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(n__div(n__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)
  div(X1,X2) -> n__div(X1,X2)
  minus(X1,X2) -> n__minus(X1,X2)
  activate(n__0()) -> |0|()
  activate(n__s(X)) -> s(activate(X))
  activate(n__div(X1,X2)) -> div(activate(X1),X2)
  activate(n__minus(X1,X2)) -> minus(X1,X2)
  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(n__div(n__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: if#(true(),X,Y) -> activate#(X)
p12: if#(false(),X,Y) -> activate#(Y)
p13: activate#(n__0()) -> |0|#()
p14: activate#(n__s(X)) -> s#(activate(X))
p15: activate#(n__s(X)) -> activate#(X)
p16: activate#(n__div(X1,X2)) -> div#(activate(X1),X2)
p17: activate#(n__div(X1,X2)) -> activate#(X1)
p18: activate#(n__minus(X1,X2)) -> minus#(X1,X2)

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(n__div(n__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: div(X1,X2) -> n__div(X1,X2)
r13: minus(X1,X2) -> n__minus(X1,X2)
r14: activate(n__0()) -> |0|()
r15: activate(n__s(X)) -> s(activate(X))
r16: activate(n__div(X1,X2)) -> div(activate(X1),X2)
r17: activate(n__minus(X1,X2)) -> minus(X1,X2)
r18: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p2, p3, p4, p5, p6, p7, p8, p9, p10, p11, p12, p15, p16, p17, p18}


-- 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))
p2: minus#(n__s(X),n__s(Y)) -> activate#(Y)
p3: activate#(n__minus(X1,X2)) -> minus#(X1,X2)
p4: minus#(n__s(X),n__s(Y)) -> activate#(X)
p5: activate#(n__div(X1,X2)) -> activate#(X1)
p6: activate#(n__div(X1,X2)) -> div#(activate(X1),X2)
p7: div#(s(X),n__s(Y)) -> activate#(Y)
p8: activate#(n__s(X)) -> activate#(X)
p9: div#(s(X),n__s(Y)) -> geq#(X,activate(Y))
p10: geq#(n__s(X),n__s(Y)) -> activate#(Y)
p11: geq#(n__s(X),n__s(Y)) -> activate#(X)
p12: geq#(n__s(X),n__s(Y)) -> geq#(activate(X),activate(Y))
p13: div#(s(X),n__s(Y)) -> if#(geq(X,activate(Y)),n__s(n__div(n__minus(X,activate(Y)),n__s(activate(Y)))),n__0())
p14: if#(false(),X,Y) -> activate#(Y)
p15: if#(true(),X,Y) -> activate#(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(n__div(n__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: div(X1,X2) -> n__div(X1,X2)
r13: minus(X1,X2) -> n__minus(X1,X2)
r14: activate(n__0()) -> |0|()
r15: activate(n__s(X)) -> s(activate(X))
r16: activate(n__div(X1,X2)) -> div(activate(X1),X2)
r17: activate(n__minus(X1,X2)) -> minus(X1,X2)
r18: activate(X) -> X

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        minus#_A(x1,x2) = max{1, x1, x2}
        n__s_A(x1) = max{15, x1}
        activate_A(x1) = max{5, x1}
        activate#_A(x1) = max{15, x1}
        n__minus_A(x1,x2) = max{x1, x2 + 4}
        n__div_A(x1,x2) = max{x1 + 28, x2 + 33}
        div#_A(x1,x2) = max{x1 + 28, x2 + 33}
        s_A(x1) = max{15, x1}
        geq#_A(x1,x2) = max{x1 + 16, x2 + 22}
        if#_A(x1,x2,x3) = max{15, x1 - 1, x2, x3}
        geq_A(x1,x2) = max{x1 + 29, x2 + 20}
        n__0_A = 2
        false_A = 1
        true_A = 1
        if_A(x1,x2,x3) = max{x2, x3 + 46}
        minus_A(x1,x2) = max{x1, x2 + 4}
        |0|_A = 3
        div_A(x1,x2) = max{x1 + 28, x2 + 33}
  
    precedence:
    
      minus# = activate# = div# = if# > n__div = div > activate = if = minus > s = geq# = geq > n__s = n__minus = n__0 = true = |0| > false
    
    partial status:
  
      pi(minus#) = []
      pi(n__s) = []
      pi(activate) = [1]
      pi(activate#) = []
      pi(n__minus) = [2]
      pi(n__div) = []
      pi(div#) = []
      pi(s) = []
      pi(geq#) = []
      pi(if#) = []
      pi(geq) = []
      pi(n__0) = []
      pi(false) = []
      pi(true) = []
      pi(if) = [2]
      pi(minus) = []
      pi(|0|) = []
      pi(div) = []

