YES

We show the termination of the TRS R:

  active(minus(|0|(),Y)) -> mark(|0|())
  active(minus(s(X),s(Y))) -> mark(minus(X,Y))
  active(geq(X,|0|())) -> mark(true())
  active(geq(|0|(),s(Y))) -> mark(false())
  active(geq(s(X),s(Y))) -> mark(geq(X,Y))
  active(div(|0|(),s(Y))) -> mark(|0|())
  active(div(s(X),s(Y))) -> mark(if(geq(X,Y),s(div(minus(X,Y),s(Y))),|0|()))
  active(if(true(),X,Y)) -> mark(X)
  active(if(false(),X,Y)) -> mark(Y)
  mark(minus(X1,X2)) -> active(minus(X1,X2))
  mark(|0|()) -> active(|0|())
  mark(s(X)) -> active(s(mark(X)))
  mark(geq(X1,X2)) -> active(geq(X1,X2))
  mark(true()) -> active(true())
  mark(false()) -> active(false())
  mark(div(X1,X2)) -> active(div(mark(X1),X2))
  mark(if(X1,X2,X3)) -> active(if(mark(X1),X2,X3))
  minus(mark(X1),X2) -> minus(X1,X2)
  minus(X1,mark(X2)) -> minus(X1,X2)
  minus(active(X1),X2) -> minus(X1,X2)
  minus(X1,active(X2)) -> minus(X1,X2)
  s(mark(X)) -> s(X)
  s(active(X)) -> s(X)
  geq(mark(X1),X2) -> geq(X1,X2)
  geq(X1,mark(X2)) -> geq(X1,X2)
  geq(active(X1),X2) -> geq(X1,X2)
  geq(X1,active(X2)) -> geq(X1,X2)
  div(mark(X1),X2) -> div(X1,X2)
  div(X1,mark(X2)) -> div(X1,X2)
  div(active(X1),X2) -> div(X1,X2)
  div(X1,active(X2)) -> div(X1,X2)
  if(mark(X1),X2,X3) -> if(X1,X2,X3)
  if(X1,mark(X2),X3) -> if(X1,X2,X3)
  if(X1,X2,mark(X3)) -> if(X1,X2,X3)
  if(active(X1),X2,X3) -> if(X1,X2,X3)
  if(X1,active(X2),X3) -> if(X1,X2,X3)
  if(X1,X2,active(X3)) -> if(X1,X2,X3)

-- SCC decomposition.

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

p1: active#(minus(|0|(),Y)) -> mark#(|0|())
p2: active#(minus(s(X),s(Y))) -> mark#(minus(X,Y))
p3: active#(minus(s(X),s(Y))) -> minus#(X,Y)
p4: active#(geq(X,|0|())) -> mark#(true())
p5: active#(geq(|0|(),s(Y))) -> mark#(false())
p6: active#(geq(s(X),s(Y))) -> mark#(geq(X,Y))
p7: active#(geq(s(X),s(Y))) -> geq#(X,Y)
p8: active#(div(|0|(),s(Y))) -> mark#(|0|())
p9: active#(div(s(X),s(Y))) -> mark#(if(geq(X,Y),s(div(minus(X,Y),s(Y))),|0|()))
p10: active#(div(s(X),s(Y))) -> if#(geq(X,Y),s(div(minus(X,Y),s(Y))),|0|())
p11: active#(div(s(X),s(Y))) -> geq#(X,Y)
p12: active#(div(s(X),s(Y))) -> s#(div(minus(X,Y),s(Y)))
p13: active#(div(s(X),s(Y))) -> div#(minus(X,Y),s(Y))
p14: active#(div(s(X),s(Y))) -> minus#(X,Y)
p15: active#(if(true(),X,Y)) -> mark#(X)
p16: active#(if(false(),X,Y)) -> mark#(Y)
p17: mark#(minus(X1,X2)) -> active#(minus(X1,X2))
p18: mark#(|0|()) -> active#(|0|())
p19: mark#(s(X)) -> active#(s(mark(X)))
p20: mark#(s(X)) -> s#(mark(X))
p21: mark#(s(X)) -> mark#(X)
p22: mark#(geq(X1,X2)) -> active#(geq(X1,X2))
p23: mark#(true()) -> active#(true())
p24: mark#(false()) -> active#(false())
p25: mark#(div(X1,X2)) -> active#(div(mark(X1),X2))
p26: mark#(div(X1,X2)) -> div#(mark(X1),X2)
p27: mark#(div(X1,X2)) -> mark#(X1)
p28: mark#(if(X1,X2,X3)) -> active#(if(mark(X1),X2,X3))
p29: mark#(if(X1,X2,X3)) -> if#(mark(X1),X2,X3)
p30: mark#(if(X1,X2,X3)) -> mark#(X1)
p31: minus#(mark(X1),X2) -> minus#(X1,X2)
p32: minus#(X1,mark(X2)) -> minus#(X1,X2)
p33: minus#(active(X1),X2) -> minus#(X1,X2)
p34: minus#(X1,active(X2)) -> minus#(X1,X2)
p35: s#(mark(X)) -> s#(X)
p36: s#(active(X)) -> s#(X)
p37: geq#(mark(X1),X2) -> geq#(X1,X2)
p38: geq#(X1,mark(X2)) -> geq#(X1,X2)
p39: geq#(active(X1),X2) -> geq#(X1,X2)
p40: geq#(X1,active(X2)) -> geq#(X1,X2)
p41: div#(mark(X1),X2) -> div#(X1,X2)
p42: div#(X1,mark(X2)) -> div#(X1,X2)
p43: div#(active(X1),X2) -> div#(X1,X2)
p44: div#(X1,active(X2)) -> div#(X1,X2)
p45: if#(mark(X1),X2,X3) -> if#(X1,X2,X3)
p46: if#(X1,mark(X2),X3) -> if#(X1,X2,X3)
p47: if#(X1,X2,mark(X3)) -> if#(X1,X2,X3)
p48: if#(active(X1),X2,X3) -> if#(X1,X2,X3)
p49: if#(X1,active(X2),X3) -> if#(X1,X2,X3)
p50: if#(X1,X2,active(X3)) -> if#(X1,X2,X3)

