YES

We show the termination of the TRS R:

  active(f(X)) -> mark(if(X,c(),f(true())))
  active(if(true(),X,Y)) -> mark(X)
  active(if(false(),X,Y)) -> mark(Y)
  mark(f(X)) -> active(f(mark(X)))
  mark(if(X1,X2,X3)) -> active(if(mark(X1),mark(X2),X3))
  mark(c()) -> active(c())
  mark(true()) -> active(true())
  mark(false()) -> active(false())
  f(mark(X)) -> f(X)
  f(active(X)) -> f(X)
  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#(f(X)) -> mark#(if(X,c(),f(true())))
p2: active#(f(X)) -> if#(X,c(),f(true()))
p3: active#(f(X)) -> f#(true())
p4: active#(if(true(),X,Y)) -> mark#(X)
p5: active#(if(false(),X,Y)) -> mark#(Y)
p6: mark#(f(X)) -> active#(f(mark(X)))
p7: mark#(f(X)) -> f#(mark(X))
p8: mark#(f(X)) -> mark#(X)
p9: mark#(if(X1,X2,X3)) -> active#(if(mark(X1),mark(X2),X3))
p10: mark#(if(X1,X2,X3)) -> if#(mark(X1),mark(X2),X3)
p11: mark#(if(X1,X2,X3)) -> mark#(X1)
p12: mark#(if(X1,X2,X3)) -> mark#(X2)
p13: mark#(c()) -> active#(c())
p14: mark#(true()) -> active#(true())
p15: mark#(false()) -> active#(false())
p16: f#(mark(X)) -> f#(X)
p17: f#(active(X)) -> f#(X)
p18: if#(mark(X1),X2,X3) -> if#(X1,X2,X3)
p19: if#(X1,mark(X2),X3) -> if#(X1,X2,X3)
p20: if#(X1,X2,mark(X3)) -> if#(X1,X2,X3)
p21: if#(active(X1),X2,X3) -> if#(X1,X2,X3)
p22: if#(X1,active(X2),X3) -> if#(X1,X2,X3)
p23: if#(X1,X2,active(X3)) -> if#(X1,X2,X3)

and R consists of:

r1: active(f(X)) -> mark(if(X,c(),f(true())))
r2: active(if(true(),X,Y)) -> mark(X)
r3: active(if(false(),X,Y)) -> mark(Y)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(if(X1,X2,X3)) -> active(if(mark(X1),mark(X2),X3))
r6: mark(c()) -> active(c())
r7: mark(true()) -> active(true())
r8: mark(false()) -> active(false())
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: if(mark(X1),X2,X3) -> if(X1,X2,X3)
r12: if(X1,mark(X2),X3) -> if(X1,X2,X3)
r13: if(X1,X2,mark(X3)) -> if(X1,X2,X3)
r14: if(active(X1),X2,X3) -> if(X1,X2,X3)
r15: if(X1,active(X2),X3) -> if(X1,X2,X3)
r16: if(X1,X2,active(X3)) -> if(X1,X2,X3)

The estimated dependency graph contains the following SCCs:

  {p1, p4, p5, p6, p8, p9, p11, p12}
  {p18, p19, p20, p21, p22, p23}
  {p16, p17}


-- Reduction pair.

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

p1: active#(f(X)) -> mark#(if(X,c(),f(true())))
p2: mark#(if(X1,X2,X3)) -> mark#(X2)
p3: mark#(if(X1,X2,X3)) -> mark#(X1)
p4: mark#(if(X1,X2,X3)) -> active#(if(mark(X1),mark(X2),X3))
p5: active#(if(false(),X,Y)) -> mark#(Y)
p6: mark#(f(X)) -> mark#(X)
p7: mark#(f(X)) -> active#(f(mark(X)))
p8: active#(if(true(),X,Y)) -> mark#(X)

and R consists of:

r1: active(f(X)) -> mark(if(X,c(),f(true())))
r2: active(if(true(),X,Y)) -> mark(X)
r3: active(if(false(),X,Y)) -> mark(Y)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(if(X1,X2,X3)) -> active(if(mark(X1),mark(X2),X3))
r6: mark(c()) -> active(c())
r7: mark(true()) -> active(true())
r8: mark(false()) -> active(false())
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: if(mark(X1),X2,X3) -> if(X1,X2,X3)
r12: if(X1,mark(X2),X3) -> if(X1,X2,X3)
r13: if(X1,X2,mark(X3)) -> if(X1,X2,X3)
r14: if(active(X1),X2,X3) -> if(X1,X2,X3)
r15: if(X1,active(X2),X3) -> if(X1,X2,X3)
r16: 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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = x1 + 2
          f_A(x1) = x1 + 5
          mark#_A(x1) = x1 + 2
          if_A(x1,x2,x3) = max{x1 + 4, x2 + 2, x3}
          c_A = 0
          true_A = 0
          mark_A(x1) = x1
          false_A = 1
          active_A(x1) = x1
    
      precedence:
      
        f = if = true > active# = mark# = c = mark = false = active
      
      partial status:
    
        pi(active#) = [1]
        pi(f) = []
        pi(mark#) = [1]
        pi(if) = []
        pi(c) = []
        pi(true) = []
        pi(mark) = []
        pi(false) = []
        pi(active) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = 55
          f_A(x1) = 54
          mark#_A(x1) = 55
          if_A(x1,x2,x3) = 34
          c_A = 60
          true_A = 60
          mark_A(x1) = 53
          false_A = 52
          active_A(x1) = 53
    
      precedence:
      
        c > active# = mark# = true = false > f = if = mark = active
      
      partial status:
    
        pi(active#) = []
        pi(f) = []
        pi(mark#) = []
        pi(if) = []
        pi(c) = []
        pi(true) = []
        pi(mark) = []
        pi(false) = []
        pi(active) = []
    

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: active#(f(X)) -> mark#(if(X,c(),f(true())))
p2: mark#(if(X1,X2,X3)) -> mark#(X2)
p3: mark#(if(X1,X2,X3)) -> mark#(X1)
p4: mark#(if(X1,X2,X3)) -> active#(if(mark(X1),mark(X2),X3))
p5: active#(if(false(),X,Y)) -> mark#(Y)
p6: mark#(f(X)) -> mark#(X)
p7: mark#(f(X)) -> active#(f(mark(X)))

and R consists of:

r1: active(f(X)) -> mark(if(X,c(),f(true())))
r2: active(if(true(),X,Y)) -> mark(X)
r3: active(if(false(),X,Y)) -> mark(Y)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(if(X1,X2,X3)) -> active(if(mark(X1),mark(X2),X3))
r6: mark(c()) -> active(c())
r7: mark(true()) -> active(true())
r8: mark(false()) -> active(false())
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: if(mark(X1),X2,X3) -> if(X1,X2,X3)
r12: if(X1,mark(X2),X3) -> if(X1,X2,X3)
r13: if(X1,X2,mark(X3)) -> if(X1,X2,X3)
r14: if(active(X1),X2,X3) -> if(X1,X2,X3)
r15: if(X1,active(X2),X3) -> if(X1,X2,X3)
r16: 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: active#(f(X)) -> mark#(if(X,c(),f(true())))
p2: mark#(f(X)) -> active#(f(mark(X)))
p3: active#(if(false(),X,Y)) -> mark#(Y)
p4: mark#(f(X)) -> mark#(X)
p5: mark#(if(X1,X2,X3)) -> active#(if(mark(X1),mark(X2),X3))
p6: mark#(if(X1,X2,X3)) -> mark#(X1)
p7: mark#(if(X1,X2,X3)) -> mark#(X2)

and R consists of:

r1: active(f(X)) -> mark(if(X,c(),f(true())))
r2: active(if(true(),X,Y)) -> mark(X)
r3: active(if(false(),X,Y)) -> mark(Y)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(if(X1,X2,X3)) -> active(if(mark(X1),mark(X2),X3))
r6: mark(c()) -> active(c())
r7: mark(true()) -> active(true())
r8: mark(false()) -> active(false())
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: if(mark(X1),X2,X3) -> if(X1,X2,X3)
r12: if(X1,mark(X2),X3) -> if(X1,X2,X3)
r13: if(X1,X2,mark(X3)) -> if(X1,X2,X3)
r14: if(active(X1),X2,X3) -> if(X1,X2,X3)
r15: if(X1,active(X2),X3) -> if(X1,X2,X3)
r16: 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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = x1 + 3
          f_A(x1) = max{16, x1 + 12}
          mark#_A(x1) = max{4, x1 + 3}
          if_A(x1,x2,x3) = max{x1 + 6, x2 + 7, x3}
          c_A = 1
          true_A = 0
          mark_A(x1) = x1
          false_A = 0
          active_A(x1) = x1
    
