YES

We show the termination of the TRS R:

  minus_active(|0|(),y) -> |0|()
  mark(|0|()) -> |0|()
  minus_active(s(x),s(y)) -> minus_active(x,y)
  mark(s(x)) -> s(mark(x))
  ge_active(x,|0|()) -> true()
  mark(minus(x,y)) -> minus_active(x,y)
  ge_active(|0|(),s(y)) -> false()
  mark(ge(x,y)) -> ge_active(x,y)
  ge_active(s(x),s(y)) -> ge_active(x,y)
  mark(div(x,y)) -> div_active(mark(x),y)
  div_active(|0|(),s(y)) -> |0|()
  mark(if(x,y,z)) -> if_active(mark(x),y,z)
  div_active(s(x),s(y)) -> if_active(ge_active(x,y),s(div(minus(x,y),s(y))),|0|())
  if_active(true(),x,y) -> mark(x)
  minus_active(x,y) -> minus(x,y)
  if_active(false(),x,y) -> mark(y)
  ge_active(x,y) -> ge(x,y)
  if_active(x,y,z) -> if(x,y,z)
  div_active(x,y) -> div(x,y)

-- SCC decomposition.

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

p1: minus_active#(s(x),s(y)) -> minus_active#(x,y)
p2: mark#(s(x)) -> mark#(x)
p3: mark#(minus(x,y)) -> minus_active#(x,y)
p4: mark#(ge(x,y)) -> ge_active#(x,y)
p5: ge_active#(s(x),s(y)) -> ge_active#(x,y)
p6: mark#(div(x,y)) -> div_active#(mark(x),y)
p7: mark#(div(x,y)) -> mark#(x)
p8: mark#(if(x,y,z)) -> if_active#(mark(x),y,z)
p9: mark#(if(x,y,z)) -> mark#(x)
p10: div_active#(s(x),s(y)) -> if_active#(ge_active(x,y),s(div(minus(x,y),s(y))),|0|())
p11: div_active#(s(x),s(y)) -> ge_active#(x,y)
p12: if_active#(true(),x,y) -> mark#(x)
p13: if_active#(false(),x,y) -> mark#(y)

and R consists of:

r1: minus_active(|0|(),y) -> |0|()
r2: mark(|0|()) -> |0|()
r3: minus_active(s(x),s(y)) -> minus_active(x,y)
r4: mark(s(x)) -> s(mark(x))
r5: ge_active(x,|0|()) -> true()
r6: mark(minus(x,y)) -> minus_active(x,y)
r7: ge_active(|0|(),s(y)) -> false()
r8: mark(ge(x,y)) -> ge_active(x,y)
r9: ge_active(s(x),s(y)) -> ge_active(x,y)
r10: mark(div(x,y)) -> div_active(mark(x),y)
r11: div_active(|0|(),s(y)) -> |0|()
r12: mark(if(x,y,z)) -> if_active(mark(x),y,z)
r13: div_active(s(x),s(y)) -> if_active(ge_active(x,y),s(div(minus(x,y),s(y))),|0|())
r14: if_active(true(),x,y) -> mark(x)
r15: minus_active(x,y) -> minus(x,y)
r16: if_active(false(),x,y) -> mark(y)
r17: ge_active(x,y) -> ge(x,y)
r18: if_active(x,y,z) -> if(x,y,z)
r19: div_active(x,y) -> div(x,y)

The estimated dependency graph contains the following SCCs:

  {p2, p6, p7, p8, p9, p10, p12, p13}
  {p1}
  {p5}


-- Reduction pair.

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

p1: if_active#(false(),x,y) -> mark#(y)
p2: mark#(if(x,y,z)) -> mark#(x)
p3: mark#(if(x,y,z)) -> if_active#(mark(x),y,z)
p4: if_active#(true(),x,y) -> mark#(x)
p5: mark#(div(x,y)) -> mark#(x)
p6: mark#(div(x,y)) -> div_active#(mark(x),y)
p7: div_active#(s(x),s(y)) -> if_active#(ge_active(x,y),s(div(minus(x,y),s(y))),|0|())
p8: mark#(s(x)) -> mark#(x)

and R consists of:

r1: minus_active(|0|(),y) -> |0|()
r2: mark(|0|()) -> |0|()
r3: minus_active(s(x),s(y)) -> minus_active(x,y)
r4: mark(s(x)) -> s(mark(x))
r5: ge_active(x,|0|()) -> true()
r6: mark(minus(x,y)) -> minus_active(x,y)
r7: ge_active(|0|(),s(y)) -> false()
r8: mark(ge(x,y)) -> ge_active(x,y)
r9: ge_active(s(x),s(y)) -> ge_active(x,y)
r10: mark(div(x,y)) -> div_active(mark(x),y)
r11: div_active(|0|(),s(y)) -> |0|()
r12: mark(if(x,y,z)) -> if_active(mark(x),y,z)
r13: div_active(s(x),s(y)) -> if_active(ge_active(x,y),s(div(minus(x,y),s(y))),|0|())
r14: if_active(true(),x,y) -> mark(x)
r15: minus_active(x,y) -> minus(x,y)
r16: if_active(false(),x,y) -> mark(y)
r17: ge_active(x,y) -> ge(x,y)
r18: if_active(x,y,z) -> if(x,y,z)
r19: div_active(x,y) -> div(x,y)