The next rules are strictly ordered:

  p9

We remove them from the problem.

-- SCC decomposition.

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

p1: minus#(n__s(X),n__s(Y)) -> minus#(activate(X),activate(Y))
p2: minus#(n__s(X),n__s(Y)) -> activate#(Y)
p3: activate#(n__minus(X1,X2)) -> minus#(X1,X2)
p4: minus#(n__s(X),n__s(Y)) -> activate#(X)
p5: activate#(n__div(X1,X2)) -> activate#(X1)
p6: activate#(n__div(X1,X2)) -> div#(activate(X1),X2)
p7: div#(s(X),n__s(Y)) -> activate#(Y)
p8: activate#(n__s(X)) -> activate#(X)
p9: geq#(n__s(X),n__s(Y)) -> activate#(Y)
p10: geq#(n__s(X),n__s(Y)) -> activate#(X)
p11: geq#(n__s(X),n__s(Y)) -> geq#(activate(X),activate(Y))
p12: div#(s(X),n__s(Y)) -> if#(geq(X,activate(Y)),n__s(n__div(n__minus(X,activate(Y)),n__s(activate(Y)))),n__0())
p13: if#(false(),X,Y) -> activate#(Y)
p14: if#(true(),X,Y) -> activate#(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(n__div(n__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: div(X1,X2) -> n__div(X1,X2)
r13: minus(X1,X2) -> n__minus(X1,X2)
r14: activate(n__0()) -> |0|()
r15: activate(n__s(X)) -> s(activate(X))
r16: activate(n__div(X1,X2)) -> div(activate(X1),X2)
r17: activate(n__minus(X1,X2)) -> minus(X1,X2)
r18: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p11}
  {p1, p2, p3, p4, p5, p6, p7, p8, p12, p13, p14}


-- 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(n__div(n__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: div(X1,X2) -> n__div(X1,X2)
r13: minus(X1,X2) -> n__minus(X1,X2)
r14: activate(n__0()) -> |0|()
r15: activate(n__s(X)) -> s(activate(X))
r16: activate(n__div(X1,X2)) -> div(activate(X1),X2)
r17: activate(n__minus(X1,X2)) -> minus(X1,X2)
r18: activate(X) -> X

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        geq#_A(x1,x2) = max{24, x1 + 16, x2 - 8}
        n__s_A(x1) = max{108, x1 + 33}
        activate_A(x1) = x1 + 25
        geq_A(x1,x2) = 8
        n__0_A = 34
        true_A = 0
        false_A = 1
        if_A(x1,x2,x3) = max{x1 + 163, x2 + 29, x3 + 26}
        minus_A(x1,x2) = 36
        |0|_A = 35
        div_A(x1,x2) = x1 + 62
        s_A(x1) = max{109, x1 + 33}
        n__div_A(x1,x2) = x1 + 62
        n__minus_A(x1,x2) = 11
  
    precedence:
    
      geq# = n__s = activate = geq = n__0 = true = false = if = minus = |0| = div = s = n__div = n__minus
    
    partial status:
  
      pi(geq#) = [1]
      pi(n__s) = []
      pi(activate) = []
      pi(geq) = []
      pi(n__0) = []
      pi(true) = []
      pi(false) = []
      pi(if) = []
      pi(minus) = []
      pi(|0|) = []
      pi(div) = []
      pi(s) = []
      pi(n__div) = []
      pi(n__minus) = []