      precedence:
      
        f = if > active# = mark# = c = true = mark = false = active
      
      partial status:
    
        pi(active#) = [1]
        pi(f) = []
        pi(mark#) = [1]
        pi(if) = []
        pi(c) = []
        pi(true) = []
        pi(mark) = []
        pi(false) = []
        pi(active) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = 15
          f_A(x1) = 0
          mark#_A(x1) = 15
          if_A(x1,x2,x3) = 16
          c_A = 59
          true_A = 59
          mark_A(x1) = 57
          false_A = 36
          active_A(x1) = 57
    
      precedence:
      
        true = mark = active > c > active# = f = mark# = if = false
      
      partial status:
    
        pi(active#) = []
        pi(f) = []
        pi(mark#) = []
        pi(if) = []
        pi(c) = []
        pi(true) = []
        pi(mark) = []
        pi(false) = []
        pi(active) = []
    

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: active#(f(X)) -> mark#(if(X,c(),f(true())))
p2: mark#(f(X)) -> active#(f(mark(X)))
p3: active#(if(false(),X,Y)) -> mark#(Y)
p4: mark#(f(X)) -> mark#(X)
p5: mark#(if(X1,X2,X3)) -> active#(if(mark(X1),mark(X2),X3))
p6: mark#(if(X1,X2,X3)) -> mark#(X1)

and R consists of:

r1: active(f(X)) -> mark(if(X,c(),f(true())))
r2: active(if(true(),X,Y)) -> mark(X)
r3: active(if(false(),X,Y)) -> mark(Y)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(if(X1,X2,X3)) -> active(if(mark(X1),mark(X2),X3))
r6: mark(c()) -> active(c())
r7: mark(true()) -> active(true())
r8: mark(false()) -> active(false())
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: if(mark(X1),X2,X3) -> if(X1,X2,X3)
r12: if(X1,mark(X2),X3) -> if(X1,X2,X3)
r13: if(X1,X2,mark(X3)) -> if(X1,X2,X3)
r14: if(active(X1),X2,X3) -> if(X1,X2,X3)
r15: if(X1,active(X2),X3) -> if(X1,X2,X3)
r16: if(X1,X2,active(X3)) -> if(X1,X2,X3)

The estimated dependency graph contains the following SCCs:

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


-- Reduction pair.

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

p1: active#(f(X)) -> mark#(if(X,c(),f(true())))
p2: mark#(if(X1,X2,X3)) -> mark#(X1)
p3: mark#(if(X1,X2,X3)) -> active#(if(mark(X1),mark(X2),X3))
p4: active#(if(false(),X,Y)) -> mark#(Y)
p5: mark#(f(X)) -> mark#(X)
p6: mark#(f(X)) -> active#(f(mark(X)))

and R consists of:

r1: active(f(X)) -> mark(if(X,c(),f(true())))
r2: active(if(true(),X,Y)) -> mark(X)
r3: active(if(false(),X,Y)) -> mark(Y)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(if(X1,X2,X3)) -> active(if(mark(X1),mark(X2),X3))
r6: mark(c()) -> active(c())
r7: mark(true()) -> active(true())
r8: mark(false()) -> active(false())
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: if(mark(X1),X2,X3) -> if(X1,X2,X3)
r12: if(X1,mark(X2),X3) -> if(X1,X2,X3)
r13: if(X1,X2,mark(X3)) -> if(X1,X2,X3)
r14: if(active(X1),X2,X3) -> if(X1,X2,X3)
r15: if(X1,active(X2),X3) -> if(X1,X2,X3)
r16: 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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = max{0, x1 - 4}
          f_A(x1) = max{2, x1}
          mark#_A(x1) = max{0, x1 - 4}
          if_A(x1,x2,x3) = max{x1, x2, x3 + 1}
          c_A = 3
          true_A = 0
          mark_A(x1) = max{3, x1}
          false_A = 5
          active_A(x1) = max{3, x1}
    
