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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      active#_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,0,0),(1,0,1,0)) x1
      f_A(x1) = ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + (1,5,0,1)
      mark#_A(x1) = (1,5,0,1)
      g_A(x1) = ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + (1,1,1,1)
      h_A(x1) = ((0,0,0,0),(0,0,0,0),(1,0,0,0),(1,1,0,0)) x1 + (0,3,1,0)
      mark_A(x1) = ((1,0,0,0),(1,0,0,0),(0,1,0,0),(0,0,0,0)) x1 + (1,5,1,0)
      active_A(x1) = x1 + (1,1,3,0)

The next rules are strictly ordered:

  p3, 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)
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 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#(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 set of usable rules consists of

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

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      active#_A(x1) = ((1,0,0,0),(0,0,0,0),(1,0,0,0),(1,1,0,0)) x1 + (0,7,5,0)
      f_A(x1) = ((1,0,0,0),(0,0,0,0),(1,1,0,0),(1,0,1,0)) x1 + (3,1,13,1)
      mark#_A(x1) = ((1,0,0,0),(1,1,0,0),(1,0,1,0),(0,0,1,1)) x1 + (1,0,0,0)
      g_A(x1) = ((0,0,0,0),(1,0,0,0),(0,1,0,0),(0,0,0,0)) x1 + (1,1,1,1)
      h_A(x1) = ((1,0,0,0),(1,1,0,0),(0,0,0,0),(0,0,0,0)) x1 + (1,1,1,0)
      mark_A(x1) = ((1,0,0,0),(0,1,0,0),(0,1,1,0),(0,0,0,1)) x1 + (0,1,3,6)
      active_A(x1) = ((1,0,0,0),(0,1,0,0),(0,1,1,0),(1,1,1,0)) x1 + (0,0,1,1)

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: 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 monotone reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      g#_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1
      mark_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + (1,1,1,1)
      active_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + (1,1,1,1)

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

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 monotone reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      h#_A(x1) = ((1,0,0,0),(0,1,0,0),(0,1,1,0),(0,1,1,1)) x1
      mark_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + (1,1,1,1)
      active_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + (1,1,1,1)

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

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 monotone reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      f#_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1
      mark_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + (1,1,1,1)
      active_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + (1,1,1,1)

The next rules are strictly ordered:

  p1, p2
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10

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