YES

We show the termination of the TRS R:

  active(f(b(),X,c())) -> mark(f(X,c(),X))
  active(c()) -> mark(b())
  active(f(X1,X2,X3)) -> f(X1,active(X2),X3)
  f(X1,mark(X2),X3) -> mark(f(X1,X2,X3))
  proper(f(X1,X2,X3)) -> f(proper(X1),proper(X2),proper(X3))
  proper(b()) -> ok(b())
  proper(c()) -> ok(c())
  f(ok(X1),ok(X2),ok(X3)) -> ok(f(X1,X2,X3))
  top(mark(X)) -> top(proper(X))
  top(ok(X)) -> top(active(X))

-- SCC decomposition.

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

p1: active#(f(b(),X,c())) -> f#(X,c(),X)
p2: active#(f(X1,X2,X3)) -> f#(X1,active(X2),X3)
p3: active#(f(X1,X2,X3)) -> active#(X2)
p4: f#(X1,mark(X2),X3) -> f#(X1,X2,X3)
p5: proper#(f(X1,X2,X3)) -> f#(proper(X1),proper(X2),proper(X3))
p6: proper#(f(X1,X2,X3)) -> proper#(X1)
p7: proper#(f(X1,X2,X3)) -> proper#(X2)
p8: proper#(f(X1,X2,X3)) -> proper#(X3)
p9: f#(ok(X1),ok(X2),ok(X3)) -> f#(X1,X2,X3)
p10: top#(mark(X)) -> top#(proper(X))
p11: top#(mark(X)) -> proper#(X)
p12: top#(ok(X)) -> top#(active(X))
p13: top#(ok(X)) -> active#(X)

and R consists of:

r1: active(f(b(),X,c())) -> mark(f(X,c(),X))
r2: active(c()) -> mark(b())
r3: active(f(X1,X2,X3)) -> f(X1,active(X2),X3)
r4: f(X1,mark(X2),X3) -> mark(f(X1,X2,X3))
r5: proper(f(X1,X2,X3)) -> f(proper(X1),proper(X2),proper(X3))
r6: proper(b()) -> ok(b())
r7: proper(c()) -> ok(c())
r8: f(ok(X1),ok(X2),ok(X3)) -> ok(f(X1,X2,X3))
r9: top(mark(X)) -> top(proper(X))
r10: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p10, p12}
  {p3}
  {p6, p7, p8}
  {p4, p9}


-- Reduction pair.

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

p1: top#(ok(X)) -> top#(active(X))
p2: top#(mark(X)) -> top#(proper(X))

and R consists of:

r1: active(f(b(),X,c())) -> mark(f(X,c(),X))
r2: active(c()) -> mark(b())
r3: active(f(X1,X2,X3)) -> f(X1,active(X2),X3)
r4: f(X1,mark(X2),X3) -> mark(f(X1,X2,X3))
r5: proper(f(X1,X2,X3)) -> f(proper(X1),proper(X2),proper(X3))
r6: proper(b()) -> ok(b())
r7: proper(c()) -> ok(c())
r8: f(ok(X1),ok(X2),ok(X3)) -> ok(f(X1,X2,X3))
r9: top(mark(X)) -> top(proper(X))
r10: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            top#_A(x1) = ((0,1),(0,0)) x1 + (1,1)
            ok_A(x1) = x1
            active_A(x1) = x1
            mark_A(x1) = ((0,0),(1,1)) x1 + (1,1)
            proper_A(x1) = x1
            f_A(x1,x2,x3) = ((0,0),(1,0)) x1 + ((0,0),(1,1)) x2 + ((0,0),(0,1)) x3 + (1,2)
            b_A() = (4,1)
            c_A() = (2,6)
    
      precedence:
      
        active = proper = f = b = c > top# = ok = mark
      
      partial status:
    
        pi(top#) = []
        pi(ok) = []
        pi(active) = [1]
        pi(mark) = []
        pi(proper) = [1]
        pi(f) = []
        pi(b) = []
        pi(c) = []
    