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))
p2: minus#(n__s(X),n__s(Y)) -> activate#(X)
p3: activate#(n__s(X)) -> activate#(X)
p4: activate#(n__div(X1,X2)) -> div#(activate(X1),X2)
p5: div#(s(X),n__s(Y)) -> if#(geq(X,activate(Y)),n__s(n__div(n__minus(X,activate(Y)),n__s(activate(Y)))),n__0())
p6: if#(true(),X,Y) -> activate#(X)
p7: activate#(n__div(X1,X2)) -> activate#(X1)
p8: activate#(n__minus(X1,X2)) -> minus#(X1,X2)
p9: minus#(n__s(X),n__s(Y)) -> activate#(Y)
p10: if#(false(),X,Y) -> activate#(Y)
p11: div#(s(X),n__s(Y)) -> 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(n__div(n__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: div(X1,X2) -> n__div(X1,X2)
r13: minus(X1,X2) -> n__minus(X1,X2)
r14: activate(n__0()) -> |0|()
r15: activate(n__s(X)) -> s(activate(X))
r16: activate(n__div(X1,X2)) -> div(activate(X1),X2)
r17: activate(n__minus(X1,X2)) -> minus(X1,X2)
r18: activate(X) -> X

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        minus#_A(x1,x2) = max{12, x1 - 1, x2 - 1}
        n__s_A(x1) = max{2, x1}
        activate_A(x1) = max{13, x1}
        activate#_A(x1) = max{0, x1 - 1}
        n__div_A(x1,x2) = max{22, x1 + 9, x2 + 9}
        div#_A(x1,x2) = max{x1 + 8, x2 + 8}
        s_A(x1) = max{13, x1}
        if#_A(x1,x2,x3) = max{1, x1 - 5, x2 - 1, x3 - 1}
        geq_A(x1,x2) = 12
        n__minus_A(x1,x2) = max{13, x1, x2}
        n__0_A = 11
        true_A = 6
        false_A = 12
        if_A(x1,x2,x3) = max{22, x1 - 19, x2, x3 + 11}
        minus_A(x1,x2) = max{13, x1, x2}
        |0|_A = 12
        div_A(x1,x2) = max{22, x1 + 9, x2 + 9}
  
    precedence:
    
      n__div = div > geq = n__minus = true = if = minus > activate = |0| > n__s = s > n__0 > minus# = activate# = div# = if# = false
    
    partial status:
  
      pi(minus#) = []
      pi(n__s) = []
      pi(activate) = [1]
      pi(activate#) = []
      pi(n__div) = []
      pi(div#) = []
      pi(s) = []
      pi(if#) = []
      pi(geq) = []
      pi(n__minus) = []
      pi(n__0) = []
      pi(true) = []
      pi(false) = []
      pi(if) = [2, 3]
      pi(minus) = []
      pi(|0|) = []
      pi(div) = []

The next rules are strictly ordered:

  p11

We remove them from the problem.

-- SCC decomposition.

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

p1: minus#(n__s(X),n__s(Y)) -> minus#(activate(X),activate(Y))
p2: minus#(n__s(X),n__s(Y)) -> activate#(X)
p3: activate#(n__s(X)) -> activate#(X)
p4: activate#(n__div(X1,X2)) -> div#(activate(X1),X2)
p5: div#(s(X),n__s(Y)) -> if#(geq(X,activate(Y)),n__s(n__div(n__minus(X,activate(Y)),n__s(activate(Y)))),n__0())
p6: if#(true(),X,Y) -> activate#(X)
p7: activate#(n__div(X1,X2)) -> activate#(X1)
p8: activate#(n__minus(X1,X2)) -> minus#(X1,X2)
p9: minus#(n__s(X),n__s(Y)) -> activate#(Y)
p10: if#(false(),X,Y) -> 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(n__div(n__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: div(X1,X2) -> n__div(X1,X2)
r13: minus(X1,X2) -> n__minus(X1,X2)
r14: activate(n__0()) -> |0|()
r15: activate(n__s(X)) -> s(activate(X))
r16: activate(n__div(X1,X2)) -> div(activate(X1),X2)
r17: activate(n__minus(X1,X2)) -> minus(X1,X2)
r18: activate(X) -> X

The estimated dependency graph contains the following SCCs:

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


-- 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))
p2: minus#(n__s(X),n__s(Y)) -> activate#(Y)
p3: activate#(n__minus(X1,X2)) -> minus#(X1,X2)
p4: minus#(n__s(X),n__s(Y)) -> activate#(X)
p5: activate#(n__div(X1,X2)) -> activate#(X1)
p6: activate#(n__div(X1,X2)) -> div#(activate(X1),X2)
p7: div#(s(X),n__s(Y)) -> if#(geq(X,activate(Y)),n__s(n__div(n__minus(X,activate(Y)),n__s(activate(Y)))),n__0())
p8: if#(false(),X,Y) -> activate#(Y)
p9: activate#(n__s(X)) -> activate#(X)
p10: if#(true(),X,Y) -> activate#(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(n__div(n__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: div(X1,X2) -> n__div(X1,X2)
r13: minus(X1,X2) -> n__minus(X1,X2)
r14: activate(n__0()) -> |0|()
r15: activate(n__s(X)) -> s(activate(X))
r16: activate(n__div(X1,X2)) -> div(activate(X1),X2)
r17: activate(n__minus(X1,X2)) -> minus(X1,X2)
r18: activate(X) -> X

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        minus#_A(x1,x2) = max{39, x1 + 30, x2 + 30}
        n__s_A(x1) = max{4, x1}
        activate_A(x1) = max{9, x1}
        activate#_A(x1) = max{32, x1 + 30}
        n__minus_A(x1,x2) = max{9, x1, x2}
        n__div_A(x1,x2) = max{42, x1 + 33, x2 + 33}
        div#_A(x1,x2) = max{x1 + 63, x2 + 63}
        s_A(x1) = max{9, x1}
        if#_A(x1,x2,x3) = max{x1 + 30, x2 + 30, x3 + 32}
        geq_A(x1,x2) = max{23, x1 + 14, x2 + 4}
        n__0_A = 36
        false_A = 3
        true_A = 1
        if_A(x1,x2,x3) = max{42, x2, x3 + 5}
        minus_A(x1,x2) = max{9, x1, x2}
        |0|_A = 36
        div_A(x1,x2) = max{42, x1 + 33, x2 + 33}
  
    precedence:
    
      minus# = n__s = activate = activate# = n__minus = n__div = div# = s = if# = geq = n__0 = false = true = if = minus = |0| = div
    
    partial status:
  
      pi(minus#) = []
      pi(n__s) = []
      pi(activate) = []
      pi(activate#) = []
      pi(n__minus) = []
      pi(n__div) = []
      pi(div#) = []
      pi(s) = []
      pi(if#) = []
      pi(geq) = []
      pi(n__0) = []
      pi(false) = []
      pi(true) = []
      pi(if) = []
      pi(minus) = []
      pi(|0|) = []
      pi(div) = []

The next rules are strictly ordered:

  p8

We remove them from the problem.

-- SCC decomposition.

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

p1: minus#(n__s(X),n__s(Y)) -> minus#(activate(X),activate(Y))
p2: minus#(n__s(X),n__s(Y)) -> activate#(Y)
p3: activate#(n__minus(X1,X2)) -> minus#(X1,X2)
p4: minus#(n__s(X),n__s(Y)) -> activate#(X)
p5: activate#(n__div(X1,X2)) -> activate#(X1)
p6: activate#(n__div(X1,X2)) -> div#(activate(X1),X2)
p7: div#(s(X),n__s(Y)) -> if#(geq(X,activate(Y)),n__s(n__div(n__minus(X,activate(Y)),n__s(activate(Y)))),n__0())
p8: activate#(n__s(X)) -> activate#(X)
p9: if#(true(),X,Y) -> activate#(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(n__div(n__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: div(X1,X2) -> n__div(X1,X2)
r13: minus(X1,X2) -> n__minus(X1,X2)
r14: activate(n__0()) -> |0|()
r15: activate(n__s(X)) -> s(activate(X))
r16: activate(n__div(X1,X2)) -> div(activate(X1),X2)
r17: activate(n__minus(X1,X2)) -> minus(X1,X2)
r18: activate(X) -> X

The estimated dependency graph contains the following SCCs:

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


-- 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))
p2: minus#(n__s(X),n__s(Y)) -> activate#(X)
p3: activate#(n__s(X)) -> activate#(X)
p4: activate#(n__div(X1,X2)) -> div#(activate(X1),X2)
p5: div#(s(X),n__s(Y)) -> if#(geq(X,activate(Y)),n__s(n__div(n__minus(X,activate(Y)),n__s(activate(Y)))),n__0())
p6: if#(true(),X,Y) -> activate#(X)
p7: activate#(n__div(X1,X2)) -> activate#(X1)
p8: activate#(n__minus(X1,X2)) -> minus#(X1,X2)
p9: minus#(n__s(X),n__s(Y)) -> 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(n__div(n__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: div(X1,X2) -> n__div(X1,X2)
r13: minus(X1,X2) -> n__minus(X1,X2)
r14: activate(n__0()) -> |0|()
r15: activate(n__s(X)) -> s(activate(X))
r16: activate(n__div(X1,X2)) -> div(activate(X1),X2)
r17: activate(n__minus(X1,X2)) -> minus(X1,X2)
r18: activate(X) -> X

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        minus#_A(x1,x2) = max{x1 + 25, x2 + 28}
        n__s_A(x1) = max{10, x1}
        activate_A(x1) = max{2, x1}
        activate#_A(x1) = x1 + 25
        n__div_A(x1,x2) = max{x1 + 24, x2 + 56}
        div#_A(x1,x2) = max{x1 + 49, x2 + 81}
        s_A(x1) = max{10, x1}
        if#_A(x1,x2,x3) = max{x1 - 1, x2 + 25, x3 + 18}
        geq_A(x1,x2) = x1 + 30
        n__minus_A(x1,x2) = max{x1, x2 + 29}
        n__0_A = 0
        true_A = 19
        if_A(x1,x2,x3) = max{x1 - 6, x2, x3 + 66}
        false_A = 9
        minus_A(x1,x2) = max{x1, x2 + 29}
        |0|_A = 1
        div_A(x1,x2) = max{x1 + 24, x2 + 56}
  
    precedence:
    
      minus# = activate# = n__div = div# = if# = geq = true = false = div > if > n__s = activate = s > n__minus = minus > |0| > n__0
    
    partial status:
  
      pi(minus#) = []
      pi(n__s) = []
      pi(activate) = [1]
      pi(activate#) = []
      pi(n__div) = []
      pi(div#) = []
      pi(s) = []
      pi(if#) = []
      pi(geq) = []
      pi(n__minus) = []
      pi(n__0) = []
      pi(true) = []
      pi(if) = [2]
      pi(false) = []
      pi(minus) = []
      pi(|0|) = []
      pi(div) = []

The next rules are strictly ordered:

  p7

We remove them from the problem.

-- SCC decomposition.

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

p1: minus#(n__s(X),n__s(Y)) -> minus#(activate(X),activate(Y))
p2: minus#(n__s(X),n__s(Y)) -> activate#(X)
p3: activate#(n__s(X)) -> activate#(X)
p4: activate#(n__div(X1,X2)) -> div#(activate(X1),X2)
p5: div#(s(X),n__s(Y)) -> if#(geq(X,activate(Y)),n__s(n__div(n__minus(X,activate(Y)),n__s(activate(Y)))),n__0())
p6: if#(true(),X,Y) -> activate#(X)
p7: activate#(n__minus(X1,X2)) -> minus#(X1,X2)
p8: minus#(n__s(X),n__s(Y)) -> 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(n__div(n__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: div(X1,X2) -> n__div(X1,X2)
r13: minus(X1,X2) -> n__minus(X1,X2)
r14: activate(n__0()) -> |0|()
r15: activate(n__s(X)) -> s(activate(X))
r16: activate(n__div(X1,X2)) -> div(activate(X1),X2)
r17: activate(n__minus(X1,X2)) -> minus(X1,X2)
r18: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7, p8}


