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:

  max/plus interpretations on natural numbers:
  
    minus#_A(x1,x2) = max{16, x1 + 5, x2 + 6}
    n__s_A(x1) = x1
    activate_A(x1) = max{8, x1}
    activate#_A(x1) = x1 + 5
    n__minus_A(x1,x2) = max{11, x1, x2 + 2}
    n__div_A(x1,x2) = max{11, x1, x2 + 3}
    div#_A(x1,x2) = max{16, x1 + 5, x2 + 8}
    s_A(x1) = max{7, x1}
    geq#_A(x1,x2) = max{14, x1 + 5, x2 + 5}
    if#_A(x1,x2,x3) = max{x1 + 1, x2 + 5, x3 + 7}
    geq_A(x1,x2) = 3
    n__0_A = 1
    false_A = 0
    true_A = 2
    if_A(x1,x2,x3) = max{8, x2, x3 + 7}
    minus_A(x1,x2) = max{11, x1, x2 + 2}
    |0|_A = 4
    div_A(x1,x2) = max{11, x1, x2 + 3}

The next rules are strictly ordered:

  p2, p7, p14

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: activate#(n__minus(X1,X2)) -> minus#(X1,X2)
p3: minus#(n__s(X),n__s(Y)) -> activate#(X)
p4: activate#(n__div(X1,X2)) -> activate#(X1)
p5: activate#(n__div(X1,X2)) -> div#(activate(X1),X2)
p6: activate#(n__s(X)) -> activate#(X)
p7: div#(s(X),n__s(Y)) -> geq#(X,activate(Y))
p8: geq#(n__s(X),n__s(Y)) -> activate#(Y)
p9: geq#(n__s(X),n__s(Y)) -> activate#(X)
p10: geq#(n__s(X),n__s(Y)) -> geq#(activate(X),activate(Y))
p11: 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())
p12: 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, p10, p11, p12}


-- 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: div#(s(X),n__s(Y)) -> geq#(X,activate(Y))
p10: geq#(n__s(X),n__s(Y)) -> geq#(activate(X),activate(Y))
p11: geq#(n__s(X),n__s(Y)) -> activate#(X)
p12: geq#(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:

  max/plus interpretations on natural numbers:
  
    minus#_A(x1,x2) = max{x1 + 2, x2 + 5}
    n__s_A(x1) = x1
    activate_A(x1) = x1
    activate#_A(x1) = x1 + 2
    n__div_A(x1,x2) = max{x1 + 2, x2 + 6}
    div#_A(x1,x2) = max{x1 + 4, x2 + 8}
    s_A(x1) = x1
    if#_A(x1,x2,x3) = max{x1 + 2, x2 + 2, x3 + 1}
    geq_A(x1,x2) = max{x1 + 1, x2 + 1}
    n__minus_A(x1,x2) = max{x1, x2 + 3}
    n__0_A = 2
    true_A = 1
    geq#_A(x1,x2) = max{8, x1 + 2, x2 + 2}
    if_A(x1,x2,x3) = max{1, x2, x3}
    false_A = 1
    minus_A(x1,x2) = max{x1, x2 + 3}
    |0|_A = 2
    div_A(x1,x2) = max{x1 + 2, x2 + 6}

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: div#(s(X),n__s(Y)) -> geq#(X,activate(Y))
p9: geq#(n__s(X),n__s(Y)) -> geq#(activate(X),activate(Y))
p10: geq#(n__s(X),n__s(Y)) -> activate#(X)
p11: geq#(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, p9, p10, p11}


-- 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__minus(X1,X2)) -> minus#(X1,X2)
p4: activate#(n__div(X1,X2)) -> div#(activate(X1),X2)
p5: div#(s(X),n__s(Y)) -> geq#(X,activate(Y))
p6: geq#(n__s(X),n__s(Y)) -> activate#(Y)
p7: activate#(n__s(X)) -> activate#(X)
p8: geq#(n__s(X),n__s(Y)) -> activate#(X)
p9: geq#(n__s(X),n__s(Y)) -> geq#(activate(X),activate(Y))
p10: 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())
p11: 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:

  max/plus interpretations on natural numbers:
  
    minus#_A(x1,x2) = x1 + 6
    n__s_A(x1) = max{9, x1}
    activate_A(x1) = max{3, x1}
    activate#_A(x1) = x1 + 6
    n__minus_A(x1,x2) = max{1, x1}
    n__div_A(x1,x2) = max{x1 + 10, x2 + 20}
    div#_A(x1,x2) = max{x1 + 16, x2 + 26}
    s_A(x1) = max{9, x1}
    geq#_A(x1,x2) = max{x1 + 16, x2 + 6}
    if#_A(x1,x2,x3) = max{x1 + 5, x2 + 6, x3 + 15}
    geq_A(x1,x2) = max{x1 + 10, x2 + 7}
    n__0_A = 0
    true_A = 1
    if_A(x1,x2,x3) = max{x2, x3 + 10}
    false_A = 1
    minus_A(x1,x2) = max{2, x1}
    |0|_A = 1
    div_A(x1,x2) = max{x1 + 10, x2 + 20}

