YES

We show the termination of the TRS R:

  active(f(X)) -> mark(g(h(f(X))))
  mark(f(X)) -> active(f(mark(X)))
  mark(g(X)) -> active(g(X))
  mark(h(X)) -> active(h(mark(X)))
  f(mark(X)) -> f(X)
  f(active(X)) -> f(X)
  g(mark(X)) -> g(X)
  g(active(X)) -> g(X)
  h(mark(X)) -> h(X)
  h(active(X)) -> h(X)

-- SCC decomposition.

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

p1: active#(f(X)) -> mark#(g(h(f(X))))
p2: active#(f(X)) -> g#(h(f(X)))
p3: active#(f(X)) -> h#(f(X))
p4: mark#(f(X)) -> active#(f(mark(X)))
p5: mark#(f(X)) -> f#(mark(X))
p6: mark#(f(X)) -> mark#(X)
p7: mark#(g(X)) -> active#(g(X))
p8: mark#(h(X)) -> active#(h(mark(X)))
p9: mark#(h(X)) -> h#(mark(X))
p10: mark#(h(X)) -> mark#(X)
p11: f#(mark(X)) -> f#(X)
p12: f#(active(X)) -> f#(X)
p13: g#(mark(X)) -> g#(X)
p14: g#(active(X)) -> g#(X)
p15: h#(mark(X)) -> h#(X)
p16: h#(active(X)) -> h#(X)

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(g(X)) -> active(g(X))
r4: mark(h(X)) -> active(h(mark(X)))
r5: f(mark(X)) -> f(X)
r6: f(active(X)) -> f(X)
r7: g(mark(X)) -> g(X)
r8: g(active(X)) -> g(X)
r9: h(mark(X)) -> h(X)
r10: h(active(X)) -> h(X)

The estimated dependency graph contains the following SCCs:

  {p1, p4, p6, p7, p8, p10}
  {p13, p14}
  {p15, p16}
  {p11, p12}


-- Reduction pair.

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

p1: active#(f(X)) -> mark#(g(h(f(X))))
p2: mark#(h(X)) -> mark#(X)
p3: mark#(h(X)) -> active#(h(mark(X)))
p4: mark#(g(X)) -> active#(g(X))
p5: mark#(f(X)) -> mark#(X)
p6: mark#(f(X)) -> active#(f(mark(X)))

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(g(X)) -> active(g(X))
r4: mark(h(X)) -> active(h(mark(X)))
r5: f(mark(X)) -> f(X)
r6: f(active(X)) -> f(X)
r7: g(mark(X)) -> g(X)
r8: g(active(X)) -> g(X)
r9: h(mark(X)) -> h(X)
r10: h(active(X)) -> h(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          active#_A(x1) = ((0,0),(1,0)) x1 + (4,1)
          f_A(x1) = (10,12)
          mark#_A(x1) = (4,11)
          g_A(x1) = (4,10)
          h_A(x1) = (6,8)
          mark_A(x1) = ((0,0),(1,0)) x1 + (21,9)
          active_A(x1) = ((1,0),(0,0)) x1 + (11,14)
  
    precedence:
    
      active# = f = mark# = g > h = mark = active
    
    partial status:
  
      pi(active#) = []
      pi(f) = []
      pi(mark#) = []
      pi(g) = []
      pi(h) = []
      pi(mark) = []
      pi(active) = []

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#(g(h(f(X))))
p2: mark#(h(X)) -> mark#(X)
p3: mark#(g(X)) -> active#(g(X))
p4: mark#(f(X)) -> mark#(X)
p5: mark#(f(X)) -> active#(f(mark(X)))

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(g(X)) -> active(g(X))
r4: mark(h(X)) -> active(h(mark(X)))
r5: f(mark(X)) -> f(X)
r6: f(active(X)) -> f(X)
r7: g(mark(X)) -> g(X)
r8: g(active(X)) -> g(X)
r9: h(mark(X)) -> h(X)
r10: h(active(X)) -> h(X)

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#(g(h(f(X))))
p2: mark#(f(X)) -> active#(f(mark(X)))
p3: mark#(f(X)) -> mark#(X)
p4: mark#(g(X)) -> active#(g(X))
p5: mark#(h(X)) -> mark#(X)

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(g(X)) -> active(g(X))
r4: mark(h(X)) -> active(h(mark(X)))
r5: f(mark(X)) -> f(X)
r6: f(active(X)) -> f(X)
r7: g(mark(X)) -> g(X)
r8: g(active(X)) -> g(X)
r9: h(mark(X)) -> h(X)
r10: h(active(X)) -> h(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          active#_A(x1) = ((0,0),(1,0)) x1 + (1,4)
          f_A(x1) = (3,8)
          mark#_A(x1) = (1,7)
          g_A(x1) = (1,6)
          h_A(x1) = (2,10)
          mark_A(x1) = x1 + (0,5)
          active_A(x1) = ((0,0),(1,0)) x1 + (1,9)
  
    precedence:
    
      mark = active > f > active# = mark# > g > h
    
    partial status:
  
      pi(active#) = []
      pi(f) = []
      pi(mark#) = []
      pi(g) = []
      pi(h) = []
      pi(mark) = []
      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#(g(h(f(X))))
p2: mark#(f(X)) -> active#(f(mark(X)))
p3: mark#(f(X)) -> mark#(X)
p4: mark#(h(X)) -> mark#(X)