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, r19

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      if_active#_A(x1,x2,x3) = x2 + ((1,0,0),(1,1,0),(0,0,0)) x3 + (0,3,5)
      false_A() = (0,4,1)
      mark#_A(x1) = x1 + (0,3,5)
      if_A(x1,x2,x3) = ((1,0,0),(1,1,0),(0,0,0)) x1 + ((1,0,0),(0,1,0),(0,1,0)) x2 + ((1,0,0),(1,1,0),(0,0,0)) x3
      mark_A(x1) = ((1,0,0),(1,1,0),(0,1,0)) x1 + (0,2,4)
      true_A() = (0,4,1)
      div_A(x1,x2) = ((1,0,0),(1,1,0),(0,0,0)) x1 + ((0,0,0),(1,0,0),(0,0,0)) x2 + (2,0,2)
      div_active#_A(x1,x2) = ((1,0,0),(0,1,0),(1,0,1)) x1 + ((0,0,0),(1,0,0),(0,0,0)) x2 + (2,0,0)
      s_A(x1) = x1 + (8,1,1)
      ge_active_A(x1,x2) = (0,5,9)
      minus_A(x1,x2) = x1 + (0,1,1)
      |0|_A() = (0,1,1)
      minus_active_A(x1,x2) = x1 + (0,2,6)
      div_active_A(x1,x2) = ((1,0,0),(1,1,0),(0,0,1)) x1 + ((0,0,0),(1,0,0),(0,0,0)) x2 + (2,1,1)
      if_active_A(x1,x2,x3) = ((1,0,0),(1,1,0),(0,1,0)) x1 + ((1,0,0),(1,1,0),(0,1,0)) x2 + ((1,0,0),(1,1,0),(0,1,0)) x3 + (0,0,1)
      ge_A(x1,x2) = (0,4,1)

The next rules are strictly ordered:

  p5, p6, p7, p8

We remove them from the problem.

-- SCC decomposition.

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

p1: if_active#(false(),x,y) -> mark#(y)
p2: mark#(if(x,y,z)) -> mark#(x)
p3: mark#(if(x,y,z)) -> if_active#(mark(x),y,z)
p4: if_active#(true(),x,y) -> mark#(x)

and R consists of:

r1: minus_active(|0|(),y) -> |0|()
r2: mark(|0|()) -> |0|()
r3: minus_active(s(x),s(y)) -> minus_active(x,y)
r4: mark(s(x)) -> s(mark(x))
r5: ge_active(x,|0|()) -> true()
r6: mark(minus(x,y)) -> minus_active(x,y)
r7: ge_active(|0|(),s(y)) -> false()
r8: mark(ge(x,y)) -> ge_active(x,y)
r9: ge_active(s(x),s(y)) -> ge_active(x,y)
r10: mark(div(x,y)) -> div_active(mark(x),y)
r11: div_active(|0|(),s(y)) -> |0|()
r12: mark(if(x,y,z)) -> if_active(mark(x),y,z)
r13: div_active(s(x),s(y)) -> if_active(ge_active(x,y),s(div(minus(x,y),s(y))),|0|())
r14: if_active(true(),x,y) -> mark(x)
r15: minus_active(x,y) -> minus(x,y)
r16: if_active(false(),x,y) -> mark(y)
r17: ge_active(x,y) -> ge(x,y)
r18: if_active(x,y,z) -> if(x,y,z)
r19: div_active(x,y) -> div(x,y)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4}


-- Reduction pair.

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

p1: if_active#(false(),x,y) -> mark#(y)
p2: mark#(if(x,y,z)) -> if_active#(mark(x),y,z)
p3: if_active#(true(),x,y) -> mark#(x)
p4: mark#(if(x,y,z)) -> mark#(x)

and R consists of:

r1: minus_active(|0|(),y) -> |0|()
r2: mark(|0|()) -> |0|()
r3: minus_active(s(x),s(y)) -> minus_active(x,y)
r4: mark(s(x)) -> s(mark(x))
r5: ge_active(x,|0|()) -> true()
r6: mark(minus(x,y)) -> minus_active(x,y)
r7: ge_active(|0|(),s(y)) -> false()
r8: mark(ge(x,y)) -> ge_active(x,y)
r9: ge_active(s(x),s(y)) -> ge_active(x,y)
r10: mark(div(x,y)) -> div_active(mark(x),y)
r11: div_active(|0|(),s(y)) -> |0|()
r12: mark(if(x,y,z)) -> if_active(mark(x),y,z)
r13: div_active(s(x),s(y)) -> if_active(ge_active(x,y),s(div(minus(x,y),s(y))),|0|())
r14: if_active(true(),x,y) -> mark(x)
r15: minus_active(x,y) -> minus(x,y)
r16: if_active(false(),x,y) -> mark(y)
r17: ge_active(x,y) -> ge(x,y)
r18: if_active(x,y,z) -> if(x,y,z)
r19: div_active(x,y) -> div(x,y)

