YES

We show the termination of the TRS R:

  minus(n__0(),Y) -> |0|()
  minus(n__s(X),n__s(Y)) -> minus(activate(X),activate(Y))
  geq(X,n__0()) -> true()
  geq(n__0(),n__s(Y)) -> false()
  geq(n__s(X),n__s(Y)) -> geq(activate(X),activate(Y))
  div(|0|(),n__s(Y)) -> |0|()
  div(s(X),n__s(Y)) -> if(geq(X,activate(Y)),n__s(div(minus(X,activate(Y)),n__s(activate(Y)))),n__0())
  if(true(),X,Y) -> activate(X)
  if(false(),X,Y) -> activate(Y)
  |0|() -> n__0()
  s(X) -> n__s(X)
  activate(n__0()) -> |0|()
  activate(n__s(X)) -> s(X)
  activate(X) -> X

-- SCC decomposition.

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

p1: minus#(n__0(),Y) -> |0|#()
p2: minus#(n__s(X),n__s(Y)) -> minus#(activate(X),activate(Y))
p3: minus#(n__s(X),n__s(Y)) -> activate#(X)
p4: minus#(n__s(X),n__s(Y)) -> activate#(Y)
p5: geq#(n__s(X),n__s(Y)) -> geq#(activate(X),activate(Y))
p6: geq#(n__s(X),n__s(Y)) -> activate#(X)
p7: geq#(n__s(X),n__s(Y)) -> activate#(Y)
p8: div#(s(X),n__s(Y)) -> if#(geq(X,activate(Y)),n__s(div(minus(X,activate(Y)),n__s(activate(Y)))),n__0())
p9: div#(s(X),n__s(Y)) -> geq#(X,activate(Y))
p10: div#(s(X),n__s(Y)) -> activate#(Y)
p11: div#(s(X),n__s(Y)) -> div#(minus(X,activate(Y)),n__s(activate(Y)))
p12: div#(s(X),n__s(Y)) -> minus#(X,activate(Y))
p13: if#(true(),X,Y) -> activate#(X)
p14: if#(false(),X,Y) -> activate#(Y)
p15: activate#(n__0()) -> |0|#()
p16: activate#(n__s(X)) -> s#(X)

and R consists of:

r1: minus(n__0(),Y) -> |0|()
r2: minus(n__s(X),n__s(Y)) -> minus(activate(X),activate(Y))
r3: geq(X,n__0()) -> true()
r4: geq(n__0(),n__s(Y)) -> false()
r5: geq(n__s(X),n__s(Y)) -> geq(activate(X),activate(Y))
r6: div(|0|(),n__s(Y)) -> |0|()
r7: div(s(X),n__s(Y)) -> if(geq(X,activate(Y)),n__s(div(minus(X,activate(Y)),n__s(activate(Y)))),n__0())
r8: if(true(),X,Y) -> activate(X)
r9: if(false(),X,Y) -> activate(Y)
r10: |0|() -> n__0()
r11: s(X) -> n__s(X)
r12: activate(n__0()) -> |0|()
r13: activate(n__s(X)) -> s(X)
r14: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p11}
  {p2}
  {p5}


-- Reduction pair.

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

p1: div#(s(X),n__s(Y)) -> div#(minus(X,activate(Y)),n__s(activate(Y)))

and R consists of:

r1: minus(n__0(),Y) -> |0|()
r2: minus(n__s(X),n__s(Y)) -> minus(activate(X),activate(Y))
r3: geq(X,n__0()) -> true()
r4: geq(n__0(),n__s(Y)) -> false()
r5: geq(n__s(X),n__s(Y)) -> geq(activate(X),activate(Y))
r6: div(|0|(),n__s(Y)) -> |0|()
r7: div(s(X),n__s(Y)) -> if(geq(X,activate(Y)),n__s(div(minus(X,activate(Y)),n__s(activate(Y)))),n__0())
r8: if(true(),X,Y) -> activate(X)
r9: if(false(),X,Y) -> activate(Y)
r10: |0|() -> n__0()
r11: s(X) -> n__s(X)
r12: activate(n__0()) -> |0|()
r13: activate(n__s(X)) -> s(X)
r14: activate(X) -> X

The set of usable rules consists of

  r1, r2, r10, r11, r12, r13, r14

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      div#_A(x1,x2) = x1
      s_A(x1) = 2
      n__s_A(x1) = 1
      minus_A(x1,x2) = 1
      activate_A(x1) = x1 + 1
      |0|_A() = 1
      n__0_A() = 1

The next rules are strictly ordered:

  p1

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

-- Reduction pair.

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

p1: minus#(n__s(X),n__s(Y)) -> minus#(activate(X),activate(Y))

and R consists of:

r1: minus(n__0(),Y) -> |0|()
r2: minus(n__s(X),n__s(Y)) -> minus(activate(X),activate(Y))
r3: geq(X,n__0()) -> true()
r4: geq(n__0(),n__s(Y)) -> false()
r5: geq(n__s(X),n__s(Y)) -> geq(activate(X),activate(Y))
r6: div(|0|(),n__s(Y)) -> |0|()
r7: div(s(X),n__s(Y)) -> if(geq(X,activate(Y)),n__s(div(minus(X,activate(Y)),n__s(activate(Y)))),n__0())
r8: if(true(),X,Y) -> activate(X)
r9: if(false(),X,Y) -> activate(Y)
r10: |0|() -> n__0()
r11: s(X) -> n__s(X)
r12: activate(n__0()) -> |0|()
r13: activate(n__s(X)) -> s(X)
r14: activate(X) -> X

The set of usable rules consists of

  r10, r11, r12, r13, r14

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      minus#_A(x1,x2) = x1 + x2
      n__s_A(x1) = x1 + 1
      activate_A(x1) = x1
      |0|_A() = 0
      n__0_A() = 0
      s_A(x1) = x1 + 1

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8, r9

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

-- Reduction pair.

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

p1: geq#(n__s(X),n__s(Y)) -> geq#(activate(X),activate(Y))

and R consists of:

r1: minus(n__0(),Y) -> |0|()
r2: minus(n__s(X),n__s(Y)) -> minus(activate(X),activate(Y))
r3: geq(X,n__0()) -> true()
r4: geq(n__0(),n__s(Y)) -> false()
r5: geq(n__s(X),n__s(Y)) -> geq(activate(X),activate(Y))
r6: div(|0|(),n__s(Y)) -> |0|()
r7: div(s(X),n__s(Y)) -> if(geq(X,activate(Y)),n__s(div(minus(X,activate(Y)),n__s(activate(Y)))),n__0())
r8: if(true(),X,Y) -> activate(X)
r9: if(false(),X,Y) -> activate(Y)
r10: |0|() -> n__0()
r11: s(X) -> n__s(X)
r12: activate(n__0()) -> |0|()
r13: activate(n__s(X)) -> s(X)
r14: activate(X) -> X

The set of usable rules consists of

  r10, r11, r12, r13, r14

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      geq#_A(x1,x2) = x1 + x2
      n__s_A(x1) = x1 + 1
      activate_A(x1) = x1
      |0|_A() = 0
      n__0_A() = 0
      s_A(x1) = x1 + 1

The next rules are strictly ordered:

  p1
  r1, r2, r3, r4, r5, r6, r7, r8, r9

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