      precedence:
      
        active# = f = mark# = if = c = true = mark = false = active
      
      partial status:
    
        pi(active#) = []
        pi(f) = []
        pi(mark#) = []
        pi(if) = []
        pi(c) = []
        pi(true) = []
        pi(mark) = []
        pi(false) = []
        pi(active) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = 97
          f_A(x1) = 91
          mark#_A(x1) = 97
          if_A(x1,x2,x3) = 3
          c_A = 56
          true_A = 58
          mark_A(x1) = 21
          false_A = 20
          active_A(x1) = 21
    
      precedence:
      
        active# = mark# > f = if = mark = active > true = false > c
      
      partial status:
    
        pi(active#) = []
        pi(f) = []
        pi(mark#) = []
        pi(if) = []
        pi(c) = []
        pi(true) = []
        pi(mark) = []
        pi(false) = []
        pi(active) = []
    

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: active#(f(X)) -> mark#(if(X,c(),f(true())))
p2: mark#(if(X1,X2,X3)) -> mark#(X1)
p3: mark#(if(X1,X2,X3)) -> active#(if(mark(X1),mark(X2),X3))
p4: mark#(f(X)) -> mark#(X)
p5: mark#(f(X)) -> active#(f(mark(X)))

and R consists of:

r1: active(f(X)) -> mark(if(X,c(),f(true())))
r2: active(if(true(),X,Y)) -> mark(X)
r3: active(if(false(),X,Y)) -> mark(Y)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(if(X1,X2,X3)) -> active(if(mark(X1),mark(X2),X3))
r6: mark(c()) -> active(c())
r7: mark(true()) -> active(true())
r8: mark(false()) -> active(false())
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: if(mark(X1),X2,X3) -> if(X1,X2,X3)
r12: if(X1,mark(X2),X3) -> if(X1,X2,X3)
r13: if(X1,X2,mark(X3)) -> if(X1,X2,X3)
r14: if(active(X1),X2,X3) -> if(X1,X2,X3)
r15: if(X1,active(X2),X3) -> if(X1,X2,X3)
r16: if(X1,X2,active(X3)) -> if(X1,X2,X3)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3, p4, p5}


-- Reduction pair.

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

p1: active#(f(X)) -> mark#(if(X,c(),f(true())))
p2: mark#(f(X)) -> active#(f(mark(X)))
p3: mark#(f(X)) -> mark#(X)
p4: mark#(if(X1,X2,X3)) -> active#(if(mark(X1),mark(X2),X3))
p5: mark#(if(X1,X2,X3)) -> mark#(X1)

and R consists of:

r1: active(f(X)) -> mark(if(X,c(),f(true())))
r2: active(if(true(),X,Y)) -> mark(X)
r3: active(if(false(),X,Y)) -> mark(Y)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(if(X1,X2,X3)) -> active(if(mark(X1),mark(X2),X3))
r6: mark(c()) -> active(c())
r7: mark(true()) -> active(true())
r8: mark(false()) -> active(false())
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: if(mark(X1),X2,X3) -> if(X1,X2,X3)
r12: if(X1,mark(X2),X3) -> if(X1,X2,X3)
r13: if(X1,X2,mark(X3)) -> if(X1,X2,X3)
r14: if(active(X1),X2,X3) -> if(X1,X2,X3)
r15: if(X1,active(X2),X3) -> if(X1,X2,X3)
r16: 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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = max{1, x1}
          f_A(x1) = x1 + 5
          mark#_A(x1) = x1
          if_A(x1,x2,x3) = max{x1 + 2, x2 + 3, x3}
          c_A = 2
          true_A = 0
          mark_A(x1) = x1
          active_A(x1) = x1
          false_A = 1
    
      precedence:
      
        f > mark# > active# > c > false > if = true = mark = active
      
      partial status:
    