and R consists of:

r1: active(minus(|0|(),Y)) -> mark(|0|())
r2: active(minus(s(X),s(Y))) -> mark(minus(X,Y))
r3: active(geq(X,|0|())) -> mark(true())
r4: active(geq(|0|(),s(Y))) -> mark(false())
r5: active(geq(s(X),s(Y))) -> mark(geq(X,Y))
r6: active(div(|0|(),s(Y))) -> mark(|0|())
r7: active(div(s(X),s(Y))) -> mark(if(geq(X,Y),s(div(minus(X,Y),s(Y))),|0|()))
r8: active(if(true(),X,Y)) -> mark(X)
r9: active(if(false(),X,Y)) -> mark(Y)
r10: mark(minus(X1,X2)) -> active(minus(X1,X2))
r11: mark(|0|()) -> active(|0|())
r12: mark(s(X)) -> active(s(mark(X)))
r13: mark(geq(X1,X2)) -> active(geq(X1,X2))
r14: mark(true()) -> active(true())
r15: mark(false()) -> active(false())
r16: mark(div(X1,X2)) -> active(div(mark(X1),X2))
r17: mark(if(X1,X2,X3)) -> active(if(mark(X1),X2,X3))
r18: minus(mark(X1),X2) -> minus(X1,X2)
r19: minus(X1,mark(X2)) -> minus(X1,X2)
r20: minus(active(X1),X2) -> minus(X1,X2)
r21: minus(X1,active(X2)) -> minus(X1,X2)
r22: s(mark(X)) -> s(X)
r23: s(active(X)) -> s(X)
r24: geq(mark(X1),X2) -> geq(X1,X2)
r25: geq(X1,mark(X2)) -> geq(X1,X2)
r26: geq(active(X1),X2) -> geq(X1,X2)
r27: geq(X1,active(X2)) -> geq(X1,X2)
r28: div(mark(X1),X2) -> div(X1,X2)
r29: div(X1,mark(X2)) -> div(X1,X2)
r30: div(active(X1),X2) -> div(X1,X2)
r31: div(X1,active(X2)) -> div(X1,X2)
r32: if(mark(X1),X2,X3) -> if(X1,X2,X3)
r33: if(X1,mark(X2),X3) -> if(X1,X2,X3)
r34: if(X1,X2,mark(X3)) -> if(X1,X2,X3)
r35: if(active(X1),X2,X3) -> if(X1,X2,X3)
r36: if(X1,active(X2),X3) -> if(X1,X2,X3)
r37: if(X1,X2,active(X3)) -> if(X1,X2,X3)

The estimated dependency graph contains the following SCCs:

  {p2, p6, p9, p15, p16, p17, p19, p21, p22, p25, p27, p28, p30}
  {p31, p32, p33, p34}
  {p37, p38, p39, p40}
  {p45, p46, p47, p48, p49, p50}
  {p35, p36}
  {p41, p42, p43, p44}


-- Reduction pair.

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

p1: mark#(if(X1,X2,X3)) -> active#(if(mark(X1),X2,X3))
p2: active#(if(false(),X,Y)) -> mark#(Y)
p3: mark#(if(X1,X2,X3)) -> mark#(X1)
p4: mark#(div(X1,X2)) -> mark#(X1)
p5: mark#(div(X1,X2)) -> active#(div(mark(X1),X2))
p6: active#(if(true(),X,Y)) -> mark#(X)
p7: mark#(geq(X1,X2)) -> active#(geq(X1,X2))
p8: active#(div(s(X),s(Y))) -> mark#(if(geq(X,Y),s(div(minus(X,Y),s(Y))),|0|()))
p9: mark#(s(X)) -> mark#(X)
p10: mark#(s(X)) -> active#(s(mark(X)))
p11: active#(geq(s(X),s(Y))) -> mark#(geq(X,Y))
p12: mark#(minus(X1,X2)) -> active#(minus(X1,X2))
p13: active#(minus(s(X),s(Y))) -> mark#(minus(X,Y))

and R consists of:

r1: active(minus(|0|(),Y)) -> mark(|0|())
r2: active(minus(s(X),s(Y))) -> mark(minus(X,Y))
r3: active(geq(X,|0|())) -> mark(true())
r4: active(geq(|0|(),s(Y))) -> mark(false())
r5: active(geq(s(X),s(Y))) -> mark(geq(X,Y))
r6: active(div(|0|(),s(Y))) -> mark(|0|())
r7: active(div(s(X),s(Y))) -> mark(if(geq(X,Y),s(div(minus(X,Y),s(Y))),|0|()))
r8: active(if(true(),X,Y)) -> mark(X)
r9: active(if(false(),X,Y)) -> mark(Y)
r10: mark(minus(X1,X2)) -> active(minus(X1,X2))
r11: mark(|0|()) -> active(|0|())
r12: mark(s(X)) -> active(s(mark(X)))
r13: mark(geq(X1,X2)) -> active(geq(X1,X2))
r14: mark(true()) -> active(true())
r15: mark(false()) -> active(false())
r16: mark(div(X1,X2)) -> active(div(mark(X1),X2))
r17: mark(if(X1,X2,X3)) -> active(if(mark(X1),X2,X3))
r18: minus(mark(X1),X2) -> minus(X1,X2)
r19: minus(X1,mark(X2)) -> minus(X1,X2)
r20: minus(active(X1),X2) -> minus(X1,X2)
r21: minus(X1,active(X2)) -> minus(X1,X2)
r22: s(mark(X)) -> s(X)
r23: s(active(X)) -> s(X)
r24: geq(mark(X1),X2) -> geq(X1,X2)
r25: geq(X1,mark(X2)) -> geq(X1,X2)
r26: geq(active(X1),X2) -> geq(X1,X2)
r27: geq(X1,active(X2)) -> geq(X1,X2)
r28: div(mark(X1),X2) -> div(X1,X2)
r29: div(X1,mark(X2)) -> div(X1,X2)
r30: div(active(X1),X2) -> div(X1,X2)
r31: div(X1,active(X2)) -> div(X1,X2)
r32: if(mark(X1),X2,X3) -> if(X1,X2,X3)
r33: if(X1,mark(X2),X3) -> if(X1,X2,X3)
r34: if(X1,X2,mark(X3)) -> if(X1,X2,X3)
r35: if(active(X1),X2,X3) -> if(X1,X2,X3)
r36: if(X1,active(X2),X3) -> if(X1,X2,X3)
r37: if(X1,X2,active(X3)) -> if(X1,X2,X3)

