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)
  active(f(X)) -> f(active(X))
  active(if(X1,X2,X3)) -> if(active(X1),X2,X3)
  active(if(X1,X2,X3)) -> if(X1,active(X2),X3)
  f(mark(X)) -> mark(f(X))
  if(mark(X1),X2,X3) -> mark(if(X1,X2,X3))
  if(X1,mark(X2),X3) -> mark(if(X1,X2,X3))
  proper(f(X)) -> f(proper(X))
  proper(if(X1,X2,X3)) -> if(proper(X1),proper(X2),proper(X3))
  proper(c()) -> ok(c())
  proper(true()) -> ok(true())
  proper(false()) -> ok(false())
  f(ok(X)) -> ok(f(X))
  if(ok(X1),ok(X2),ok(X3)) -> ok(if(X1,X2,X3))
  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(X)) -> if#(X,c(),f(true()))
p2: active#(f(X)) -> f#(true())
p3: active#(f(X)) -> f#(active(X))
p4: active#(f(X)) -> active#(X)
p5: active#(if(X1,X2,X3)) -> if#(active(X1),X2,X3)
p6: active#(if(X1,X2,X3)) -> active#(X1)
p7: active#(if(X1,X2,X3)) -> if#(X1,active(X2),X3)
p8: active#(if(X1,X2,X3)) -> active#(X2)
p9: f#(mark(X)) -> f#(X)
p10: if#(mark(X1),X2,X3) -> if#(X1,X2,X3)
p11: if#(X1,mark(X2),X3) -> if#(X1,X2,X3)
p12: proper#(f(X)) -> f#(proper(X))
p13: proper#(f(X)) -> proper#(X)
p14: proper#(if(X1,X2,X3)) -> if#(proper(X1),proper(X2),proper(X3))
p15: proper#(if(X1,X2,X3)) -> proper#(X1)
p16: proper#(if(X1,X2,X3)) -> proper#(X2)
p17: proper#(if(X1,X2,X3)) -> proper#(X3)
p18: f#(ok(X)) -> f#(X)
p19: if#(ok(X1),ok(X2),ok(X3)) -> if#(X1,X2,X3)
p20: top#(mark(X)) -> top#(proper(X))
p21: top#(mark(X)) -> proper#(X)
p22: top#(ok(X)) -> top#(active(X))
p23: top#(ok(X)) -> active#(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: active(f(X)) -> f(active(X))
r5: active(if(X1,X2,X3)) -> if(active(X1),X2,X3)
r6: active(if(X1,X2,X3)) -> if(X1,active(X2),X3)
r7: f(mark(X)) -> mark(f(X))
r8: if(mark(X1),X2,X3) -> mark(if(X1,X2,X3))
r9: if(X1,mark(X2),X3) -> mark(if(X1,X2,X3))
r10: proper(f(X)) -> f(proper(X))
r11: proper(if(X1,X2,X3)) -> if(proper(X1),proper(X2),proper(X3))
r12: proper(c()) -> ok(c())
r13: proper(true()) -> ok(true())
r14: proper(false()) -> ok(false())
r15: f(ok(X)) -> ok(f(X))
r16: if(ok(X1),ok(X2),ok(X3)) -> ok(if(X1,X2,X3))
r17: top(mark(X)) -> top(proper(X))
r18: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p20, p22}
  {p4, p6, p8}
  {p13, p15, p16, p17}
  {p10, p11, p19}
  {p9, p18}


-- 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(X)) -> mark(if(X,c(),f(true())))
r2: active(if(true(),X,Y)) -> mark(X)
r3: active(if(false(),X,Y)) -> mark(Y)
r4: active(f(X)) -> f(active(X))
r5: active(if(X1,X2,X3)) -> if(active(X1),X2,X3)
r6: active(if(X1,X2,X3)) -> if(X1,active(X2),X3)
r7: f(mark(X)) -> mark(f(X))
r8: if(mark(X1),X2,X3) -> mark(if(X1,X2,X3))
r9: if(X1,mark(X2),X3) -> mark(if(X1,X2,X3))
r10: proper(f(X)) -> f(proper(X))
r11: proper(if(X1,X2,X3)) -> if(proper(X1),proper(X2),proper(X3))
r12: proper(c()) -> ok(c())
r13: proper(true()) -> ok(true())
r14: proper(false()) -> ok(false())
r15: f(ok(X)) -> ok(f(X))
r16: if(ok(X1),ok(X2),ok(X3)) -> ok(if(X1,X2,X3))
r17: top(mark(X)) -> top(proper(X))
r18: 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