        pi(active#) = [1]
        pi(f) = []
        pi(mark#) = [1]
        pi(if) = []
        pi(c) = []
        pi(true) = []
        pi(mark) = []
        pi(active) = []
        pi(false) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = max{7, x1}
          f_A(x1) = 93
          mark#_A(x1) = x1 + 38
          if_A(x1,x2,x3) = 7
          c_A = 99
          true_A = 10
          mark_A(x1) = 37
          active_A(x1) = 37
          false_A = 11
    
      precedence:
      
        mark# > f = c = mark = active > if > active# = true = false
      
      partial status:
    
        pi(active#) = [1]
        pi(f) = []
        pi(mark#) = []
        pi(if) = []
        pi(c) = []
        pi(true) = []
        pi(mark) = []
        pi(active) = []
        pi(false) = []
    

The next rules are strictly ordered:

  p3, p5

We remove them from the problem.

-- SCC decomposition.

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

p1: active#(f(X)) -> mark#(if(X,c(),f(true())))
p2: mark#(f(X)) -> active#(f(mark(X)))
p3: mark#(if(X1,X2,X3)) -> active#(if(mark(X1),mark(X2),X3))

and R consists of:

r1: active(f(X)) -> mark(if(X,c(),f(true())))
r2: active(if(true(),X,Y)) -> mark(X)
r3: active(if(false(),X,Y)) -> mark(Y)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(if(X1,X2,X3)) -> active(if(mark(X1),mark(X2),X3))
r6: mark(c()) -> active(c())
r7: mark(true()) -> active(true())
r8: mark(false()) -> active(false())
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: if(mark(X1),X2,X3) -> if(X1,X2,X3)
r12: if(X1,mark(X2),X3) -> if(X1,X2,X3)
r13: if(X1,X2,mark(X3)) -> if(X1,X2,X3)
r14: if(active(X1),X2,X3) -> if(X1,X2,X3)
r15: if(X1,active(X2),X3) -> if(X1,X2,X3)
r16: if(X1,X2,active(X3)) -> if(X1,X2,X3)

The estimated dependency graph contains the following SCCs:

  {p1, p2, p3}


-- Reduction pair.

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

p1: active#(f(X)) -> mark#(if(X,c(),f(true())))
p2: mark#(if(X1,X2,X3)) -> active#(if(mark(X1),mark(X2),X3))
p3: mark#(f(X)) -> active#(f(mark(X)))

and R consists of:

r1: active(f(X)) -> mark(if(X,c(),f(true())))
r2: active(if(true(),X,Y)) -> mark(X)
r3: active(if(false(),X,Y)) -> mark(Y)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(if(X1,X2,X3)) -> active(if(mark(X1),mark(X2),X3))
r6: mark(c()) -> active(c())
r7: mark(true()) -> active(true())
r8: mark(false()) -> active(false())
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: if(mark(X1),X2,X3) -> if(X1,X2,X3)
r12: if(X1,mark(X2),X3) -> if(X1,X2,X3)
r13: if(X1,X2,mark(X3)) -> if(X1,X2,X3)
r14: if(active(X1),X2,X3) -> if(X1,X2,X3)
r15: if(X1,active(X2),X3) -> if(X1,X2,X3)
r16: 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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = max{19, x1 - 10}
          f_A(x1) = x1 + 13
          mark#_A(x1) = x1 + 6
          if_A(x1,x2,x3) = max{13, x2 + 5, x3}
          c_A = 1
          true_A = 0
          mark_A(x1) = x1
          active_A(x1) = x1
          false_A = 1
    
      precedence:
      
        active# = f = mark# = if = c = true = mark = active = false
      
      partial status:
    
        pi(active#) = []
        pi(f) = []
        pi(mark#) = []
        pi(if) = []
        pi(c) = []
        pi(true) = []
        pi(mark) = []
        pi(active) = []
        pi(false) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = 10
          f_A(x1) = 17
          mark#_A(x1) = max{1, x1 - 1}
          if_A(x1,x2,x3) = 11
          c_A = 8
          true_A = 16
          mark_A(x1) = 28
          active_A(x1) = 28
          false_A = 9
    
      precedence:
      
        f > false > true > active# = mark# = if = c > mark = active
      
      partial status:
    
        pi(active#) = []
        pi(f) = []
        pi(mark#) = []
        pi(if) = []
        pi(c) = []
        pi(true) = []
        pi(mark) = []
        pi(active) = []
        pi(false) = []
    

The next rules are strictly ordered:

  p3

We remove them from the problem.