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#(X)
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)) -> geq#(X,activate(Y))
p6: geq#(n__s(X),n__s(Y)) -> activate#(Y)
p7: activate#(n__s(X)) -> activate#(X)
p8: geq#(n__s(X),n__s(Y)) -> geq#(activate(X),activate(Y))
p9: 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())
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 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#(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: div#(s(X),n__s(Y)) -> geq#(X,activate(Y))
p9: geq#(n__s(X),n__s(Y)) -> geq#(activate(X),activate(Y))
p10: geq#(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:

  max/plus interpretations on natural numbers:
  
    minus#_A(x1,x2) = x1 + 1
    n__s_A(x1) = x1
    activate_A(x1) = x1
    activate#_A(x1) = x1 + 1
    n__div_A(x1,x2) = max{x1 + 5, x2 + 5}
    div#_A(x1,x2) = max{x1 + 6, x2 + 6}
    s_A(x1) = x1
    if#_A(x1,x2,x3) = max{x1 + 5, x2 + 1, x3 + 2}
    geq_A(x1,x2) = max{x1 + 1, x2 + 1}
    n__minus_A(x1,x2) = x1
    n__0_A = 3
    true_A = 1
    geq#_A(x1,x2) = max{x1 + 4, x2 + 3}
    if_A(x1,x2,x3) = max{x1 + 4, x2, x3}
    false_A = 1
    minus_A(x1,x2) = x1
    |0|_A = 3
    div_A(x1,x2) = max{x1 + 5, x2 + 5}

The next rules are strictly ordered:

  p8, p10

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: 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 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#(X)
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 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:

  max/plus interpretations on natural numbers:
  
    minus#_A(x1,x2) = max{1, x1 - 59}
    n__s_A(x1) = max{43, x1 + 18}
    activate_A(x1) = max{32, x1 + 16}
    activate#_A(x1) = max{1, x1 - 42}
    n__minus_A(x1,x2) = max{4, x1 - 17}
    n__div_A(x1,x2) = max{62, x1 + 46}
    div#_A(x1,x2) = max{3, x1 - 12}
    s_A(x1) = max{51, x1 + 18}
    if#_A(x1,x2,x3) = max{x1 + 2, x2 - 42, x3 + 2}
    geq_A(x1,x2) = 0
    n__0_A = 3
    true_A = 0
    if_A(x1,x2,x3) = max{32, x2 + 17, x3 + 31}
    false_A = 0
    minus_A(x1,x2) = max{22, x1 - 9}
    |0|_A = 22
    div_A(x1,x2) = max{62, x1 + 46}

The next rules are strictly ordered:

  p5

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__minus(X1,X2)) -> minus#(X1,X2)
p4: activate#(n__div(X1,X2)) -> div#(activate(X1),X2)
p5: if#(true(),X,Y) -> activate#(X)
p6: 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, 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))
p2: minus#(n__s(X),n__s(Y)) -> activate#(X)
p3: activate#(n__s(X)) -> activate#(X)
p4: 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:

  max/plus interpretations on natural numbers:
  
    minus#_A(x1,x2) = x1
    n__s_A(x1) = x1
    activate_A(x1) = x1
    activate#_A(x1) = max{0, x1 - 4}
    n__minus_A(x1,x2) = x1 + 5
    geq_A(x1,x2) = max{x1 + 1, x2 + 1}
    n__0_A = 5
    true_A = 1
    false_A = 2
    if_A(x1,x2,x3) = max{x2, x3}
    minus_A(x1,x2) = x1 + 5
    |0|_A = 5
    div_A(x1,x2) = 5
    s_A(x1) = x1
    n__div_A(x1,x2) = 5

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#(X)
p3: 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}
  {p3}


-- 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:

  max/plus interpretations on natural numbers:
  
    minus#_A(x1,x2) = max{5, x1, x2}
    n__s_A(x1) = x1 + 8
    activate_A(x1) = x1 + 5
    geq_A(x1,x2) = x1 + 5
    n__0_A = 4
    true_A = 5
    false_A = 9
    if_A(x1,x2,x3) = max{x1 + 8, x2 + 5, x3 + 5}
    minus_A(x1,x2) = 5
    |0|_A = 5
    div_A(x1,x2) = max{6, x1 + 5}
    s_A(x1) = max{13, x1 + 8}
    n__div_A(x1,x2) = x1 + 5
    n__minus_A(x1,x2) = 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: 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:

  max/plus interpretations on natural numbers:
  
    activate#_A(x1) = x1
    n__s_A(x1) = x1 + 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(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:

  max/plus interpretations on natural numbers:
  
    geq#_A(x1,x2) = max{0, x1 - 15, x2 - 9}
    n__s_A(x1) = max{27, x1 + 16}
    activate_A(x1) = max{11, x1}
    geq_A(x1,x2) = max{1, x1}
    n__0_A = 8
    true_A = 1
    false_A = 8
    if_A(x1,x2,x3) = max{x1 + 22, x2, x3 + 15}
    minus_A(x1,x2) = 9
    |0|_A = 9
    div_A(x1,x2) = max{22, x1 + 11}
    s_A(x1) = max{27, x1 + 16}
    n__div_A(x1,x2) = max{22, x1 + 11}
    n__minus_A(x1,x2) = 0

The next rules are strictly ordered:

  p1

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