-- 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))
p2: minus#(n__s(X),n__s(Y)) -> activate#(Y)
p3: activate#(n__minus(X1,X2)) -> minus#(X1,X2)
p4: minus#(n__s(X),n__s(Y)) -> activate#(X)
p5: activate#(n__div(X1,X2)) -> div#(activate(X1),X2)
p6: div#(s(X),n__s(Y)) -> if#(geq(X,activate(Y)),n__s(n__div(n__minus(X,activate(Y)),n__s(activate(Y)))),n__0())
p7: if#(true(),X,Y) -> activate#(X)
p8: activate#(n__s(X)) -> activate#(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(n__div(n__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: div(X1,X2) -> n__div(X1,X2)
r13: minus(X1,X2) -> n__minus(X1,X2)
r14: activate(n__0()) -> |0|()
r15: activate(n__s(X)) -> s(activate(X))
r16: activate(n__div(X1,X2)) -> div(activate(X1),X2)
r17: activate(n__minus(X1,X2)) -> minus(X1,X2)
r18: activate(X) -> X

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        minus#_A(x1,x2) = max{x1 + 5, x2 + 6}
        n__s_A(x1) = max{17, x1}
        activate_A(x1) = x1
        activate#_A(x1) = x1 + 2
        n__minus_A(x1,x2) = max{x1 + 8, x2 + 7}
        n__div_A(x1,x2) = x2 + 16
        div#_A(x1,x2) = x2 + 18
        s_A(x1) = max{17, x1}
        if#_A(x1,x2,x3) = max{14, x1 + 3, x2 + 2, x3 + 2}
        geq_A(x1,x2) = max{2, x2 - 1}
        n__0_A = 5
        true_A = 3
        if_A(x1,x2,x3) = max{7, x1 - 2, x2, x3 + 6}
        false_A = 0
        minus_A(x1,x2) = max{x1 + 8, x2 + 7}
        |0|_A = 5
        div_A(x1,x2) = x2 + 16
  
    precedence:
    
      activate = n__div = geq = if = div > n__s = s = minus > activate# = div# = if# = true > minus# = n__0 = |0| > n__minus > false
    
    partial status:
  
      pi(minus#) = []
      pi(n__s) = []
      pi(activate) = []
      pi(activate#) = []
      pi(n__minus) = [1, 2]
      pi(n__div) = []
      pi(div#) = []
      pi(s) = []
      pi(if#) = []
      pi(geq) = []
      pi(n__0) = []
      pi(true) = []
      pi(if) = []
      pi(false) = []
      pi(minus) = []
      pi(|0|) = []
      pi(div) = []

The next rules are strictly ordered:

  p4

We remove them from the problem.

-- SCC decomposition.

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

p1: minus#(n__s(X),n__s(Y)) -> minus#(activate(X),activate(Y))
p2: minus#(n__s(X),n__s(Y)) -> activate#(Y)
p3: activate#(n__minus(X1,X2)) -> minus#(X1,X2)
p4: activate#(n__div(X1,X2)) -> div#(activate(X1),X2)
p5: div#(s(X),n__s(Y)) -> if#(geq(X,activate(Y)),n__s(n__div(n__minus(X,activate(Y)),n__s(activate(Y)))),n__0())
p6: if#(true(),X,Y) -> activate#(X)
p7: activate#(n__s(X)) -> activate#(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(n__div(n__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: div(X1,X2) -> n__div(X1,X2)
r13: minus(X1,X2) -> n__minus(X1,X2)
r14: activate(n__0()) -> |0|()
r15: activate(n__s(X)) -> s(activate(X))
r16: activate(n__div(X1,X2)) -> div(activate(X1),X2)
r17: activate(n__minus(X1,X2)) -> minus(X1,X2)
r18: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5, p6, p7}


-- 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))
p2: minus#(n__s(X),n__s(Y)) -> activate#(Y)
p3: activate#(n__s(X)) -> activate#(X)
p4: activate#(n__div(X1,X2)) -> div#(activate(X1),X2)
p5: div#(s(X),n__s(Y)) -> if#(geq(X,activate(Y)),n__s(n__div(n__minus(X,activate(Y)),n__s(activate(Y)))),n__0())
p6: if#(true(),X,Y) -> activate#(X)
p7: activate#(n__minus(X1,X2)) -> minus#(X1,X2)

