YES

We show the termination of the TRS R:

  active(f(f(a()))) -> mark(c(f(g(f(a())))))
  mark(f(X)) -> active(f(mark(X)))
  mark(a()) -> active(a())
  mark(c(X)) -> active(c(X))
  mark(g(X)) -> active(g(mark(X)))
  f(mark(X)) -> f(X)
  f(active(X)) -> f(X)
  c(mark(X)) -> c(X)
  c(active(X)) -> c(X)
  g(mark(X)) -> g(X)
  g(active(X)) -> g(X)

-- SCC decomposition.

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

p1: active#(f(f(a()))) -> mark#(c(f(g(f(a())))))
p2: active#(f(f(a()))) -> c#(f(g(f(a()))))
p3: active#(f(f(a()))) -> f#(g(f(a())))
p4: active#(f(f(a()))) -> g#(f(a()))
p5: mark#(f(X)) -> active#(f(mark(X)))
p6: mark#(f(X)) -> f#(mark(X))
p7: mark#(f(X)) -> mark#(X)
p8: mark#(a()) -> active#(a())
p9: mark#(c(X)) -> active#(c(X))
p10: mark#(g(X)) -> active#(g(mark(X)))
p11: mark#(g(X)) -> g#(mark(X))
p12: mark#(g(X)) -> mark#(X)
p13: f#(mark(X)) -> f#(X)
p14: f#(active(X)) -> f#(X)
p15: c#(mark(X)) -> c#(X)
p16: c#(active(X)) -> c#(X)
p17: g#(mark(X)) -> g#(X)
p18: g#(active(X)) -> g#(X)

and R consists of:

r1: active(f(f(a()))) -> mark(c(f(g(f(a())))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(a()) -> active(a())
r4: mark(c(X)) -> active(c(X))
r5: mark(g(X)) -> active(g(mark(X)))
r6: f(mark(X)) -> f(X)
r7: f(active(X)) -> f(X)
r8: c(mark(X)) -> c(X)
r9: c(active(X)) -> c(X)
r10: g(mark(X)) -> g(X)
r11: g(active(X)) -> g(X)

The estimated dependency graph contains the following SCCs:

  {p7, p12}
  {p1, p9}
  {p13, p14}
  {p17, p18}
  {p15, p16}


-- Reduction pair.

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

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

and R consists of:

r1: active(f(f(a()))) -> mark(c(f(g(f(a())))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(a()) -> active(a())
r4: mark(c(X)) -> active(c(X))
r5: mark(g(X)) -> active(g(mark(X)))
r6: f(mark(X)) -> f(X)
r7: f(active(X)) -> f(X)
r8: c(mark(X)) -> c(X)
r9: c(active(X)) -> c(X)
r10: g(mark(X)) -> g(X)
r11: g(active(X)) -> g(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      mark#_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1
      g_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

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(f(a()))) -> mark#(c(f(g(f(a())))))
p2: mark#(c(X)) -> active#(c(X))

and R consists of:

r1: active(f(f(a()))) -> mark(c(f(g(f(a())))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(a()) -> active(a())
r4: mark(c(X)) -> active(c(X))
r5: mark(g(X)) -> active(g(mark(X)))
r6: f(mark(X)) -> f(X)
r7: f(active(X)) -> f(X)
r8: c(mark(X)) -> c(X)
r9: c(active(X)) -> c(X)
r10: g(mark(X)) -> g(X)
r11: g(active(X)) -> g(X)

The set of usable rules consists of

  r8, r9

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),(0,0,0,0)) x1 + (0,2,0,0)
      f_A(x1) = ((0,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x1 + (3,0,1,1)
      a_A() = (1,1,1,0)
      mark#_A(x1) = ((0,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x1 + (2,0,0,1)
      c_A(x1) = (1,1,1,1)
      g_A(x1) = ((0,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x1 + (1,0,1,1)
      mark_A(x1) = ((0,0,0,0),(1,0,0,0),(0,0,0,0),(1,1,0,0)) x1 + (1,1,1,1)
      active_A(x1) = ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + (1,1,1,1)

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(f(a()))) -> mark(c(f(g(f(a())))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(a()) -> active(a())
r4: mark(c(X)) -> active(c(X))
r5: mark(g(X)) -> active(g(mark(X)))
r6: f(mark(X)) -> f(X)
r7: f(active(X)) -> f(X)
r8: c(mark(X)) -> c(X)
r9: c(active(X)) -> c(X)
r10: g(mark(X)) -> g(X)
r11: g(active(X)) -> g(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, r11

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(f(a()))) -> mark(c(f(g(f(a())))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(a()) -> active(a())
r4: mark(c(X)) -> active(c(X))
r5: mark(g(X)) -> active(g(mark(X)))
r6: f(mark(X)) -> f(X)
r7: f(active(X)) -> f(X)
r8: c(mark(X)) -> c(X)
r9: c(active(X)) -> c(X)
r10: g(mark(X)) -> g(X)
r11: g(active(X)) -> g(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, r11

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

and R consists of:

r1: active(f(f(a()))) -> mark(c(f(g(f(a())))))
r2: mark(f(X)) -> active(f(mark(X)))
r3: mark(a()) -> active(a())
r4: mark(c(X)) -> active(c(X))
r5: mark(g(X)) -> active(g(mark(X)))
r6: f(mark(X)) -> f(X)
r7: f(active(X)) -> f(X)
r8: c(mark(X)) -> c(X)
r9: c(active(X)) -> c(X)
r10: g(mark(X)) -> g(X)
r11: g(active(X)) -> g(X)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      c#_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, r11

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