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, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32, r33, r34, r35, r36, r37

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = max{12, x1 - 139}
          if_A(x1,x2,x3) = max{774, x1 + 172, x2 + 18, x3 + 170}
          active#_A(x1) = max{10, x1 - 157}
          mark_A(x1) = x1 + 3
          false_A = 1
          div_A(x1,x2) = max{718, x1 + 653}
          true_A = 5
          geq_A(x1,x2) = max{341, x1 + 152}
          s_A(x1) = max{207, x1 + 18}
          minus_A(x1,x2) = max{10, x1 - 37}
          |0|_A = 6
          active_A(x1) = max{4, x1}
    
      precedence:
      
        mark = false = div = minus = |0| = active > geq > if = active# > mark# > true = s
      
      partial status:
    
        pi(mark#) = []
        pi(if) = [1]
        pi(active#) = []
        pi(mark) = []
        pi(false) = []
        pi(div) = []
        pi(true) = []
        pi(geq) = [1]
        pi(s) = []
        pi(minus) = []
        pi(|0|) = []
        pi(active) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = 56
          if_A(x1,x2,x3) = 70
          active#_A(x1) = 56
          mark_A(x1) = 11
          false_A = 55
          div_A(x1,x2) = 65
          true_A = 266
          geq_A(x1,x2) = 68
          s_A(x1) = 35
          minus_A(x1,x2) = 194
          |0|_A = 0
          active_A(x1) = 11
    
      precedence:
      
        false = minus > mark# = if = active# = mark = div = true = geq = s = active > |0|
      
      partial status:
    
        pi(mark#) = []
        pi(if) = []
        pi(active#) = []
        pi(mark) = []
        pi(false) = []
        pi(div) = []
        pi(true) = []
        pi(geq) = []
        pi(s) = []
        pi(minus) = []
        pi(|0|) = []
        pi(active) = []
    

The next rules are strictly ordered:

  p1, p2, p3, p5, p10, p11

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(div(X1,X2)) -> mark#(X1)
p2: active#(if(true(),X,Y)) -> mark#(X)
p3: mark#(geq(X1,X2)) -> active#(geq(X1,X2))
p4: active#(div(s(X),s(Y))) -> mark#(if(geq(X,Y),s(div(minus(X,Y),s(Y))),|0|()))
p5: mark#(s(X)) -> mark#(X)
p6: mark#(minus(X1,X2)) -> active#(minus(X1,X2))
p7: active#(minus(s(X),s(Y))) -> mark#(minus(X,Y))

and R consists of:

r1: active(minus(|0|(),Y)) -> mark(|0|())
r2: active(minus(s(X),s(Y))) -> mark(minus(X,Y))
r3: active(geq(X,|0|())) -> mark(true())
r4: active(geq(|0|(),s(Y))) -> mark(false())
r5: active(geq(s(X),s(Y))) -> mark(geq(X,Y))
r6: active(div(|0|(),s(Y))) -> mark(|0|())
r7: active(div(s(X),s(Y))) -> mark(if(geq(X,Y),s(div(minus(X,Y),s(Y))),|0|()))
r8: active(if(true(),X,Y)) -> mark(X)
r9: active(if(false(),X,Y)) -> mark(Y)
r10: mark(minus(X1,X2)) -> active(minus(X1,X2))
r11: mark(|0|()) -> active(|0|())
r12: mark(s(X)) -> active(s(mark(X)))
r13: mark(geq(X1,X2)) -> active(geq(X1,X2))
r14: mark(true()) -> active(true())
r15: mark(false()) -> active(false())
r16: mark(div(X1,X2)) -> active(div(mark(X1),X2))
r17: mark(if(X1,X2,X3)) -> active(if(mark(X1),X2,X3))
r18: minus(mark(X1),X2) -> minus(X1,X2)
r19: minus(X1,mark(X2)) -> minus(X1,X2)
r20: minus(active(X1),X2) -> minus(X1,X2)
r21: minus(X1,active(X2)) -> minus(X1,X2)
r22: s(mark(X)) -> s(X)
r23: s(active(X)) -> s(X)
r24: geq(mark(X1),X2) -> geq(X1,X2)
r25: geq(X1,mark(X2)) -> geq(X1,X2)
r26: geq(active(X1),X2) -> geq(X1,X2)
r27: geq(X1,active(X2)) -> geq(X1,X2)
r28: div(mark(X1),X2) -> div(X1,X2)
r29: div(X1,mark(X2)) -> div(X1,X2)
r30: div(active(X1),X2) -> div(X1,X2)
r31: div(X1,active(X2)) -> div(X1,X2)
r32: if(mark(X1),X2,X3) -> if(X1,X2,X3)
r33: if(X1,mark(X2),X3) -> if(X1,X2,X3)
r34: if(X1,X2,mark(X3)) -> if(X1,X2,X3)
r35: if(active(X1),X2,X3) -> if(X1,X2,X3)
r36: if(X1,active(X2),X3) -> if(X1,X2,X3)
r37: if(X1,X2,active(X3)) -> if(X1,X2,X3)

