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:

  max/plus interpretations on natural numbers:
  
    activate#_A(x1) = x1 + 3
    n__add_A(x1,x2) = max{2, x1, x2 + 1}
    add#_A(x1,x2) = max{x1 + 3, x2 + 3}
    activate_A(x1) = x1
    s_A(x1) = x1
    n__from_A(x1) = max{2, x1}
    from#_A(x1) = x1 + 3
    n__first_A(x1,x2) = max{x1 + 2, x2}
    first#_A(x1,x2) = max{x1 + 5, x2 + 3}
    cons_A(x1,x2) = max{2, x1, x2}
    |0|_A = 0
    add_A(x1,x2) = max{2, x1, x2 + 1}
    first_A(x1,x2) = max{x1 + 2, x2}
    nil_A = 0
    from_A(x1) = max{2, x1}
    n__s_A(x1) = x1

The next rules are strictly ordered:

  p6, p10

We remove them from the problem.

-- SCC decomposition.

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)) -> first#(activate(X1),activate(X2))
p7: first#(s(X),cons(Y,Z)) -> activate#(Z)
p8: activate#(n__add(X1,X2)) -> activate#(X1)
p9: first#(s(X),cons(Y,Z)) -> activate#(Y)
p10: add#(s(X),Y) -> activate#(X)
p11: 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 estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8, p9, p10, p11}


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

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:

  max/plus interpretations on natural numbers:
  
    activate#_A(x1) = max{6, x1}
    n__add_A(x1,x2) = max{x1 + 9, x2 + 6}
    add#_A(x1,x2) = max{x1 + 6, x2 + 6}
    activate_A(x1) = x1
    |0|_A = 0
    n__first_A(x1,x2) = max{x1 + 8, x2 + 8}
    first#_A(x1,x2) = max{7, x1 + 2, x2 + 2}
    s_A(x1) = x1
    cons_A(x1,x2) = max{x1 + 5, x2}
    n__from_A(x1) = x1 + 5
    from#_A(x1) = max{6, x1 + 5}
    add_A(x1,x2) = max{x1 + 9, x2 + 6}
    first_A(x1,x2) = max{x1 + 8, x2 + 8}
    nil_A = 5
    from_A(x1) = x1 + 5
    n__s_A(x1) = x1

The next rules are strictly ordered:

  p3, p4, p5, p6, p9

We remove them from the problem.

-- SCC decomposition.

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

p1: activate#(n__add(X1,X2)) -> add#(activate(X1),X2)
p2: add#(|0|(),X) -> activate#(X)
p3: activate#(n__from(X)) -> from#(X)
p4: from#(X) -> activate#(X)
p5: add#(s(X),Y) -> activate#(X)
p6: add#(s(X),Y) -> activate#(Y)

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, p2, p3, p4, p5, p6}


-- 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: add#(s(X),Y) -> activate#(X)
p6: 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:

  max/plus interpretations on natural numbers:
  
    activate#_A(x1) = max{2, x1 - 1}
    n__add_A(x1,x2) = max{x1 + 8, x2 + 3}
    add#_A(x1,x2) = max{x1 + 1, x2 + 2}
    activate_A(x1) = x1
    s_A(x1) = max{5, x1 - 2}
    n__from_A(x1) = x1 + 3
    from#_A(x1) = x1 + 2
    |0|_A = 1
    add_A(x1,x2) = max{x1 + 8, x2 + 3}
    first_A(x1,x2) = 3
    nil_A = 0
    cons_A(x1,x2) = 3
    n__first_A(x1,x2) = 3
    from_A(x1) = x1 + 3
    n__s_A(x1) = max{5, x1 - 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__add(X1,X2)) -> add#(activate(X1),X2)
p2: activate#(n__from(X)) -> from#(X)
p3: from#(X) -> activate#(X)
p4: add#(s(X),Y) -> activate#(X)
p5: 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 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: activate#(n__add(X1,X2)) -> add#(activate(X1),X2)
p2: add#(|0|(),X) -> activate#(X)
p3: activate#(n__from(X)) -> from#(X)
p4: from#(X) -> activate#(X)
p5: add#(s(X),Y) -> 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:

  max/plus interpretations on natural numbers:
  
    activate#_A(x1) = x1 + 1
    n__add_A(x1,x2) = max{x1 + 2, x2 + 2}
    add#_A(x1,x2) = max{x1 + 1, x2 + 3}
    activate_A(x1) = x1
    |0|_A = 0
    n__from_A(x1) = x1 + 4
    from#_A(x1) = x1 + 5
    s_A(x1) = max{2, x1}
    add_A(x1,x2) = max{x1 + 2, x2 + 2}
    first_A(x1,x2) = 6
    nil_A = 2
    cons_A(x1,x2) = max{4, x2 - 2}
    n__first_A(x1,x2) = 6
    from_A(x1) = x1 + 4
    n__s_A(x1) = max{2, x1}

The next rules are strictly ordered:

  p2, p4

We remove them from the problem.

-- SCC decomposition.

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

p1: activate#(n__add(X1,X2)) -> add#(activate(X1),X2)
p2: activate#(n__from(X)) -> from#(X)
p3: add#(s(X),Y) -> 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 estimated dependency graph contains the following SCCs:

  {p1, p3}


-- 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#(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:

  max/plus interpretations on natural numbers:
  
    activate#_A(x1) = x1 + 1
    n__add_A(x1,x2) = max{6, x1 + 2, x2 + 1}
    add#_A(x1,x2) = max{7, x1 + 3, x2 + 1}
    activate_A(x1) = max{4, x1}
    s_A(x1) = max{2, x1}
    add_A(x1,x2) = max{6, x1 + 2, x2 + 1}
    |0|_A = 2
    first_A(x1,x2) = x2
    nil_A = 0
    cons_A(x1,x2) = 7
    n__first_A(x1,x2) = x2
    from_A(x1) = 7
    n__from_A(x1) = 7
    n__s_A(x1) = max{1, x1}

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__add(X1,X2)) -> add#(activate(X1),X2)

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:

  (no SCCs)