    2. weighted path order
    
      base order:
    
        matrix interpretations:
        
          carrier: N^2
          order: standard order
          interpretations:
            top#_A(x1) = (4,7)
            ok_A(x1) = (5,6)
            active_A(x1) = ((1,1),(0,0)) x1 + (3,0)
            mark_A(x1) = (11,0)
            proper_A(x1) = ((1,1),(1,1)) x1 + (1,1)
            f_A(x1,x2,x3) = (2,6)
            b_A() = (12,1)
            c_A() = (2,2)
    
      precedence:
      
        proper = b > active > f > top# = ok = mark = c
      
      partial status:
    
        pi(top#) = []
        pi(ok) = []
        pi(active) = []
        pi(mark) = []
        pi(proper) = [1]
        pi(f) = []
        pi(b) = []
        pi(c) = []
    

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: top#(ok(X)) -> top#(active(X))

and R consists of:

r1: active(f(b(),X,c())) -> mark(f(X,c(),X))
r2: active(c()) -> mark(b())
r3: active(f(X1,X2,X3)) -> f(X1,active(X2),X3)
r4: f(X1,mark(X2),X3) -> mark(f(X1,X2,X3))
r5: proper(f(X1,X2,X3)) -> f(proper(X1),proper(X2),proper(X3))
r6: proper(b()) -> ok(b())
r7: proper(c()) -> ok(c())
r8: f(ok(X1),ok(X2),ok(X3)) -> ok(f(X1,X2,X3))
r9: top(mark(X)) -> top(proper(X))
r10: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: top#(ok(X)) -> top#(active(X))

and R consists of:

r1: active(f(b(),X,c())) -> mark(f(X,c(),X))
r2: active(c()) -> mark(b())
r3: active(f(X1,X2,X3)) -> f(X1,active(X2),X3)
r4: f(X1,mark(X2),X3) -> mark(f(X1,X2,X3))
r5: proper(f(X1,X2,X3)) -> f(proper(X1),proper(X2),proper(X3))
r6: proper(b()) -> ok(b())
r7: proper(c()) -> ok(c())
r8: f(ok(X1),ok(X2),ok(X3)) -> ok(f(X1,X2,X3))
r9: top(mark(X)) -> top(proper(X))
r10: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  r1, r2, r3, r4, r8

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          top#_A(x1) = max{0, x1 - 11}
          ok_A(x1) = max{18, x1 + 16}
          active_A(x1) = max{10, x1 + 4}
          f_A(x1,x2,x3) = max{17, x1 + 15, x2 + 11, x3 + 15}
          mark_A(x1) = max{9, x1}
          b_A = 3
          c_A = 4
    
      precedence:
      
        active = f > ok = c > b > mark > top#
      
      partial status:
    
        pi(top#) = []
        pi(ok) = [1]
        pi(active) = [1]
        pi(f) = [2, 3]
        pi(mark) = [1]
        pi(b) = []
        pi(c) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          top#_A(x1) = 1
          ok_A(x1) = max{11, x1 + 10}
          active_A(x1) = max{2, x1}
          f_A(x1,x2,x3) = max{10, x2 + 5, x3 + 8}
          mark_A(x1) = max{13, x1 + 12}
          b_A = 3
          c_A = 4
    
      precedence:
      
        top# = ok = active = f = mark = b = c
      
      partial status:
    
        pi(top#) = []
        pi(ok) = [1]
        pi(active) = [1]
        pi(f) = [2, 3]
        pi(mark) = [1]
        pi(b) = []
        pi(c) = []
    

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: active#(f(X1,X2,X3)) -> active#(X2)

and R consists of:

r1: active(f(b(),X,c())) -> mark(f(X,c(),X))
r2: active(c()) -> mark(b())
r3: active(f(X1,X2,X3)) -> f(X1,active(X2),X3)
r4: f(X1,mark(X2),X3) -> mark(f(X1,X2,X3))
r5: proper(f(X1,X2,X3)) -> f(proper(X1),proper(X2),proper(X3))
r6: proper(b()) -> ok(b())
r7: proper(c()) -> ok(c())
r8: f(ok(X1),ok(X2),ok(X3)) -> ok(f(X1,X2,X3))
r9: top(mark(X)) -> top(proper(X))
r10: top(ok(X)) -> top(active(X))