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: mark#(div(X1,X2)) -> mark#(X1)
p2: mark#(minus(X1,X2)) -> active#(minus(X1,X2))
p3: active#(minus(s(X),s(Y))) -> mark#(minus(X,Y))
p4: mark#(s(X)) -> mark#(X)
p5: mark#(geq(X1,X2)) -> active#(geq(X1,X2))
p6: active#(div(s(X),s(Y))) -> mark#(if(geq(X,Y),s(div(minus(X,Y),s(Y))),|0|()))
p7: active#(if(true(),X,Y)) -> mark#(X)

and R consists of:

r1: active(minus(|0|(),Y)) -> mark(|0|())
r2: active(minus(s(X),s(Y))) -> mark(minus(X,Y))
r3: active(geq(X,|0|())) -> mark(true())
r4: active(geq(|0|(),s(Y))) -> mark(false())
r5: active(geq(s(X),s(Y))) -> mark(geq(X,Y))
r6: active(div(|0|(),s(Y))) -> mark(|0|())
r7: active(div(s(X),s(Y))) -> mark(if(geq(X,Y),s(div(minus(X,Y),s(Y))),|0|()))
r8: active(if(true(),X,Y)) -> mark(X)
r9: active(if(false(),X,Y)) -> mark(Y)
r10: mark(minus(X1,X2)) -> active(minus(X1,X2))
r11: mark(|0|()) -> active(|0|())
r12: mark(s(X)) -> active(s(mark(X)))
r13: mark(geq(X1,X2)) -> active(geq(X1,X2))
r14: mark(true()) -> active(true())
r15: mark(false()) -> active(false())
r16: mark(div(X1,X2)) -> active(div(mark(X1),X2))
r17: mark(if(X1,X2,X3)) -> active(if(mark(X1),X2,X3))
r18: minus(mark(X1),X2) -> minus(X1,X2)
r19: minus(X1,mark(X2)) -> minus(X1,X2)
r20: minus(active(X1),X2) -> minus(X1,X2)
r21: minus(X1,active(X2)) -> minus(X1,X2)
r22: s(mark(X)) -> s(X)
r23: s(active(X)) -> s(X)
r24: geq(mark(X1),X2) -> geq(X1,X2)
r25: geq(X1,mark(X2)) -> geq(X1,X2)
r26: geq(active(X1),X2) -> geq(X1,X2)
r27: geq(X1,active(X2)) -> geq(X1,X2)
r28: div(mark(X1),X2) -> div(X1,X2)
r29: div(X1,mark(X2)) -> div(X1,X2)
r30: div(active(X1),X2) -> div(X1,X2)
r31: div(X1,active(X2)) -> div(X1,X2)
r32: if(mark(X1),X2,X3) -> if(X1,X2,X3)
r33: if(X1,mark(X2),X3) -> if(X1,X2,X3)
r34: if(X1,X2,mark(X3)) -> if(X1,X2,X3)
r35: if(active(X1),X2,X3) -> if(X1,X2,X3)
r36: if(X1,active(X2),X3) -> if(X1,X2,X3)
r37: if(X1,X2,active(X3)) -> if(X1,X2,X3)

The set of usable rules consists of

  r18, r19, r20, r21, r22, r23, r24, r25, r26, r27, r28, r29, r30, r31, r32, r33, r34, r35, r36, r37

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = x1 + 29
          div_A(x1,x2) = max{107, x1 + 85}
          minus_A(x1,x2) = max{5, x1 - 31}
          active#_A(x1) = max{31, x1 + 28}
          s_A(x1) = max{66, x1 + 13}
          geq_A(x1,x2) = 2
          if_A(x1,x2,x3) = max{122, x1 - 122, x2 + 2, x3 + 84}
          |0|_A = 12
          true_A = 124
          mark_A(x1) = x1 + 66
          active_A(x1) = max{122, x1 + 121}
    
      precedence:
      
        div > mark# = minus = s > geq = |0| > active# = true > if = mark = active
      
      partial status:
    
        pi(mark#) = [1]
        pi(div) = [1]
        pi(minus) = []
        pi(active#) = [1]
        pi(s) = [1]
        pi(geq) = []
        pi(if) = [2, 3]
        pi(|0|) = []
        pi(true) = []
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          mark#_A(x1) = max{3, x1 - 18}
          div_A(x1,x2) = x1 + 48
          minus_A(x1,x2) = 21
          active#_A(x1) = max{3, x1 - 19}
          s_A(x1) = max{41, x1 + 15}
          geq_A(x1,x2) = 101
          if_A(x1,x2,x3) = max{88, x2 + 2, x3 + 43}
          |0|_A = 0
          true_A = 42
          mark_A(x1) = max{26, x1 + 10}
          active_A(x1) = max{26, x1 + 10}
    
      precedence:
      
        minus > active# = s = geq = if = true > mark# = div = |0| = mark = active
      
      partial status:
    
        pi(mark#) = []
        pi(div) = []
        pi(minus) = []
        pi(active#) = []
        pi(s) = [1]
        pi(geq) = []
        pi(if) = [2, 3]
        pi(|0|) = []
        pi(true) = []
        pi(mark) = [1]
        pi(active) = [1]
    

The next rules are strictly ordered:

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

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#(mark(X1),X2) -> minus#(X1,X2)
p2: minus#(X1,active(X2)) -> minus#(X1,X2)
p3: minus#(active(X1),X2) -> minus#(X1,X2)
p4: minus#(X1,mark(X2)) -> minus#(X1,X2)

and R consists of:

r1: active(minus(|0|(),Y)) -> mark(|0|())
r2: active(minus(s(X),s(Y))) -> mark(minus(X,Y))
r3: active(geq(X,|0|())) -> mark(true())
r4: active(geq(|0|(),s(Y))) -> mark(false())
r5: active(geq(s(X),s(Y))) -> mark(geq(X,Y))
r6: active(div(|0|(),s(Y))) -> mark(|0|())
r7: active(div(s(X),s(Y))) -> mark(if(geq(X,Y),s(div(minus(X,Y),s(Y))),|0|()))
r8: active(if(true(),X,Y)) -> mark(X)
r9: active(if(false(),X,Y)) -> mark(Y)
r10: mark(minus(X1,X2)) -> active(minus(X1,X2))
r11: mark(|0|()) -> active(|0|())
r12: mark(s(X)) -> active(s(mark(X)))
r13: mark(geq(X1,X2)) -> active(geq(X1,X2))
r14: mark(true()) -> active(true())
r15: mark(false()) -> active(false())
r16: mark(div(X1,X2)) -> active(div(mark(X1),X2))
r17: mark(if(X1,X2,X3)) -> active(if(mark(X1),X2,X3))
r18: minus(mark(X1),X2) -> minus(X1,X2)
r19: minus(X1,mark(X2)) -> minus(X1,X2)
r20: minus(active(X1),X2) -> minus(X1,X2)
r21: minus(X1,active(X2)) -> minus(X1,X2)
r22: s(mark(X)) -> s(X)
r23: s(active(X)) -> s(X)
r24: geq(mark(X1),X2) -> geq(X1,X2)
r25: geq(X1,mark(X2)) -> geq(X1,X2)
r26: geq(active(X1),X2) -> geq(X1,X2)
r27: geq(X1,active(X2)) -> geq(X1,X2)
r28: div(mark(X1),X2) -> div(X1,X2)
r29: div(X1,mark(X2)) -> div(X1,X2)
r30: div(active(X1),X2) -> div(X1,X2)
r31: div(X1,active(X2)) -> div(X1,X2)
r32: if(mark(X1),X2,X3) -> if(X1,X2,X3)
r33: if(X1,mark(X2),X3) -> if(X1,X2,X3)
r34: if(X1,X2,mark(X3)) -> if(X1,X2,X3)
r35: if(active(X1),X2,X3) -> if(X1,X2,X3)
r36: if(X1,active(X2),X3) -> if(X1,X2,X3)
r37: if(X1,X2,active(X3)) -> if(X1,X2,X3)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          minus#_A(x1,x2) = max{4, x1 + 1, x2 + 3}
          mark_A(x1) = max{3, x1 + 2}
          active_A(x1) = max{1, x1}
    
      precedence:
      
        minus# = mark = active
      
      partial status:
    
        pi(minus#) = [1, 2]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          minus#_A(x1,x2) = max{x1 + 1, x2 + 1}
          mark_A(x1) = x1
          active_A(x1) = x1
    
      precedence:
      
        minus# = mark = active
      
      partial status:
    
        pi(minus#) = [1, 2]
        pi(mark) = [1]
        pi(active) = [1]
    

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: geq#(mark(X1),X2) -> geq#(X1,X2)
p2: geq#(X1,active(X2)) -> geq#(X1,X2)
p3: geq#(active(X1),X2) -> geq#(X1,X2)
p4: geq#(X1,mark(X2)) -> geq#(X1,X2)

and R consists of:

r1: active(minus(|0|(),Y)) -> mark(|0|())
r2: active(minus(s(X),s(Y))) -> mark(minus(X,Y))
r3: active(geq(X,|0|())) -> mark(true())
r4: active(geq(|0|(),s(Y))) -> mark(false())
r5: active(geq(s(X),s(Y))) -> mark(geq(X,Y))
r6: active(div(|0|(),s(Y))) -> mark(|0|())
r7: active(div(s(X),s(Y))) -> mark(if(geq(X,Y),s(div(minus(X,Y),s(Y))),|0|()))
r8: active(if(true(),X,Y)) -> mark(X)
r9: active(if(false(),X,Y)) -> mark(Y)
r10: mark(minus(X1,X2)) -> active(minus(X1,X2))
r11: mark(|0|()) -> active(|0|())
r12: mark(s(X)) -> active(s(mark(X)))
r13: mark(geq(X1,X2)) -> active(geq(X1,X2))
r14: mark(true()) -> active(true())
r15: mark(false()) -> active(false())
r16: mark(div(X1,X2)) -> active(div(mark(X1),X2))
r17: mark(if(X1,X2,X3)) -> active(if(mark(X1),X2,X3))
r18: minus(mark(X1),X2) -> minus(X1,X2)
r19: minus(X1,mark(X2)) -> minus(X1,X2)
r20: minus(active(X1),X2) -> minus(X1,X2)
r21: minus(X1,active(X2)) -> minus(X1,X2)
r22: s(mark(X)) -> s(X)
r23: s(active(X)) -> s(X)
r24: geq(mark(X1),X2) -> geq(X1,X2)
r25: geq(X1,mark(X2)) -> geq(X1,X2)
r26: geq(active(X1),X2) -> geq(X1,X2)
r27: geq(X1,active(X2)) -> geq(X1,X2)
r28: div(mark(X1),X2) -> div(X1,X2)
r29: div(X1,mark(X2)) -> div(X1,X2)
r30: div(active(X1),X2) -> div(X1,X2)
r31: div(X1,active(X2)) -> div(X1,X2)
r32: if(mark(X1),X2,X3) -> if(X1,X2,X3)
r33: if(X1,mark(X2),X3) -> if(X1,X2,X3)
r34: if(X1,X2,mark(X3)) -> if(X1,X2,X3)
r35: if(active(X1),X2,X3) -> if(X1,X2,X3)
r36: if(X1,active(X2),X3) -> if(X1,X2,X3)
r37: if(X1,X2,active(X3)) -> if(X1,X2,X3)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          geq#_A(x1,x2) = max{2, x1 + 1, x2 + 1}
          mark_A(x1) = max{1, x1}
          active_A(x1) = max{1, x1}
    
