y) -> s(n__add(activate(X),activate(Y)))
r7: first(|0|(),X) -> nil()
r8: first(s(X),cons(Y,Z)) -> cons(activate(Y),n__first(activate(X),activate(Z)))
r9: from(X) -> cons(activate(X),n__from(n__s(activate(X))))
r10: add(X1,X2) -> n__add(X1,X2)
r11: first(X1,X2) -> n__first(X1,X2)
r12: from(X) -> n__from(X)
r13: s(X) -> n__s(X)
r14: activate(n__add(X1,X2)) -> add(activate(X1),X2)
r15: activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
r16: activate(n__from(X)) -> from(X)
r17: activate(n__s(X)) -> s(X)
r18: 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#(X1)

and R consists of:

r1: and(true(),X) -> activate(X)
r2: and(false(),Y) -> false()
r3: if(true(),X,Y) -> activate(X)
r4: if(false(),X,Y) -> activate(Y)
r5: add(|0|(),X) -> activate(X)
r6: add(s(X),Y) -> s(n__add(activate(X),activate(Y)))
r7: first(|0|(),X) -> nil()
r8: first(s(X),cons(Y,Z)) -> cons(activate(Y),n__first(activate(X),activate(Z)))
r9: from(X) -> cons(activate(X),n__from(n__s(activate(X))))
r10: add(X1,X2) -> n__add(X1,X2)
r11: first(X1,X2) -> n__first(X1,X2)
r12: from(X) -> n__from(X)
r13: s(X) -> n__s(X)
r14: activate(n__add(X1,X2)) -> add(activate(X1),X2)
r15: activate(n__first(X1,X2)) -> first(activate(X1),activate(X2))
r16: activate(n__from(X)) -> from(X)
r17: activate(n__s(X)) -> s(X)
r18: activate(X) -> X

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          activate#_A(x1) = max{2, x1 + 1}
          n__first_A(x1,x2) = max{x1, x2 + 1}
    
      precedence:
      
        activate# = n__first
      
      partial status:
    
        pi(activate#) = [1]
        pi(n__first) = [1, 2]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          activate#_A(x1) = max{2, x1 - 1}
          n__first_A(x1,x2) = 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.