YES

We show the termination of the TRS R:

  active(f(X)) -> mark(if(X,c(),f(true())))
  active(if(true(),X,Y)) -> mark(X)
  active(if(false(),X,Y)) -> mark(Y)
  mark(f(X)) -> active(f(mark(X)))
  mark(if(X1,X2,X3)) -> active(if(mark(X1),mark(X2),X3))
  mark(c()) -> active(c())
  mark(true()) -> active(true())
  mark(false()) -> active(false())
  f(mark(X)) -> f(X)
  f(active(X)) -> f(X)
  if(mark(X1),X2,X3) -> if(X1,X2,X3)
  if(X1,mark(X2),X3) -> if(X1,X2,X3)
  if(X1,X2,mark(X3)) -> if(X1,X2,X3)
  if(active(X1),X2,X3) -> if(X1,X2,X3)
  if(X1,active(X2),X3) -> if(X1,X2,X3)
  if(X1,X2,active(X3)) -> if(X1,X2,X3)

-- SCC decomposition.

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

p1: active#(f(X)) -> mark#(if(X,c(),f(true())))
p2: active#(f(X)) -> if#(X,c(),f(true()))
p3: active#(f(X)) -> f#(true())
p4: active#(if(true(),X,Y)) -> mark#(X)
p5: active#(if(false(),X,Y)) -> mark#(Y)
p6: mark#(f(X)) -> active#(f(mark(X)))
p7: mark#(f(X)) -> f#(mark(X))
p8: mark#(f(X)) -> mark#(X)
p9: mark#(if(X1,X2,X3)) -> active#(if(mark(X1),mark(X2),X3))
p10: mark#(if(X1,X2,X3)) -> if#(mark(X1),mark(X2),X3)
p11: mark#(if(X1,X2,X3)) -> mark#(X1)
p12: mark#(if(X1,X2,X3)) -> mark#(X2)
p13: mark#(c()) -> active#(c())
p14: mark#(true()) -> active#(true())
p15: mark#(false()) -> active#(false())
p16: f#(mark(X)) -> f#(X)
p17: f#(active(X)) -> f#(X)
p18: if#(mark(X1),X2,X3) -> if#(X1,X2,X3)
p19: if#(X1,mark(X2),X3) -> if#(X1,X2,X3)
p20: if#(X1,X2,mark(X3)) -> if#(X1,X2,X3)
p21: if#(active(X1),X2,X3) -> if#(X1,X2,X3)
p22: if#(X1,active(X2),X3) -> if#(X1,X2,X3)
p23: if#(X1,X2,active(X3)) -> if#(X1,X2,X3)

and R consists of:

r1: active(f(X)) -> mark(if(X,c(),f(true())))
r2: active(if(true(),X,Y)) -> mark(X)
r3: active(if(false(),X,Y)) -> mark(Y)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(if(X1,X2,X3)) -> active(if(mark(X1),mark(X2),X3))
r6: mark(c()) -> active(c())
r7: mark(true()) -> active(true())
r8: mark(false()) -> active(false())
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: if(mark(X1),X2,X3) -> if(X1,X2,X3)
r12: if(X1,mark(X2),X3) -> if(X1,X2,X3)
r13: if(X1,X2,mark(X3)) -> if(X1,X2,X3)
r14: if(active(X1),X2,X3) -> if(X1,X2,X3)
r15: if(X1,active(X2),X3) -> if(X1,X2,X3)
r16: if(X1,X2,active(X3)) -> if(X1,X2,X3)

The estimated dependency graph contains the following SCCs:

  {p1, p4, p5, p6, p8, p9, p11, p12}
  {p18, p19, p20, p21, p22, p23}
  {p16, p17}


-- Reduction pair.

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

p1: active#(f(X)) -> mark#(if(X,c(),f(true())))
p2: mark#(if(X1,X2,X3)) -> mark#(X2)
p3: mark#(if(X1,X2,X3)) -> mark#(X1)
p4: mark#(if(X1,X2,X3)) -> active#(if(mark(X1),mark(X2),X3))
p5: active#(if(false(),X,Y)) -> mark#(Y)
p6: mark#(f(X)) -> mark#(X)
p7: mark#(f(X)) -> active#(f(mark(X)))
p8: active#(if(true(),X,Y)) -> mark#(X)

and R consists of:

r1: active(f(X)) -> mark(if(X,c(),f(true())))
r2: active(if(true(),X,Y)) -> mark(X)
r3: active(if(false(),X,Y)) -> mark(Y)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(if(X1,X2,X3)) -> active(if(mark(X1),mark(X2),X3))
r6: mark(c()) -> active(c())
r7: mark(true()) -> active(true())
r8: mark(false()) -> active(false())
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: if(mark(X1),X2,X3) -> if(X1,X2,X3)
r12: if(X1,mark(X2),X3) -> if(X1,X2,X3)
r13: if(X1,X2,mark(X3)) -> if(X1,X2,X3)
r14: if(active(X1),X2,X3) -> if(X1,X2,X3)
r15: if(X1,active(X2),X3) -> if(X1,X2,X3)
r16: if(X1,X2,active(X3)) -> if(X1,X2,X3)

The set of usable rules consists of

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

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      active#_A(x1) = ((1,0,0),(0,1,0),(0,1,0)) x1
      f_A(x1) = ((1,0,0),(1,1,0),(1,0,1)) x1 + (1,18,1)
      mark#_A(x1) = ((1,0,0),(1,1,0),(1,1,0)) x1 + (0,1,9)
      if_A(x1,x2,x3) = ((1,0,0),(0,1,0),(1,0,1)) x1 + ((1,0,0),(1,1,0),(0,0,1)) x2 + ((1,0,0),(0,0,0),(0,1,0)) x3 + (0,7,1)
      c_A() = (0,0,1)
      true_A() = (0,3,1)
      mark_A(x1) = ((1,0,0),(1,1,0),(1,1,1)) x1 + (0,0,5)
      false_A() = (1,1,1)
      active_A(x1) = x1 + (0,0,1)

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6, p7, p8

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

and R consists of:

r1: active(f(X)) -> mark(if(X,c(),f(true())))
r2: active(if(true(),X,Y)) -> mark(X)
r3: active(if(false(),X,Y)) -> mark(Y)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(if(X1,X2,X3)) -> active(if(mark(X1),mark(X2),X3))
r6: mark(c()) -> active(c())
r7: mark(true()) -> active(true())
r8: mark(false()) -> active(false())
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: if(mark(X1),X2,X3) -> if(X1,X2,X3)
r12: if(X1,mark(X2),X3) -> if(X1,X2,X3)
r13: if(X1,X2,mark(X3)) -> if(X1,X2,X3)
r14: if(active(X1),X2,X3) -> if(X1,X2,X3)
r15: if(X1,active(X2),X3) -> if(X1,X2,X3)
r16: if(X1,X2,active(X3)) -> if(X1,X2,X3)

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      if#_A(x1,x2,x3) = ((1,0,0),(1,1,0),(1,1,1)) x1 + ((1,0,0),(1,1,0),(1,1,1)) x2 + ((1,0,0),(1,1,0),(1,1,0)) x3
      mark_A(x1) = ((1,0,0),(1,1,0),(1,1,1)) x1 + (1,1,1)
      active_A(x1) = ((1,0,0),(1,1,0),(1,1,1)) x1 + (1,1,1)

The next rules are strictly ordered:

  p1, p2, p3, p4, p5, p6

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(if(X,c(),f(true())))
r2: active(if(true(),X,Y)) -> mark(X)
r3: active(if(false(),X,Y)) -> mark(Y)
r4: mark(f(X)) -> active(f(mark(X)))
r5: mark(if(X1,X2,X3)) -> active(if(mark(X1),mark(X2),X3))
r6: mark(c()) -> active(c())
r7: mark(true()) -> active(true())
r8: mark(false()) -> active(false())
r9: f(mark(X)) -> f(X)
r10: f(active(X)) -> f(X)
r11: if(mark(X1),X2,X3) -> if(X1,X2,X3)
r12: if(X1,mark(X2),X3) -> if(X1,X2,X3)
r13: if(X1,X2,mark(X3)) -> if(X1,X2,X3)
r14: if(active(X1),X2,X3) -> if(X1,X2,X3)
r15: if(X1,active(X2),X3) -> if(X1,X2,X3)
r16: if(X1,X2,active(X3)) -> if(X1,X2,X3)

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  matrix interpretations:
  
    carrier: N^3
    order: lexicographic order
    interpretations:
      f#_A(x1) = ((1,0,0),(1,1,0),(1,1,1)) x1
      mark_A(x1) = ((1,0,0),(1,1,0),(1,1,1)) x1 + (1,1,1)
      active_A(x1) = ((1,0,0),(1,1,0),(1,1,1)) x1 + (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

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