      precedence:
      
        geq# = mark = active
      
      partial status:
    
        pi(geq#) = [1, 2]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          geq#_A(x1,x2) = max{x1 + 1, x2 - 1}
          mark_A(x1) = x1
          active_A(x1) = x1
    
      precedence:
      
        geq# = mark = active
      
      partial status:
    
        pi(geq#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    

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: if#(mark(X1),X2,X3) -> if#(X1,X2,X3)
p2: if#(X1,X2,active(X3)) -> if#(X1,X2,X3)
p3: if#(X1,active(X2),X3) -> if#(X1,X2,X3)
p4: if#(active(X1),X2,X3) -> if#(X1,X2,X3)
p5: if#(X1,X2,mark(X3)) -> if#(X1,X2,X3)
p6: if#(X1,mark(X2),X3) -> if#(X1,X2,X3)

and R consists of:

r1: active(minus(|0|(),Y)) -> mark(|0|())
r2: active(minus(s(X),s(Y))) -> mark(minus(X,Y))
r3: active(geq(X,|0|())) -> mark(true())
r4: active(geq(|0|(),s(Y))) -> mark(false())
r5: active(geq(s(X),s(Y))) -> mark(geq(X,Y))
r6: active(div(|0|(),s(Y))) -> mark(|0|())
r7: active(div(s(X),s(Y))) -> mark(if(geq(X,Y),s(div(minus(X,Y),s(Y))),|0|()))
r8: active(if(true(),X,Y)) -> mark(X)
r9: active(if(false(),X,Y)) -> mark(Y)
r10: mark(minus(X1,X2)) -> active(minus(X1,X2))
r11: mark(|0|()) -> active(|0|())
r12: mark(s(X)) -> active(s(mark(X)))
r13: mark(geq(X1,X2)) -> active(geq(X1,X2))
r14: mark(true()) -> active(true())
r15: mark(false()) -> active(false())
r16: mark(div(X1,X2)) -> active(div(mark(X1),X2))
r17: mark(if(X1,X2,X3)) -> active(if(mark(X1),X2,X3))
r18: minus(mark(X1),X2) -> minus(X1,X2)
r19: minus(X1,mark(X2)) -> minus(X1,X2)
r20: minus(active(X1),X2) -> minus(X1,X2)
r21: minus(X1,active(X2)) -> minus(X1,X2)
r22: s(mark(X)) -> s(X)
r23: s(active(X)) -> s(X)
r24: geq(mark(X1),X2) -> geq(X1,X2)
r25: geq(X1,mark(X2)) -> geq(X1,X2)
r26: geq(active(X1),X2) -> geq(X1,X2)
r27: geq(X1,active(X2)) -> geq(X1,X2)
r28: div(mark(X1),X2) -> div(X1,X2)
r29: div(X1,mark(X2)) -> div(X1,X2)
r30: div(active(X1),X2) -> div(X1,X2)
r31: div(X1,active(X2)) -> div(X1,X2)
r32: if(mark(X1),X2,X3) -> if(X1,X2,X3)
r33: if(X1,mark(X2),X3) -> if(X1,X2,X3)
r34: if(X1,X2,mark(X3)) -> if(X1,X2,X3)
r35: if(active(X1),X2,X3) -> if(X1,X2,X3)
r36: if(X1,active(X2),X3) -> if(X1,X2,X3)
r37: if(X1,X2,active(X3)) -> if(X1,X2,X3)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if#_A(x1,x2,x3) = max{3, x1 + 1, x2 + 1, x3 + 1}
          mark_A(x1) = max{2, x1 + 1}
          active_A(x1) = max{1, x1}
    
      precedence:
      
        if# = mark = active
      
      partial status:
    
        pi(if#) = [1, 2]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if#_A(x1,x2,x3) = max{2, x1 + 1, x2 + 1}
          mark_A(x1) = max{1, x1}
          active_A(x1) = x1 + 1
    
      precedence:
      
        if# = mark = active
      
      partial status:
    
        pi(if#) = [1, 2]
        pi(mark) = [1]
        pi(active) = []
    

The next rules are strictly ordered:

  p1, p3, p4, p6

We remove them from the problem.

-- SCC decomposition.

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

p1: if#(X1,X2,active(X3)) -> if#(X1,X2,X3)
p2: if#(X1,X2,mark(X3)) -> if#(X1,X2,X3)

and R consists of:

r1: active(minus(|0|(),Y)) -> mark(|0|())
r2: active(minus(s(X),s(Y))) -> mark(minus(X,Y))
r3: active(geq(X,|0|())) -> mark(true())
r4: active(geq(|0|(),s(Y))) -> mark(false())
r5: active(geq(s(X),s(Y))) -> mark(geq(X,Y))
r6: active(div(|0|(),s(Y))) -> mark(|0|())
r7: active(div(s(X),s(Y))) -> mark(if(geq(X,Y),s(div(minus(X,Y),s(Y))),|0|()))
r8: active(if(true(),X,Y)) -> mark(X)
r9: active(if(false(),X,Y)) -> mark(Y)
r10: mark(minus(X1,X2)) -> active(minus(X1,X2))
r11: mark(|0|()) -> active(|0|())
r12: mark(s(X)) -> active(s(mark(X)))
r13: mark(geq(X1,X2)) -> active(geq(X1,X2))
r14: mark(true()) -> active(true())
r15: mark(false()) -> active(false())
r16: mark(div(X1,X2)) -> active(div(mark(X1),X2))
r17: mark(if(X1,X2,X3)) -> active(if(mark(X1),X2,X3))
r18: minus(mark(X1),X2) -> minus(X1,X2)
r19: minus(X1,mark(X2)) -> minus(X1,X2)
r20: minus(active(X1),X2) -> minus(X1,X2)
r21: minus(X1,active(X2)) -> minus(X1,X2)
r22: s(mark(X)) -> s(X)
r23: s(active(X)) -> s(X)
r24: geq(mark(X1),X2) -> geq(X1,X2)
r25: geq(X1,mark(X2)) -> geq(X1,X2)
r26: geq(active(X1),X2) -> geq(X1,X2)
r27: geq(X1,active(X2)) -> geq(X1,X2)
r28: div(mark(X1),X2) -> div(X1,X2)
r29: div(X1,mark(X2)) -> div(X1,X2)
r30: div(active(X1),X2) -> div(X1,X2)
r31: div(X1,active(X2)) -> div(X1,X2)
r32: if(mark(X1),X2,X3) -> if(X1,X2,X3)
r33: if(X1,mark(X2),X3) -> if(X1,X2,X3)
r34: if(X1,X2,mark(X3)) -> if(X1,X2,X3)
r35: if(active(X1),X2,X3) -> if(X1,X2,X3)
r36: if(X1,active(X2),X3) -> if(X1,X2,X3)
r37: if(X1,X2,active(X3)) -> if(X1,X2,X3)

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: if#(X1,X2,active(X3)) -> if#(X1,X2,X3)
p2: if#(X1,X2,mark(X3)) -> if#(X1,X2,X3)

and R consists of:

r1: active(minus(|0|(),Y)) -> mark(|0|())
r2: active(minus(s(X),s(Y))) -> mark(minus(X,Y))
r3: active(geq(X,|0|())) -> mark(true())
r4: active(geq(|0|(),s(Y))) -> mark(false())
r5: active(geq(s(X),s(Y))) -> mark(geq(X,Y))
r6: active(div(|0|(),s(Y))) -> mark(|0|())
r7: active(div(s(X),s(Y))) -> mark(if(geq(X,Y),s(div(minus(X,Y),s(Y))),|0|()))
r8: active(if(true(),X,Y)) -> mark(X)
r9: active(if(false(),X,Y)) -> mark(Y)
r10: mark(minus(X1,X2)) -> active(minus(X1,X2))
r11: mark(|0|()) -> active(|0|())
r12: mark(s(X)) -> active(s(mark(X)))
r13: mark(geq(X1,X2)) -> active(geq(X1,X2))
r14: mark(true()) -> active(true())
r15: mark(false()) -> active(false())
r16: mark(div(X1,X2)) -> active(div(mark(X1),X2))
r17: mark(if(X1,X2,X3)) -> active(if(mark(X1),X2,X3))
r18: minus(mark(X1),X2) -> minus(X1,X2)
r19: minus(X1,mark(X2)) -> minus(X1,X2)
r20: minus(active(X1),X2) -> minus(X1,X2)
r21: minus(X1,active(X2)) -> minus(X1,X2)
r22: s(mark(X)) -> s(X)
r23: s(active(X)) -> s(X)
r24: geq(mark(X1),X2) -> geq(X1,X2)
r25: geq(X1,mark(X2)) -> geq(X1,X2)
r26: geq(active(X1),X2) -> geq(X1,X2)
r27: geq(X1,active(X2)) -> geq(X1,X2)
r28: div(mark(X1),X2) -> div(X1,X2)
r29: div(X1,mark(X2)) -> div(X1,X2)
r30: div(active(X1),X2) -> div(X1,X2)
r31: div(X1,active(X2)) -> div(X1,X2)
r32: if(mark(X1),X2,X3) -> if(X1,X2,X3)
r33: if(X1,mark(X2),X3) -> if(X1,X2,X3)
r34: if(X1,X2,mark(X3)) -> if(X1,X2,X3)
r35: if(active(X1),X2,X3) -> if(X1,X2,X3)
r36: if(X1,active(X2),X3) -> if(X1,X2,X3)
r37: if(X1,X2,active(X3)) -> if(X1,X2,X3)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if#_A(x1,x2,x3) = max{2, x1, x2 + 1, x3 + 1}
          active_A(x1) = max{1, x1}
          mark_A(x1) = max{1, x1}
    
      precedence:
      
        if# = active = mark
      
      partial status:
    
        pi(if#) = [1, 3]
        pi(active) = [1]
        pi(mark) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          if#_A(x1,x2,x3) = max{0, x3 - 2}
          active_A(x1) = max{3, x1 + 1}
          mark_A(x1) = x1
    
      precedence:
      
        if# = active = mark
      
      partial status:
    
        pi(if#) = []
        pi(active) = [1]
        pi(mark) = [1]
    