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(g(X)) -> active(g(X))
r4: mark(h(X)) -> active(h(mark(X)))
r5: f(mark(X)) -> f(X)
r6: f(active(X)) -> f(X)
r7: g(mark(X)) -> g(X)
r8: g(active(X)) -> g(X)
r9: h(mark(X)) -> h(X)
r10: h(active(X)) -> h(X)

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: active#(f(X)) -> mark#(g(h(f(X))))
p2: mark#(h(X)) -> mark#(X)
p3: mark#(f(X)) -> mark#(X)
p4: mark#(f(X)) -> active#(f(mark(X)))

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(g(X)) -> active(g(X))
r4: mark(h(X)) -> active(h(mark(X)))
r5: f(mark(X)) -> f(X)
r6: f(active(X)) -> f(X)
r7: g(mark(X)) -> g(X)
r8: g(active(X)) -> g(X)
r9: h(mark(X)) -> h(X)
r10: h(active(X)) -> h(X)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          active#_A(x1) = ((0,0),(1,0)) x1 + (5,1)
          f_A(x1) = ((1,0),(0,0)) x1 + (8,8)
          mark#_A(x1) = ((1,0),(1,1)) x1 + (2,1)
          g_A(x1) = (2,4)
          h_A(x1) = ((1,0),(0,0)) x1 + (10,10)
          mark_A(x1) = ((1,0),(1,0)) x1 + (1,1)
          active_A(x1) = ((1,0),(0,0)) x1 + (0,3)
  
    precedence:
    
      g = h > f = mark = active > active# = mark#
    
    partial status:
  
      pi(active#) = []
      pi(f) = []
      pi(mark#) = [1]
      pi(g) = []
      pi(h) = []
      pi(mark) = []
      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#(g(h(f(X))))
p2: mark#(h(X)) -> mark#(X)
p3: mark#(f(X)) -> mark#(X)

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(g(X)) -> active(g(X))
r4: mark(h(X)) -> active(h(mark(X)))
r5: f(mark(X)) -> f(X)
r6: f(active(X)) -> f(X)
r7: g(mark(X)) -> g(X)
r8: g(active(X)) -> g(X)
r9: h(mark(X)) -> h(X)
r10: h(active(X)) -> h(X)

The estimated dependency graph contains the following SCCs:

  {p2, p3}


-- Reduction pair.

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

p1: mark#(h(X)) -> mark#(X)
p2: mark#(f(X)) -> mark#(X)

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(g(X)) -> active(g(X))
r4: mark(h(X)) -> active(h(mark(X)))
r5: f(mark(X)) -> f(X)
r6: f(active(X)) -> f(X)
r7: g(mark(X)) -> g(X)
r8: g(active(X)) -> g(X)
r9: h(mark(X)) -> h(X)
r10: h(active(X)) -> h(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          mark#_A(x1) = ((1,0),(0,0)) x1 + (1,1)
          h_A(x1) = x1 + (2,2)
          f_A(x1) = ((1,0),(0,0)) x1 + (2,2)
  
    precedence:
    
      mark# = h = f
    
    partial status:
  
      pi(mark#) = []
      pi(h) = [1]
      pi(f) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: mark#(f(X)) -> mark#(X)

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(g(X)) -> active(g(X))
r4: mark(h(X)) -> active(h(mark(X)))
r5: f(mark(X)) -> f(X)
r6: f(active(X)) -> f(X)
r7: g(mark(X)) -> g(X)
r8: g(active(X)) -> g(X)
r9: h(mark(X)) -> h(X)
r10: h(active(X)) -> h(X)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: mark#(f(X)) -> mark#(X)

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(g(X)) -> active(g(X))
r4: mark(h(X)) -> active(h(mark(X)))
r5: f(mark(X)) -> f(X)
r6: f(active(X)) -> f(X)
r7: g(mark(X)) -> g(X)
r8: g(active(X)) -> g(X)
r9: h(mark(X)) -> h(X)
r10: h(active(X)) -> h(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          mark#_A(x1) = ((1,0),(1,0)) x1 + (2,2)
          f_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      mark# = f
    
    partial status:
  
      pi(mark#) = []
      pi(f) = []

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

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(g(X)) -> active(g(X))
r4: mark(h(X)) -> active(h(mark(X)))
r5: f(mark(X)) -> f(X)
r6: f(active(X)) -> f(X)
r7: g(mark(X)) -> g(X)
r8: g(active(X)) -> g(X)
r9: h(mark(X)) -> h(X)
r10: h(active(X)) -> h(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          g#_A(x1) = ((1,0),(0,0)) x1 + (1,1)
          mark_A(x1) = x1 + (2,2)
          active_A(x1) = ((1,0),(0,0)) x1 + (2,2)
  
    precedence:
    
      g# = mark = active
    
    partial status:
  
      pi(g#) = []
      pi(mark) = [1]
      pi(active) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: g#(active(X)) -> g#(X)

