YES

We show the termination of the TRS R:

  active(f(a(),X,X)) -> mark(f(X,b(),b()))
  active(b()) -> mark(a())
  mark(f(X1,X2,X3)) -> active(f(X1,mark(X2),X3))
  mark(a()) -> active(a())
  mark(b()) -> active(b())
  f(mark(X1),X2,X3) -> f(X1,X2,X3)
  f(X1,mark(X2),X3) -> f(X1,X2,X3)
  f(X1,X2,mark(X3)) -> f(X1,X2,X3)
  f(active(X1),X2,X3) -> f(X1,X2,X3)
  f(X1,active(X2),X3) -> f(X1,X2,X3)
  f(X1,X2,active(X3)) -> f(X1,X2,X3)

-- SCC decomposition.

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

p1: active#(f(a(),X,X)) -> mark#(f(X,b(),b()))
p2: active#(f(a(),X,X)) -> f#(X,b(),b())
p3: active#(b()) -> mark#(a())
p4: mark#(f(X1,X2,X3)) -> active#(f(X1,mark(X2),X3))
p5: mark#(f(X1,X2,X3)) -> f#(X1,mark(X2),X3)
p6: mark#(f(X1,X2,X3)) -> mark#(X2)
p7: mark#(a()) -> active#(a())
p8: mark#(b()) -> active#(b())
p9: f#(mark(X1),X2,X3) -> f#(X1,X2,X3)
p10: f#(X1,mark(X2),X3) -> f#(X1,X2,X3)
p11: f#(X1,X2,mark(X3)) -> f#(X1,X2,X3)
p12: f#(active(X1),X2,X3) -> f#(X1,X2,X3)
p13: f#(X1,active(X2),X3) -> f#(X1,X2,X3)
p14: f#(X1,X2,active(X3)) -> f#(X1,X2,X3)

and R consists of:

r1: active(f(a(),X,X)) -> mark(f(X,b(),b()))
r2: active(b()) -> mark(a())
r3: mark(f(X1,X2,X3)) -> active(f(X1,mark(X2),X3))
r4: mark(a()) -> active(a())
r5: mark(b()) -> active(b())
r6: f(mark(X1),X2,X3) -> f(X1,X2,X3)
r7: f(X1,mark(X2),X3) -> f(X1,X2,X3)
r8: f(X1,X2,mark(X3)) -> f(X1,X2,X3)
r9: f(active(X1),X2,X3) -> f(X1,X2,X3)
r10: f(X1,active(X2),X3) -> f(X1,X2,X3)
r11: f(X1,X2,active(X3)) -> f(X1,X2,X3)

The estimated dependency graph contains the following SCCs:

  {p1, p4, p6}
  {p9, p10, p11, p12, p13, p14}


-- Reduction pair.

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

p1: active#(f(a(),X,X)) -> mark#(f(X,b(),b()))
p2: mark#(f(X1,X2,X3)) -> mark#(X2)
p3: mark#(f(X1,X2,X3)) -> active#(f(X1,mark(X2),X3))

and R consists of:

r1: active(f(a(),X,X)) -> mark(f(X,b(),b()))
r2: active(b()) -> mark(a())
r3: mark(f(X1,X2,X3)) -> active(f(X1,mark(X2),X3))
r4: mark(a()) -> active(a())
r5: mark(b()) -> active(b())
r6: f(mark(X1),X2,X3) -> f(X1,X2,X3)
r7: f(X1,mark(X2),X3) -> f(X1,X2,X3)
r8: f(X1,X2,mark(X3)) -> f(X1,X2,X3)
r9: f(active(X1),X2,X3) -> f(X1,X2,X3)
r10: f(X1,active(X2),X3) -> f(X1,X2,X3)
r11: f(X1,X2,active(X3)) -> f(X1,X2,X3)

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      active#_A(x1) = ((1,0,0,0),(1,1,0,0),(0,0,1,0),(0,0,0,1)) x1
      f_A(x1,x2,x3) = ((1,0,0,0),(1,1,0,0),(1,0,1,0),(0,1,0,0)) x1 + ((1,0,0,0),(1,0,0,0),(1,0,0,0),(1,0,0,0)) x2 + ((1,0,0,0),(1,1,0,0),(1,1,1,0),(0,0,0,0)) x3 + (1,0,0,1)
      a_A() = (0,1,5,1)
      mark#_A(x1) = ((1,0,0,0),(1,1,0,0),(0,1,1,0),(0,1,1,0)) x1 + (0,0,1,2)
      b_A() = (0,1,1,1)
      mark_A(x1) = x1 + (0,1,1,1)
      active_A(x1) = x1 + (0,1,1,1)

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#(mark(X1),X2,X3) -> f#(X1,X2,X3)
p2: f#(X1,X2,active(X3)) -> f#(X1,X2,X3)
p3: f#(X1,active(X2),X3) -> f#(X1,X2,X3)
p4: f#(active(X1),X2,X3) -> f#(X1,X2,X3)
p5: f#(X1,X2,mark(X3)) -> f#(X1,X2,X3)
p6: f#(X1,mark(X2),X3) -> f#(X1,X2,X3)

and R consists of:

r1: active(f(a(),X,X)) -> mark(f(X,b(),b()))
r2: active(b()) -> mark(a())
r3: mark(f(X1,X2,X3)) -> active(f(X1,mark(X2),X3))
r4: mark(a()) -> active(a())
r5: mark(b()) -> active(b())
r6: f(mark(X1),X2,X3) -> f(X1,X2,X3)
r7: f(X1,mark(X2),X3) -> f(X1,X2,X3)
r8: f(X1,X2,mark(X3)) -> f(X1,X2,X3)
r9: f(active(X1),X2,X3) -> f(X1,X2,X3)
r10: f(X1,active(X2),X3) -> f(X1,X2,X3)
r11: f(X1,X2,active(X3)) -> f(X1,X2,X3)

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,x2,x3) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,0,1,1)) x2 + ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x3
      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, p3, p4, p5, p6
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11

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