-- SCC decomposition.

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

p1: active#(f(X)) -> mark#(if(X,c(),f(true())))
p2: mark#(if(X1,X2,X3)) -> active#(if(mark(X1),mark(X2),X3))

and R consists of:

r1: active(f(X)) -> mark(if(X,c(),f(true())))
r2: active(if(true(),X,Y)) -> mark(X)
r3: active(if(false(),X,Y)) -> mark(Y)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(if(X1,X2,X3)) -> active(if(mark(X1),mark(X2),X3))
r6: mark(c()) -> active(c())
r7: mark(true()) -> active(true())
r8: mark(false()) -> active(false())
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: if(mark(X1),X2,X3) -> if(X1,X2,X3)
r12: if(X1,mark(X2),X3) -> if(X1,X2,X3)
r13: if(X1,X2,mark(X3)) -> if(X1,X2,X3)
r14: if(active(X1),X2,X3) -> if(X1,X2,X3)
r15: if(X1,active(X2),X3) -> if(X1,X2,X3)
r16: 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: active#(f(X)) -> mark#(if(X,c(),f(true())))
p2: mark#(if(X1,X2,X3)) -> active#(if(mark(X1),mark(X2),X3))

and R consists of:

r1: active(f(X)) -> mark(if(X,c(),f(true())))
r2: active(if(true(),X,Y)) -> mark(X)
r3: active(if(false(),X,Y)) -> mark(Y)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(if(X1,X2,X3)) -> active(if(mark(X1),mark(X2),X3))
r6: mark(c()) -> active(c())
r7: mark(true()) -> active(true())
r8: mark(false()) -> active(false())
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: if(mark(X1),X2,X3) -> if(X1,X2,X3)
r12: if(X1,mark(X2),X3) -> if(X1,X2,X3)
r13: if(X1,X2,mark(X3)) -> if(X1,X2,X3)
r14: if(active(X1),X2,X3) -> if(X1,X2,X3)
r15: if(X1,active(X2),X3) -> if(X1,X2,X3)
r16: 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

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = x1
          f_A(x1) = x1 + 4
          mark#_A(x1) = max{1, x1}
          if_A(x1,x2,x3) = max{x1, x2 + 4, x3}
          c_A = 0
          true_A = 0
          mark_A(x1) = x1
          active_A(x1) = x1
          false_A = 1
    
      precedence:
      
        f > c > mark# = false > active# = if = true > mark = active
      
      partial status:
    
        pi(active#) = [1]
        pi(f) = []
        pi(mark#) = []
        pi(if) = []
        pi(c) = []
        pi(true) = []
        pi(mark) = []
        pi(active) = []
        pi(false) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = 11
          f_A(x1) = 25
          mark#_A(x1) = 10
          if_A(x1,x2,x3) = 5
          c_A = 12
          true_A = 19
          mark_A(x1) = 0
          active_A(x1) = 0
          false_A = 1
    
      precedence:
      
        active# > mark# > f = c > if > false > true = mark = active
      
      partial status:
    
        pi(active#) = []
        pi(f) = []
        pi(mark#) = []
        pi(if) = []
        pi(c) = []
        pi(true) = []
        pi(mark) = []
        pi(active) = []
        pi(false) = []
    

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: active#(f(X)) -> mark#(if(X,c(),f(true())))

and R consists of:

r1: active(f(X)) -> mark(if(X,c(),f(true())))
r2: active(if(true(),X,Y)) -> mark(X)
r3: active(if(false(),X,Y)) -> mark(Y)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(if(X1,X2,X3)) -> active(if(mark(X1),mark(X2),X3))
r6: mark(c()) -> active(c())
r7: mark(true()) -> active(true())
r8: mark(false()) -> active(false())
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: if(mark(X1),X2,X3) -> if(X1,X2,X3)
r12: if(X1,mark(X2),X3) -> if(X1,X2,X3)
r13: if(X1,X2,mark(X3)) -> if(X1,X2,X3)
r14: if(active(X1),X2,X3) -> if(X1,X2,X3)
r15: if(X1,active(X2),X3) -> if(X1,X2,X3)
r16: if(X1,X2,active(X3)) -> if(X1,X2,X3)