The next rules are strictly ordered:

  p1, p2

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: s#(mark(X)) -> s#(X)
p2: s#(active(X)) -> s#(X)

and R consists of:

r1: active(minus(|0|(),Y)) -> mark(|0|())
r2: active(minus(s(X),s(Y))) -> mark(minus(X,Y))
r3: active(geq(X,|0|())) -> mark(true())
r4: active(geq(|0|(),s(Y))) -> mark(false())
r5: active(geq(s(X),s(Y))) -> mark(geq(X,Y))
r6: active(div(|0|(),s(Y))) -> mark(|0|())
r7: active(div(s(X),s(Y))) -> mark(if(geq(X,Y),s(div(minus(X,Y),s(Y))),|0|()))
r8: active(if(true(),X,Y)) -> mark(X)
r9: active(if(false(),X,Y)) -> mark(Y)
r10: mark(minus(X1,X2)) -> active(minus(X1,X2))
r11: mark(|0|()) -> active(|0|())
r12: mark(s(X)) -> active(s(mark(X)))
r13: mark(geq(X1,X2)) -> active(geq(X1,X2))
r14: mark(true()) -> active(true())
r15: mark(false()) -> active(false())
r16: mark(div(X1,X2)) -> active(div(mark(X1),X2))
r17: mark(if(X1,X2,X3)) -> active(if(mark(X1),X2,X3))
r18: minus(mark(X1),X2) -> minus(X1,X2)
r19: minus(X1,mark(X2)) -> minus(X1,X2)
r20: minus(active(X1),X2) -> minus(X1,X2)
r21: minus(X1,active(X2)) -> minus(X1,X2)
r22: s(mark(X)) -> s(X)
r23: s(active(X)) -> s(X)
r24: geq(mark(X1),X2) -> geq(X1,X2)
r25: geq(X1,mark(X2)) -> geq(X1,X2)
r26: geq(active(X1),X2) -> geq(X1,X2)
r27: geq(X1,active(X2)) -> geq(X1,X2)
r28: div(mark(X1),X2) -> div(X1,X2)
r29: div(X1,mark(X2)) -> div(X1,X2)
r30: div(active(X1),X2) -> div(X1,X2)
r31: div(X1,active(X2)) -> div(X1,X2)
r32: if(mark(X1),X2,X3) -> if(X1,X2,X3)
r33: if(X1,mark(X2),X3) -> if(X1,X2,X3)
r34: if(X1,X2,mark(X3)) -> if(X1,X2,X3)
r35: if(active(X1),X2,X3) -> if(X1,X2,X3)
r36: if(X1,active(X2),X3) -> if(X1,X2,X3)
r37: if(X1,X2,active(X3)) -> if(X1,X2,X3)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          s#_A(x1) = max{6, x1 + 4}
          mark_A(x1) = max{4, x1 + 3}
          active_A(x1) = max{2, x1 + 1}
    
      precedence:
      
        s# = mark = active
      
      partial status:
    
        pi(s#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          s#_A(x1) = x1 + 1
          mark_A(x1) = x1
          active_A(x1) = x1
    
      precedence:
      
        s# = mark = active
      
      partial status:
    
        pi(s#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    

The next rules are strictly ordered:

  p1, p2

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: div#(mark(X1),X2) -> div#(X1,X2)
p2: div#(X1,active(X2)) -> div#(X1,X2)
p3: div#(active(X1),X2) -> div#(X1,X2)
p4: div#(X1,mark(X2)) -> div#(X1,X2)

and R consists of:

r1: active(minus(|0|(),Y)) -> mark(|0|())
r2: active(minus(s(X),s(Y))) -> mark(minus(X,Y))
r3: active(geq(X,|0|())) -> mark(true())
r4: active(geq(|0|(),s(Y))) -> mark(false())
r5: active(geq(s(X),s(Y))) -> mark(geq(X,Y))
r6: active(div(|0|(),s(Y))) -> mark(|0|())
r7: active(div(s(X),s(Y))) -> mark(if(geq(X,Y),s(div(minus(X,Y),s(Y))),|0|()))
r8: active(if(true(),X,Y)) -> mark(X)
r9: active(if(false(),X,Y)) -> mark(Y)
r10: mark(minus(X1,X2)) -> active(minus(X1,X2))
r11: mark(|0|()) -> active(|0|())
r12: mark(s(X)) -> active(s(mark(X)))
r13: mark(geq(X1,X2)) -> active(geq(X1,X2))
r14: mark(true()) -> active(true())
r15: mark(false()) -> active(false())
r16: mark(div(X1,X2)) -> active(div(mark(X1),X2))
r17: mark(if(X1,X2,X3)) -> active(if(mark(X1),X2,X3))
r18: minus(mark(X1),X2) -> minus(X1,X2)
r19: minus(X1,mark(X2)) -> minus(X1,X2)
r20: minus(active(X1),X2) -> minus(X1,X2)
r21: minus(X1,active(X2)) -> minus(X1,X2)
r22: s(mark(X)) -> s(X)
r23: s(active(X)) -> s(X)
r24: geq(mark(X1),X2) -> geq(X1,X2)
r25: geq(X1,mark(X2)) -> geq(X1,X2)
r26: geq(active(X1),X2) -> geq(X1,X2)
r27: geq(X1,active(X2)) -> geq(X1,X2)
r28: div(mark(X1),X2) -> div(X1,X2)
r29: div(X1,mark(X2)) -> div(X1,X2)
r30: div(active(X1),X2) -> div(X1,X2)
r31: div(X1,active(X2)) -> div(X1,X2)
r32: if(mark(X1),X2,X3) -> if(X1,X2,X3)
r33: if(X1,mark(X2),X3) -> if(X1,X2,X3)
r34: if(X1,X2,mark(X3)) -> if(X1,X2,X3)
r35: if(active(X1),X2,X3) -> if(X1,X2,X3)
r36: if(X1,active(X2),X3) -> if(X1,X2,X3)
r37: if(X1,X2,active(X3)) -> if(X1,X2,X3)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          div#_A(x1,x2) = max{2, x1 + 1, x2 + 1}
          mark_A(x1) = max{1, x1}
          active_A(x1) = max{1, x1}
    
      precedence:
      
        div# = mark = active
      
      partial status:
    
        pi(div#) = [1, 2]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          div#_A(x1,x2) = max{x1 + 1, x2 - 1}
          mark_A(x1) = x1
          active_A(x1) = x1
    
      precedence:
      
        div# = mark = active
      
      partial status:
    
        pi(div#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    

The next rules are strictly ordered:

  p1, p2, p3, p4

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