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      top#_A(x1) = ((1,0,0,0),(0,0,0,0),(0,1,0,0),(0,0,1,0)) x1
      ok_A(x1) = x1
      active_A(x1) = x1 + (0,0,0,1)
      mark_A(x1) = x1 + (0,0,1,0)
      proper_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,1,1,0)) x1
      f_A(x1) = ((1,0,0,0),(1,1,0,0),(1,0,1,0),(1,0,0,0)) x1 + (1,2,4,1)
      if_A(x1,x2,x3) = ((1,0,0,0),(1,1,0,0),(0,0,1,0),(0,0,0,1)) x1 + ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,1,0,0)) x2 + x3 + (0,0,2,1)
      c_A() = (0,1,0,1)
      true_A() = (0,0,1,1)
      false_A() = (1,1,1,1)

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

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

p1: top#(ok(X)) -> top#(active(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: active(f(X)) -> f(active(X))
r5: active(if(X1,X2,X3)) -> if(active(X1),X2,X3)
r6: active(if(X1,X2,X3)) -> if(X1,active(X2),X3)
r7: f(mark(X)) -> mark(f(X))
r8: if(mark(X1),X2,X3) -> mark(if(X1,X2,X3))
r9: if(X1,mark(X2),X3) -> mark(if(X1,X2,X3))
r10: proper(f(X)) -> f(proper(X))
r11: proper(if(X1,X2,X3)) -> if(proper(X1),proper(X2),proper(X3))
r12: proper(c()) -> ok(c())
r13: proper(true()) -> ok(true())
r14: proper(false()) -> ok(false())
r15: f(ok(X)) -> ok(f(X))
r16: if(ok(X1),ok(X2),ok(X3)) -> ok(if(X1,X2,X3))
r17: top(mark(X)) -> top(proper(X))
r18: top(ok(X)) -> top(active(X))

The estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: top#(ok(X)) -> top#(active(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: active(f(X)) -> f(active(X))
r5: active(if(X1,X2,X3)) -> if(active(X1),X2,X3)
r6: active(if(X1,X2,X3)) -> if(X1,active(X2),X3)
r7: f(mark(X)) -> mark(f(X))
r8: if(mark(X1),X2,X3) -> mark(if(X1,X2,X3))
r9: if(X1,mark(X2),X3) -> mark(if(X1,X2,X3))
r10: proper(f(X)) -> f(proper(X))
r11: proper(if(X1,X2,X3)) -> if(proper(X1),proper(X2),proper(X3))
r12: proper(c()) -> ok(c())
r13: proper(true()) -> ok(true())
r14: proper(false()) -> ok(false())
r15: f(ok(X)) -> ok(f(X))
r16: if(ok(X1),ok(X2),ok(X3)) -> ok(if(X1,X2,X3))
r17: top(mark(X)) -> top(proper(X))
r18: top(ok(X)) -> top(active(X))

The set of usable rules consists of

  r1, r2, r3, r4, r5, r6, r7, r8, r9, r15, r16

Take the reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      top#_A(x1) = ((0,0,0,0),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1
      ok_A(x1) = ((1,0,0,0),(0,1,0,0),(1,0,0,0),(1,1,0,0)) x1 + (2,1,1,1)
      active_A(x1) = ((1,0,0,0),(1,1,0,0),(0,0,1,0),(0,0,0,1)) x1 + (1,2,1,1)
      f_A(x1) = ((1,0,0,0),(1,1,0,0),(0,0,0,0),(0,1,0,0)) x1 + (2,2,1,1)
      mark_A(x1) = ((0,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x1 + (1,0,5,4)
      if_A(x1,x2,x3) = x1 + ((1,0,0,0),(1,0,0,0),(0,1,0,0),(0,0,0,0)) x2 + ((1,0,0,0),(0,1,0,0),(1,0,1,0),(1,1,1,1)) x3 + (5,2,2,1)
      c_A() = (1,1,1,1)
      true_A() = (1,1,3,1)
      false_A() = (1,1,1,1)