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:
        
          active#_A(x1) = max{2, x1 + 1}
          f_A(x1,x2,x3) = max{x1, x2, x3}
    
      precedence:
      
        active# = f
      
      partial status:
    
        pi(active#) = [1]
        pi(f) = [1, 2, 3]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = x1 + 3
          f_A(x1,x2,x3) = max{x1 + 2, x2 + 2, x3 + 2}
    
      precedence:
      
        active# = f
      
      partial status:
    
        pi(active#) = []
        pi(f) = [3]
    

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: proper#(f(X1,X2,X3)) -> proper#(X3)
p2: proper#(f(X1,X2,X3)) -> proper#(X2)
p3: proper#(f(X1,X2,X3)) -> proper#(X1)

and R consists of:

r1: active(f(b(),X,c())) -> mark(f(X,c(),X))
r2: active(c()) -> mark(b())
r3: active(f(X1,X2,X3)) -> f(X1,active(X2),X3)
r4: f(X1,mark(X2),X3) -> mark(f(X1,X2,X3))
r5: proper(f(X1,X2,X3)) -> f(proper(X1),proper(X2),proper(X3))
r6: proper(b()) -> ok(b())
r7: proper(c()) -> ok(c())
r8: f(ok(X1),ok(X2),ok(X3)) -> ok(f(X1,X2,X3))
r9: top(mark(X)) -> top(proper(X))
r10: top(ok(X)) -> top(active(X))

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:
        
          proper#_A(x1) = max{2, x1 + 1}
          f_A(x1,x2,x3) = max{x1, x2, x3}
    
      precedence:
      
        proper# = f
      
      partial status:
    
        pi(proper#) = [1]
        pi(f) = [1, 2, 3]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          proper#_A(x1) = x1 + 1
          f_A(x1,x2,x3) = max{x1, x2, x3}
    
      precedence:
      
        proper# = f
      
      partial status:
    
        pi(proper#) = [1]
        pi(f) = [1, 2, 3]
    

The next rules are strictly ordered:

  p1, p2, p3

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#(X1,mark(X2),X3) -> f#(X1,X2,X3)
p2: f#(ok(X1),ok(X2),ok(X3)) -> f#(X1,X2,X3)

and R consists of:

r1: active(f(b(),X,c())) -> mark(f(X,c(),X))
r2: active(c()) -> mark(b())
r3: active(f(X1,X2,X3)) -> f(X1,active(X2),X3)
r4: f(X1,mark(X2),X3) -> mark(f(X1,X2,X3))
r5: proper(f(X1,X2,X3)) -> f(proper(X1),proper(X2),proper(X3))
r6: proper(b()) -> ok(b())
r7: proper(c()) -> ok(c())
r8: f(ok(X1),ok(X2),ok(X3)) -> ok(f(X1,X2,X3))
r9: top(mark(X)) -> top(proper(X))
r10: top(ok(X)) -> top(active(X))

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:
        
          f#_A(x1,x2,x3) = max{2, x1 - 1, x2 + 1, x3 - 1}
          mark_A(x1) = max{1, x1}
          ok_A(x1) = max{1, x1}
    
      precedence:
      
        f# = mark = ok
      
      partial status:
    
        pi(f#) = [2]
        pi(mark) = [1]
        pi(ok) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2,x3) = max{0, x2 - 2}
          mark_A(x1) = max{3, x1 + 1}
          ok_A(x1) = max{3, x1 + 1}
    
      precedence:
      
        f# = mark = ok
      
      partial status:
    
        pi(f#) = []
        pi(mark) = [1]
        pi(ok) = [1]
    

The next rules are strictly ordered:

  p1, p2

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