YES

We show the termination of the TRS R:

  active(f(f(X))) -> mark(c(f(g(f(X)))))
  active(c(X)) -> mark(d(X))
  active(h(X)) -> mark(c(d(X)))
  active(f(X)) -> f(active(X))
  active(h(X)) -> h(active(X))
  f(mark(X)) -> mark(f(X))
  h(mark(X)) -> mark(h(X))
  proper(f(X)) -> f(proper(X))
  proper(c(X)) -> c(proper(X))
  proper(g(X)) -> g(proper(X))
  proper(d(X)) -> d(proper(X))
  proper(h(X)) -> h(proper(X))
  f(ok(X)) -> ok(f(X))
  c(ok(X)) -> ok(c(X))
  g(ok(X)) -> ok(g(X))
  d(ok(X)) -> ok(d(X))
  h(ok(X)) -> ok(h(X))
  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(f(X))) -> c#(f(g(f(X))))
p2: active#(f(f(X))) -> f#(g(f(X)))
p3: active#(f(f(X))) -> g#(f(X))
p4: active#(c(X)) -> d#(X)
p5: active#(h(X)) -> c#(d(X))
p6: active#(h(X)) -> d#(X)
p7: active#(f(X)) -> f#(active(X))
p8: active#(f(X)) -> active#(X)
p9: active#(h(X)) -> h#(active(X))
p10: active#(h(X)) -> active#(X)
p11: f#(mark(X)) -> f#(X)
p12: h#(mark(X)) -> h#(X)
p13: proper#(f(X)) -> f#(proper(X))
p14: proper#(f(X)) -> proper#(X)
p15: proper#(c(X)) -> c#(proper(X))
p16: proper#(c(X)) -> proper#(X)
p17: proper#(g(X)) -> g#(proper(X))
p18: proper#(g(X)) -> proper#(X)
p19: proper#(d(X)) -> d#(proper(X))
p20: proper#(d(X)) -> proper#(X)
p21: proper#(h(X)) -> h#(proper(X))
p22: proper#(h(X)) -> proper#(X)
p23: f#(ok(X)) -> f#(X)
p24: c#(ok(X)) -> c#(X)
p25: g#(ok(X)) -> g#(X)
p26: d#(ok(X)) -> d#(X)
p27: h#(ok(X)) -> h#(X)
p28: top#(mark(X)) -> top#(proper(X))
p29: top#(mark(X)) -> proper#(X)
p30: top#(ok(X)) -> top#(active(X))
p31: top#(ok(X)) -> active#(X)

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p28, p30}
  {p8, p10}
  {p14, p16, p18, p20, p22}
  {p24}
  {p11, p23}
  {p25}
  {p26}
  {p12, p27}


-- 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(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15, r16, r17

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      top#_A(x1) = ((1,0,0,0),(0,1,0,0),(0,1,1,0),(0,1,1,1)) x1
      ok_A(x1) = ((1,0,0,0),(0,0,0,0),(1,0,0,0),(1,1,0,0)) x1 + (0,11,1,7)
      active_A(x1) = ((1,0,0,0),(0,0,0,0),(1,0,0,0),(0,1,0,0)) x1 + (0,1,3,6)
      mark_A(x1) = ((1,0,0,0),(0,0,0,0),(1,0,0,0),(1,1,0,0)) x1 + (3,11,4,1)
      proper_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(1,0,0,0)) x1 + (1,6,7,0)
      f_A(x1) = ((1,0,0,0),(0,1,0,0),(1,0,0,0),(0,1,0,0)) x1 + (5,0,7,0)
      h_A(x1) = ((1,0,0,0),(0,1,0,0),(1,1,0,0),(1,0,0,0)) x1 + (9,0,0,7)
      c_A(x1) = (5,12,7,25)
      g_A(x1) = x1 + (0,1,1,0)
      d_A(x1) = (1,17,3,26)

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

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      active#_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,1,1,1)) x1
      h_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + (1,1,1,1)
      f_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, r11, r12, r13, r14, r15, r16, r17, r18, r19

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#(h(X)) -> proper#(X)
p2: proper#(d(X)) -> proper#(X)
p3: proper#(g(X)) -> proper#(X)
p4: proper#(c(X)) -> proper#(X)
p5: proper#(f(X)) -> proper#(X)

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      proper#_A(x1) = ((1,0,0,0),(1,0,0,0),(1,1,0,0),(1,1,1,0)) x1
      h_A(x1) = ((1,0,0,0),(1,0,0,0),(1,1,0,0),(1,1,1,0)) x1 + (1,1,1,1)
      d_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + (1,1,1,1)
      g_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,0,0),(1,1,1,0)) x1 + (1,1,1,1)
      c_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + (1,1,1,1)
      f_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, p3, p4, p5

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: c#(ok(X)) -> c#(X)

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      c#_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,1,1,1)) x1
      ok_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,0)) x1 + (1,0,0,1)

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#(ok(X)) -> f#(X)

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(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),(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)
      ok_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, r11, r12, r13, r14, r15, r16, r17, r18, r19

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#(ok(X)) -> g#(X)

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

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

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: d#(ok(X)) -> d#(X)

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      d#_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,1,1,1)) x1
      ok_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,0)) x1 + (1,0,0,1)

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#(ok(X)) -> h#(X)

and R consists of:

r1: active(f(f(X))) -> mark(c(f(g(f(X)))))
r2: active(c(X)) -> mark(d(X))
r3: active(h(X)) -> mark(c(d(X)))
r4: active(f(X)) -> f(active(X))
r5: active(h(X)) -> h(active(X))
r6: f(mark(X)) -> mark(f(X))
r7: h(mark(X)) -> mark(h(X))
r8: proper(f(X)) -> f(proper(X))
r9: proper(c(X)) -> c(proper(X))
r10: proper(g(X)) -> g(proper(X))
r11: proper(d(X)) -> d(proper(X))
r12: proper(h(X)) -> h(proper(X))
r13: f(ok(X)) -> ok(f(X))
r14: c(ok(X)) -> ok(c(X))
r15: g(ok(X)) -> ok(g(X))
r16: d(ok(X)) -> ok(d(X))
r17: h(ok(X)) -> ok(h(X))
r18: top(mark(X)) -> top(proper(X))
r19: top(ok(X)) -> top(active(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),(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)
      ok_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, r11, r12, r13, r14, r15, r16, r17, r18, r19

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