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(g(X)) -> active(g(X))
r4: mark(h(X)) -> active(h(mark(X)))
r5: f(mark(X)) -> f(X)
r6: f(active(X)) -> f(X)
r7: g(mark(X)) -> g(X)
r8: g(active(X)) -> g(X)
r9: h(mark(X)) -> h(X)
r10: h(active(X)) -> h(X)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: g#(active(X)) -> g#(X)

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(g(X)) -> active(g(X))
r4: mark(h(X)) -> active(h(mark(X)))
r5: f(mark(X)) -> f(X)
r6: f(active(X)) -> f(X)
r7: g(mark(X)) -> g(X)
r8: g(active(X)) -> g(X)
r9: h(mark(X)) -> h(X)
r10: h(active(X)) -> h(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          g#_A(x1) = ((1,0),(1,0)) x1 + (2,2)
          active_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      g# = active
    
    partial status:
  
      pi(g#) = []
      pi(active) = []

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

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(g(X)) -> active(g(X))
r4: mark(h(X)) -> active(h(mark(X)))
r5: f(mark(X)) -> f(X)
r6: f(active(X)) -> f(X)
r7: g(mark(X)) -> g(X)
r8: g(active(X)) -> g(X)
r9: h(mark(X)) -> h(X)
r10: h(active(X)) -> h(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          h#_A(x1) = ((1,0),(0,0)) x1 + (1,1)
          mark_A(x1) = x1 + (2,2)
          active_A(x1) = ((1,0),(0,0)) x1 + (2,2)
  
    precedence:
    
      h# = mark = active
    
    partial status:
  
      pi(h#) = []
      pi(mark) = [1]
      pi(active) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: h#(active(X)) -> h#(X)

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(g(X)) -> active(g(X))
r4: mark(h(X)) -> active(h(mark(X)))
r5: f(mark(X)) -> f(X)
r6: f(active(X)) -> f(X)
r7: g(mark(X)) -> g(X)
r8: g(active(X)) -> g(X)
r9: h(mark(X)) -> h(X)
r10: h(active(X)) -> h(X)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: h#(active(X)) -> h#(X)

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(g(X)) -> active(g(X))
r4: mark(h(X)) -> active(h(mark(X)))
r5: f(mark(X)) -> f(X)
r6: f(active(X)) -> f(X)
r7: g(mark(X)) -> g(X)
r8: g(active(X)) -> g(X)
r9: h(mark(X)) -> h(X)
r10: h(active(X)) -> h(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          h#_A(x1) = ((1,0),(1,0)) x1 + (2,2)
          active_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      h# = active
    
    partial status:
  
      pi(h#) = []
      pi(active) = []

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

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(g(X)) -> active(g(X))
r4: mark(h(X)) -> active(h(mark(X)))
r5: f(mark(X)) -> f(X)
r6: f(active(X)) -> f(X)
r7: g(mark(X)) -> g(X)
r8: g(active(X)) -> g(X)
r9: h(mark(X)) -> h(X)
r10: h(active(X)) -> h(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          f#_A(x1) = ((1,0),(0,0)) x1 + (1,1)
          mark_A(x1) = x1 + (2,2)
          active_A(x1) = ((1,0),(0,0)) x1 + (2,2)
  
    precedence:
    
      f# = mark = active
    
    partial status:
  
      pi(f#) = []
      pi(mark) = [1]
      pi(active) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: f#(active(X)) -> f#(X)

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(g(X)) -> active(g(X))
r4: mark(h(X)) -> active(h(mark(X)))
r5: f(mark(X)) -> f(X)
r6: f(active(X)) -> f(X)
r7: g(mark(X)) -> g(X)
r8: g(active(X)) -> g(X)
r9: h(mark(X)) -> h(X)
r10: h(active(X)) -> h(X)

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: f#(active(X)) -> f#(X)

and R consists of:

r1: active(f(X)) -> mark(g(h(f(X))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(g(X)) -> active(g(X))
r4: mark(h(X)) -> active(h(mark(X)))
r5: f(mark(X)) -> f(X)
r6: f(active(X)) -> f(X)
r7: g(mark(X)) -> g(X)
r8: g(active(X)) -> g(X)
r9: h(mark(X)) -> h(X)
r10: h(active(X)) -> h(X)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: lexicographic order
        interpretations:
          f#_A(x1) = ((1,0),(1,0)) x1 + (2,2)
          active_A(x1) = ((1,0),(0,0)) x1 + (1,1)
  
    precedence:
    
      f# = active
    
    partial status:
  
      pi(f#) = []
      pi(active) = []

The next rules are strictly ordered:

  p1

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