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: active#(if(X1,X2,X3)) -> active#(X2)
p2: active#(if(X1,X2,X3)) -> active#(X1)
p3: active#(f(X)) -> active#(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: active(f(X)) -> f(active(X))
r5: active(if(X1,X2,X3)) -> if(active(X1),X2,X3)
r6: active(if(X1,X2,X3)) -> if(X1,active(X2),X3)
r7: f(mark(X)) -> mark(f(X))
r8: if(mark(X1),X2,X3) -> mark(if(X1,X2,X3))
r9: if(X1,mark(X2),X3) -> mark(if(X1,X2,X3))
r10: proper(f(X)) -> f(proper(X))
r11: proper(if(X1,X2,X3)) -> if(proper(X1),proper(X2),proper(X3))
r12: proper(c()) -> ok(c())
r13: proper(true()) -> ok(true())
r14: proper(false()) -> ok(false())
r15: f(ok(X)) -> ok(f(X))
r16: if(ok(X1),ok(X2),ok(X3)) -> ok(if(X1,X2,X3))
r17: top(mark(X)) -> top(proper(X))
r18: 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:
      active#_A(x1) = ((1,0,0,0),(1,1,0,0),(1,0,1,0),(1,1,1,1)) x1
      if_A(x1,x2,x3) = ((1,0,0,0),(1,0,0,0),(1,1,0,0),(1,1,1,0)) x1 + ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x2 + ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x3 + (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

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#(if(X1,X2,X3)) -> proper#(X3)
p2: proper#(if(X1,X2,X3)) -> proper#(X2)
p3: proper#(if(X1,X2,X3)) -> proper#(X1)
p4: proper#(f(X)) -> proper#(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: active(f(X)) -> f(active(X))
r5: active(if(X1,X2,X3)) -> if(active(X1),X2,X3)
r6: active(if(X1,X2,X3)) -> if(X1,active(X2),X3)
r7: f(mark(X)) -> mark(f(X))
r8: if(mark(X1),X2,X3) -> mark(if(X1,X2,X3))
r9: if(X1,mark(X2),X3) -> mark(if(X1,X2,X3))
r10: proper(f(X)) -> f(proper(X))
r11: proper(if(X1,X2,X3)) -> if(proper(X1),proper(X2),proper(X3))
r12: proper(c()) -> ok(c())
r13: proper(true()) -> ok(true())
r14: proper(false()) -> ok(false())
r15: f(ok(X)) -> ok(f(X))
r16: if(ok(X1),ok(X2),ok(X3)) -> ok(if(X1,X2,X3))
r17: top(mark(X)) -> top(proper(X))
r18: 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),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1
      if_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),(1,0,0,0),(1,1,0,0),(1,1,1,0)) x2 + ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x3 + (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

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#(ok(X1),ok(X2),ok(X3)) -> if#(X1,X2,X3)
p3: 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: active(f(X)) -> f(active(X))
r5: active(if(X1,X2,X3)) -> if(active(X1),X2,X3)
r6: active(if(X1,X2,X3)) -> if(X1,active(X2),X3)
r7: f(mark(X)) -> mark(f(X))
r8: if(mark(X1),X2,X3) -> mark(if(X1,X2,X3))
r9: if(X1,mark(X2),X3) -> mark(if(X1,X2,X3))
r10: proper(f(X)) -> f(proper(X))
r11: proper(if(X1,X2,X3)) -> if(proper(X1),proper(X2),proper(X3))
r12: proper(c()) -> ok(c())
r13: proper(true()) -> ok(true())
r14: proper(false()) -> ok(false())
r15: f(ok(X)) -> ok(f(X))
r16: if(ok(X1),ok(X2),ok(X3)) -> ok(if(X1,X2,X3))
r17: top(mark(X)) -> top(proper(X))
r18: 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:
      if#_A(x1,x2,x3) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(0,0,1,0)) x1 + ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) 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)
      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, 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(X)) -> f#(X)
p2: f#(ok(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: active(f(X)) -> f(active(X))
r5: active(if(X1,X2,X3)) -> if(active(X1),X2,X3)
r6: active(if(X1,X2,X3)) -> if(X1,active(X2),X3)
r7: f(mark(X)) -> mark(f(X))
r8: if(mark(X1),X2,X3) -> mark(if(X1,X2,X3))
r9: if(X1,mark(X2),X3) -> mark(if(X1,X2,X3))
r10: proper(f(X)) -> f(proper(X))
r11: proper(if(X1,X2,X3)) -> if(proper(X1),proper(X2),proper(X3))
r12: proper(c()) -> ok(c())
r13: proper(true()) -> ok(true())
r14: proper(false()) -> ok(false())
r15: f(ok(X)) -> ok(f(X))
r16: if(ok(X1),ok(X2),ok(X3)) -> ok(if(X1,X2,X3))
r17: top(mark(X)) -> top(proper(X))
r18: 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

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