YES

We show the termination of the TRS R:

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

-- SCC decomposition.

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

p1: a__f#(a(),X,X) -> a__f#(X,a__b(),b())
p2: a__f#(a(),X,X) -> a__b#()
p3: mark#(f(X1,X2,X3)) -> a__f#(X1,mark(X2),X3)
p4: mark#(f(X1,X2,X3)) -> mark#(X2)
p5: mark#(b()) -> a__b#()

and R consists of:

r1: a__f(a(),X,X) -> a__f(X,a__b(),b())
r2: a__b() -> a()
r3: mark(f(X1,X2,X3)) -> a__f(X1,mark(X2),X3)
r4: mark(b()) -> a__b()
r5: mark(a()) -> a()
r6: a__f(X1,X2,X3) -> f(X1,X2,X3)
r7: a__b() -> b()

The estimated dependency graph contains the following SCCs:

  {p4}
  {p1}


-- Reduction pair.

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

p1: mark#(f(X1,X2,X3)) -> mark#(X2)

and R consists of:

r1: a__f(a(),X,X) -> a__f(X,a__b(),b())
r2: a__b() -> a()
r3: mark(f(X1,X2,X3)) -> a__f(X1,mark(X2),X3)
r4: mark(b()) -> a__b()
r5: mark(a()) -> a()
r6: a__f(X1,X2,X3) -> f(X1,X2,X3)
r7: a__b() -> b()

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      mark#_A(x1) = x1
      f_A(x1,x2,x3) = x2 + 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: a__f#(a(),X,X) -> a__f#(X,a__b(),b())

and R consists of:

r1: a__f(a(),X,X) -> a__f(X,a__b(),b())
r2: a__b() -> a()
r3: mark(f(X1,X2,X3)) -> a__f(X1,mark(X2),X3)
r4: mark(b()) -> a__b()
r5: mark(a()) -> a()
r6: a__f(X1,X2,X3) -> f(X1,X2,X3)
r7: a__b() -> b()

The set of usable rules consists of

  r2, r7

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^1
    order: standard order
    interpretations:
      a__f#_A(x1,x2,x3) = x1 + x3
      a_A() = 1
      a__b_A() = 1
      b_A() = 0

The next rules are strictly ordered:

  p1

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