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, r19

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      if_active#_A(x1,x2,x3) = ((1,0,0),(0,1,0),(1,1,1)) x1 + x2 + x3
      false_A() = (8,1,7)
      mark#_A(x1) = x1 + (7,3,15)
      if_A(x1,x2,x3) = ((1,0,0),(0,0,0),(1,1,0)) x1 + x2 + ((1,0,0),(0,0,0),(1,1,0)) x3 + (1,4,0)
      mark_A(x1) = ((1,0,0),(0,0,0),(0,1,0)) x1 + (2,6,6)
      true_A() = (8,4,4)
      minus_active_A(x1,x2) = (2,8,0)
      |0|_A() = (0,7,14)
      s_A(x1) = ((0,0,0),(1,0,0),(0,1,0)) x1 + (0,2,1)
      ge_active_A(x1,x2) = (9,5,5)
      div_active_A(x1,x2) = (11,8,15)
      if_active_A(x1,x2,x3) = ((1,0,0),(0,0,0),(0,1,0)) x1 + x2 + x3 + (1,5,1)
      div_A(x1,x2) = ((0,0,0),(0,0,0),(1,0,0)) x1 + ((0,0,0),(0,0,0),(1,0,0)) x2 + (10,1,0)
      minus_A(x1,x2) = (1,9,0)
      ge_A(x1,x2) = (9,0,0)

The next rules are strictly ordered:

  p1, p2, p3, p4

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_active#(s(x),s(y)) -> minus_active#(x,y)

and R consists of:

r1: minus_active(|0|(),y) -> |0|()
r2: mark(|0|()) -> |0|()
r3: minus_active(s(x),s(y)) -> minus_active(x,y)
r4: mark(s(x)) -> s(mark(x))
r5: ge_active(x,|0|()) -> true()
r6: mark(minus(x,y)) -> minus_active(x,y)
r7: ge_active(|0|(),s(y)) -> false()
r8: mark(ge(x,y)) -> ge_active(x,y)
r9: ge_active(s(x),s(y)) -> ge_active(x,y)
r10: mark(div(x,y)) -> div_active(mark(x),y)
r11: div_active(|0|(),s(y)) -> |0|()
r12: mark(if(x,y,z)) -> if_active(mark(x),y,z)
r13: div_active(s(x),s(y)) -> if_active(ge_active(x,y),s(div(minus(x,y),s(y))),|0|())
r14: if_active(true(),x,y) -> mark(x)
r15: minus_active(x,y) -> minus(x,y)
r16: if_active(false(),x,y) -> mark(y)
r17: ge_active(x,y) -> ge(x,y)
r18: if_active(x,y,z) -> if(x,y,z)
r19: div_active(x,y) -> div(x,y)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      minus_active#_A(x1,x2) = ((0,0,0),(0,0,0),(1,0,0)) x1
      s_A(x1) = ((1,0,0),(0,1,0),(1,1,1)) x1 + (1,1,1)

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: ge_active#(s(x),s(y)) -> ge_active#(x,y)

and R consists of:

r1: minus_active(|0|(),y) -> |0|()
r2: mark(|0|()) -> |0|()
r3: minus_active(s(x),s(y)) -> minus_active(x,y)
r4: mark(s(x)) -> s(mark(x))
r5: ge_active(x,|0|()) -> true()
r6: mark(minus(x,y)) -> minus_active(x,y)
r7: ge_active(|0|(),s(y)) -> false()
r8: mark(ge(x,y)) -> ge_active(x,y)
r9: ge_active(s(x),s(y)) -> ge_active(x,y)
r10: mark(div(x,y)) -> div_active(mark(x),y)
r11: div_active(|0|(),s(y)) -> |0|()
r12: mark(if(x,y,z)) -> if_active(mark(x),y,z)
r13: div_active(s(x),s(y)) -> if_active(ge_active(x,y),s(div(minus(x,y),s(y))),|0|())
r14: if_active(true(),x,y) -> mark(x)
r15: minus_active(x,y) -> minus(x,y)
r16: if_active(false(),x,y) -> mark(y)
r17: ge_active(x,y) -> ge(x,y)
r18: if_active(x,y,z) -> if(x,y,z)
r19: div_active(x,y) -> div(x,y)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      ge_active#_A(x1,x2) = ((0,0,0),(0,0,0),(1,0,0)) x1
      s_A(x1) = ((1,0,0),(0,1,0),(1,1,1)) x1 + (1,1,1)

The next rules are strictly ordered:

  p1

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