The estimated dependency graph contains the following SCCs:

  (no SCCs)

-- 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(f(X)) -> mark(if(X,c(),f(true())))
r2: active(if(true(),X,Y)) -> mark(X)
r3: active(if(false(),X,Y)) -> mark(Y)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(if(X1,X2,X3)) -> active(if(mark(X1),mark(X2),X3))
r6: mark(c()) -> active(c())
r7: mark(true()) -> active(true())
r8: mark(false()) -> active(false())
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: if(mark(X1),X2,X3) -> if(X1,X2,X3)
r12: if(X1,mark(X2),X3) -> if(X1,X2,X3)
r13: if(X1,X2,mark(X3)) -> if(X1,X2,X3)
r14: if(active(X1),X2,X3) -> if(X1,X2,X3)
r15: if(X1,active(X2),X3) -> if(X1,X2,X3)
r16: 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(f(X)) -> mark(if(X,c(),f(true())))
r2: active(if(true(),X,Y)) -> mark(X)
r3: active(if(false(),X,Y)) -> mark(Y)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(if(X1,X2,X3)) -> active(if(mark(X1),mark(X2),X3))
r6: mark(c()) -> active(c())
r7: mark(true()) -> active(true())
r8: mark(false()) -> active(false())
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: if(mark(X1),X2,X3) -> if(X1,X2,X3)
r12: if(X1,mark(X2),X3) -> if(X1,X2,X3)
r13: if(X1,X2,mark(X3)) -> if(X1,X2,X3)
r14: if(active(X1),X2,X3) -> if(X1,X2,X3)
r15: if(X1,active(X2),X3) -> if(X1,X2,X3)
r16: 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(f(X)) -> mark(if(X,c(),f(true())))
r2: active(if(true(),X,Y)) -> mark(X)
r3: active(if(false(),X,Y)) -> mark(Y)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(if(X1,X2,X3)) -> active(if(mark(X1),mark(X2),X3))
r6: mark(c()) -> active(c())
r7: mark(true()) -> active(true())
r8: mark(false()) -> active(false())
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: if(mark(X1),X2,X3) -> if(X1,X2,X3)
r12: if(X1,mark(X2),X3) -> if(X1,X2,X3)
r13: if(X1,X2,mark(X3)) -> if(X1,X2,X3)
r14: if(active(X1),X2,X3) -> if(X1,X2,X3)
r15: if(X1,active(X2),X3) -> if(X1,X2,X3)
r16: 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: f#(mark(X)) -> f#(X)
p2: f#(active(X)) -> f#(X)

and R consists of:

r1: active(f(X)) -> mark(if(X,c(),f(true())))
r2: active(if(true(),X,Y)) -> mark(X)
r3: active(if(false(),X,Y)) -> mark(Y)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(if(X1,X2,X3)) -> active(if(mark(X1),mark(X2),X3))
r6: mark(c()) -> active(c())
r7: mark(true()) -> active(true())
r8: mark(false()) -> active(false())
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: if(mark(X1),X2,X3) -> if(X1,X2,X3)
r12: if(X1,mark(X2),X3) -> if(X1,X2,X3)
r13: if(X1,X2,mark(X3)) -> if(X1,X2,X3)
r14: if(active(X1),X2,X3) -> if(X1,X2,X3)
r15: if(X1,active(X2),X3) -> if(X1,X2,X3)
r16: 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:
        
          f#_A(x1) = x1 + 2
          mark_A(x1) = x1
          active_A(x1) = x1 + 2
    
      precedence:
      
        mark > active > f#
      
      partial status:
    
        pi(f#) = [1]
        pi(mark) = [1]
        pi(active) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = x1 + 1
          mark_A(x1) = x1
          active_A(x1) = x1
    
      precedence:
      
        f# = mark = active
      
      partial status:
    
        pi(f#) = [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.