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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        div#_A(x1,x2) = max{0, x1 - 3}
        s_A(x1) = x1 + 11
        n__s_A(x1) = x1 + 10
        minus_A(x1,x2) = x1 + 9
        activate_A(x1) = max{7, x1 + 6}
        |0|_A = 8
        n__0_A = 8
  
    precedence:
    
      div# = s = n__s = minus = activate = |0| = n__0
    
    partial status:
  
      pi(div#) = []
      pi(s) = [1]
      pi(n__s) = [1]
      pi(minus) = [1]
      pi(activate) = [1]
      pi(|0|) = []
      pi(n__0) = []

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 reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        minus#_A(x1,x2) = max{0, x1 - 6, x2 - 2}
        n__s_A(x1) = x1 + 5
        activate_A(x1) = x1 + 4
        |0|_A = 5
        n__0_A = 4
        s_A(x1) = x1 + 6
  
    precedence:
    
      minus# = n__s = activate = |0| = n__0 = s
    
    partial status:
  
      pi(minus#) = []
      pi(n__s) = [1]
      pi(activate) = [1]
      pi(|0|) = []
      pi(n__0) = []
      pi(s) = [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: 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 reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        geq#_A(x1,x2) = max{0, x1 - 6, x2 - 2}
        n__s_A(x1) = x1 + 5
        activate_A(x1) = x1 + 4
        |0|_A = 5
        n__0_A = 4
        s_A(x1) = x1 + 6
  
    precedence:
    
      geq# = n__s = activate = |0| = n__0 = s
    
    partial status:
  
      pi(geq#) = []
      pi(n__s) = [1]
      pi(activate) = [1]
      pi(|0|) = []
      pi(n__0) = []
      pi(s) = [1]

The next rules are strictly ordered:

  p1

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