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(n__div(n__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: div(X1,X2) -> n__div(X1,X2)
r13: minus(X1,X2) -> n__minus(X1,X2)
r14: activate(n__0()) -> |0|()
r15: activate(n__s(X)) -> s(activate(X))
r16: activate(n__div(X1,X2)) -> div(activate(X1),X2)
r17: activate(n__minus(X1,X2)) -> minus(X1,X2)
r18: activate(X) -> X

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        minus#_A(x1,x2) = x2 + 6
        n__s_A(x1) = x1
        activate_A(x1) = x1
        activate#_A(x1) = max{0, x1 - 4}
        n__div_A(x1,x2) = x2 + 19
        div#_A(x1,x2) = x2 + 15
        s_A(x1) = x1
        if#_A(x1,x2,x3) = max{15, x2 - 4, x3 + 11}
        geq_A(x1,x2) = max{x1 + 25, x2 + 25}
        n__minus_A(x1,x2) = max{x1 + 3, x2 + 11}
        n__0_A = 4
        true_A = 1
        if_A(x1,x2,x3) = max{19, x2, x3 + 15}
        false_A = 3
        minus_A(x1,x2) = max{x1 + 3, x2 + 11}
        |0|_A = 4
        div_A(x1,x2) = x2 + 19
  
    precedence:
    
      n__div = div > if > minus# = activate = n__minus = n__0 = minus = |0| > n__s = activate# = div# = s = if# = geq = true = false
    
    partial status:
  
      pi(minus#) = []
      pi(n__s) = []
      pi(activate) = [1]
      pi(activate#) = []
      pi(n__div) = []
      pi(div#) = []
      pi(s) = []
      pi(if#) = []
      pi(geq) = []
      pi(n__minus) = []
      pi(n__0) = []
      pi(true) = []
      pi(if) = [2]
      pi(false) = []
      pi(minus) = []
      pi(|0|) = []
      pi(div) = []

The next rules are strictly ordered:

  p7

We remove them from the problem.

-- SCC decomposition.

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

p1: minus#(n__s(X),n__s(Y)) -> minus#(activate(X),activate(Y))
p2: minus#(n__s(X),n__s(Y)) -> activate#(Y)
p3: activate#(n__s(X)) -> activate#(X)
p4: activate#(n__div(X1,X2)) -> div#(activate(X1),X2)
p5: div#(s(X),n__s(Y)) -> if#(geq(X,activate(Y)),n__s(n__div(n__minus(X,activate(Y)),n__s(activate(Y)))),n__0())
p6: if#(true(),X,Y) -> activate#(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(n__div(n__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: div(X1,X2) -> n__div(X1,X2)
r13: minus(X1,X2) -> n__minus(X1,X2)
r14: activate(n__0()) -> |0|()
r15: activate(n__s(X)) -> s(activate(X))
r16: activate(n__div(X1,X2)) -> div(activate(X1),X2)
r17: activate(n__minus(X1,X2)) -> minus(X1,X2)
r18: activate(X) -> X

The estimated dependency graph contains the following SCCs:

  {p1}
  {p3, p4, p5, p6}


-- 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(n__div(n__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: div(X1,X2) -> n__div(X1,X2)
r13: minus(X1,X2) -> n__minus(X1,X2)
r14: activate(n__0()) -> |0|()
r15: activate(n__s(X)) -> s(activate(X))
r16: activate(n__div(X1,X2)) -> div(activate(X1),X2)
r17: activate(n__minus(X1,X2)) -> minus(X1,X2)
r18: activate(X) -> X

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        minus#_A(x1,x2) = max{x1 + 2, x2 + 2}
        n__s_A(x1) = max{100, x1 + 53}
        activate_A(x1) = max{54, x1 + 37}
        geq_A(x1,x2) = max{117, x1 + 47, x2 + 2}
        n__0_A = 2
        true_A = 0
        false_A = 1
        if_A(x1,x2,x3) = max{71, x2 + 40, x3 + 68}
        minus_A(x1,x2) = 38
        |0|_A = 37
        div_A(x1,x2) = max{36, x1 + 18}
        s_A(x1) = max{137, x1 + 53}
        n__div_A(x1,x2) = max{35, x1 + 18}
        n__minus_A(x1,x2) = 0
  
    precedence:
    
      activate = minus = n__minus > geq = true = false > n__s = s > minus# > div = n__div > n__0 = if = |0|
    
    partial status:
  
      pi(minus#) = [2]
      pi(n__s) = []
      pi(activate) = [1]
      pi(geq) = [2]
      pi(n__0) = []
      pi(true) = []
      pi(false) = []
      pi(if) = [2, 3]
      pi(minus) = []
      pi(|0|) = []
      pi(div) = []
      pi(s) = []
      pi(n__div) = []
      pi(n__minus) = []

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: activate#(n__s(X)) -> activate#(X)
p2: activate#(n__div(X1,X2)) -> div#(activate(X1),X2)
p3: div#(s(X),n__s(Y)) -> if#(geq(X,activate(Y)),n__s(n__div(n__minus(X,activate(Y)),n__s(activate(Y)))),n__0())
p4: if#(true(),X,Y) -> activate#(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(n__div(n__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: div(X1,X2) -> n__div(X1,X2)
r13: minus(X1,X2) -> n__minus(X1,X2)
r14: activate(n__0()) -> |0|()
r15: activate(n__s(X)) -> s(activate(X))
r16: activate(n__div(X1,X2)) -> div(activate(X1),X2)
r17: activate(n__minus(X1,X2)) -> minus(X1,X2)
r18: activate(X) -> X

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17, r18

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        activate#_A(x1) = x1 + 6
        n__s_A(x1) = max{64, x1 + 6}
        n__div_A(x1,x2) = x1 + 95
        div#_A(x1,x2) = x1 + 91
        activate_A(x1) = x1 + 4
        s_A(x1) = max{68, x1 + 6}
        if#_A(x1,x2,x3) = max{x2 + 7, x3 + 6}
        geq_A(x1,x2) = max{x1 + 77, x2 - 7}
        n__minus_A(x1,x2) = 25
        n__0_A = 9
        true_A = 1
        if_A(x1,x2,x3) = max{x2 + 37, x3 + 14}
        false_A = 65
        minus_A(x1,x2) = 26
        |0|_A = 12
        div_A(x1,x2) = x1 + 95
  
    precedence:
    
      activate = |0| > n__div = s = if = false = div > div# = if# > n__s > activate# > n__minus = n__0 = minus > geq > true
    
    partial status:
  
      pi(activate#) = [1]
      pi(n__s) = []
      pi(n__div) = []
      pi(div#) = []
      pi(activate) = []
      pi(s) = []
      pi(if#) = []
      pi(geq) = []
      pi(n__minus) = []
      pi(n__0) = []
      pi(true) = []
      pi(if) = []
      pi(false) = []
      pi(minus) = []
      pi(|0|) = []
      pi(div) = []

The next rules are strictly ordered:

  p4

We remove them from the problem.

-- SCC decomposition.

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

p1: activate#(n__s(X)) -> activate#(X)
p2: activate#(n__div(X1,X2)) -> div#(activate(X1),X2)
p3: div#(s(X),n__s(Y)) -> if#(geq(X,activate(Y)),n__s(n__div(n__minus(X,activate(Y)),n__s(activate(Y)))),n__0())

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(n__div(n__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: div(X1,X2) -> n__div(X1,X2)
r13: minus(X1,X2) -> n__minus(X1,X2)
r14: activate(n__0()) -> |0|()
r15: activate(n__s(X)) -> s(activate(X))
r16: activate(n__div(X1,X2)) -> div(activate(X1),X2)
r17: activate(n__minus(X1,X2)) -> minus(X1,X2)
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__s(X)) -> activate#(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(n__div(n__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: div(X1,X2) -> n__div(X1,X2)
r13: minus(X1,X2) -> n__minus(X1,X2)
r14: activate(n__0()) -> |0|()
r15: activate(n__s(X)) -> s(activate(X))
r16: activate(n__div(X1,X2)) -> div(activate(X1),X2)
r17: activate(n__minus(X1,X2)) -> minus(X1,X2)
r18: activate(X) -> X

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        activate#_A(x1) = x1 + 2
        n__s_A(x1) = x1 + 2
  
    precedence:
    
      activate# = n__s
    
    partial status:
  
      pi(activate#) = [1]
      pi(n__s) = [1]

The next rules are strictly ordered:

  p1

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