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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        top#_A(x1) = ((1,1),(0,0)) x1
        ok_A(x1) = x1 + (0,1)
        active_A(x1) = x1 + (0,1)
        mark_A(x1) = x1 + (0,1)
        proper_A(x1) = x1 + (0,1)
        f_A(x1) = x1 + (1,0)
        if_A(x1,x2,x3) = x1 + x2 + x3 + (0,1)
        c_A() = (0,0)
        true_A() = (0,0)
        false_A() = (1,0)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        top#_A(x1) = ((1,1),(0,1)) x1
        ok_A(x1) = x1
        active_A(x1) = x1
        mark_A(x1) = x1 + (1,2)
        proper_A(x1) = x1 + (3,0)
        f_A(x1) = x1 + (2,3)
        if_A(x1,x2,x3) = x1 + x2 + (2,0)
        c_A() = (1,1)
        true_A() = (1,2)
        false_A() = (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        top#_A(x1) = ((0,1),(0,1)) x1
        ok_A(x1) = ((0,0),(1,0)) x1 + (2,1)
        active_A(x1) = ((1,1),(1,0)) x1 + (1,1)
        mark_A(x1) = (9,5)
        proper_A(x1) = (1,1)
        f_A(x1) = (9,10)
        if_A(x1,x2,x3) = (3,4)
        c_A() = (1,1)
        true_A() = (1,1)
        false_A() = (1,0)
    

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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        top#_A(x1) = ((1,1),(1,1)) x1
        ok_A(x1) = ((1,1),(1,1)) x1 + (1,3)
        active_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        f_A(x1) = ((1,1),(1,1)) x1 + (1,2)
        mark_A(x1) = (1,1)
        if_A(x1,x2,x3) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + ((1,1),(1,1)) x3 + (1,2)
        c_A() = (1,1)
        true_A() = (1,1)
        false_A() = (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        top#_A(x1) = ((1,0),(1,1)) x1
        ok_A(x1) = ((1,1),(1,0)) x1 + (1,7)
        active_A(x1) = ((1,0),(1,0)) x1 + (4,5)
        f_A(x1) = x1 + (5,11)
        mark_A(x1) = (8,12)
        if_A(x1,x2,x3) = ((0,1),(0,0)) x1 + x2 + (2,8)
        c_A() = (1,1)
        true_A() = (1,1)
        false_A() = (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        top#_A(x1) = x1
        ok_A(x1) = x1 + (2,3)
        active_A(x1) = (3,1)
        f_A(x1) = x1 + (1,1)
        mark_A(x1) = (2,1)
        if_A(x1,x2,x3) = ((0,0),(1,0)) x2 + (1,2)
        c_A() = (1,1)
        true_A() = (1,1)
        false_A() = (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 monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        active#_A(x1) = ((1,1),(1,0)) x1
        if_A(x1,x2,x3) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + ((1,1),(1,1)) x3 + (1,1)
        f_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        active#_A(x1) = ((1,0),(1,1)) x1
        if_A(x1,x2,x3) = ((1,1),(0,1)) x1 + ((1,1),(1,1)) x2 + ((1,1),(1,1)) x3 + (1,1)
        f_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        active#_A(x1) = ((1,1),(1,0)) x1
        if_A(x1,x2,x3) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + ((1,1),(1,1)) x3 + (1,1)
        f_A(x1) = ((1,1),(0,1)) x1 + (1,1)
    

The next rules are strictly ordered:

  p1, p2, p3
  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.

-- 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 monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        proper#_A(x1) = ((1,1),(1,1)) x1
        if_A(x1,x2,x3) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + ((1,1),(1,1)) x3 + (1,1)
        f_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        proper#_A(x1) = ((1,1),(1,1)) x1
        if_A(x1,x2,x3) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + ((1,1),(1,1)) x3 + (1,1)
        f_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        proper#_A(x1) = ((1,1),(1,1)) x1
        if_A(x1,x2,x3) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + ((1,1),(1,1)) x3 + (1,1)
        f_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    

The next rules are strictly ordered:

  p1, p2, p3, p4
  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.

-- 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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        if#_A(x1,x2,x3) = ((1,1),(0,0)) x1 + ((1,1),(0,1)) x2 + ((1,1),(0,0)) x3
        mark_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        ok_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        if#_A(x1,x2,x3) = ((1,1),(1,1)) x1 + ((1,1),(1,1)) x2 + ((1,1),(1,1)) x3
        mark_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        ok_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        if#_A(x1,x2,x3) = ((0,0),(1,0)) x1 + ((1,1),(1,1)) x2 + ((0,1),(0,1)) x3
        mark_A(x1) = ((0,1),(1,0)) x1 + (1,1)
        ok_A(x1) = ((0,1),(1,1)) x1 + (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:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1) = ((1,1),(1,0)) x1
        mark_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        ok_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1) = ((1,1),(1,0)) x1
        mark_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        ok_A(x1) = ((1,1),(1,1)) x1 + (1,1)
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        f#_A(x1) = ((1,1),(1,1)) x1
        mark_A(x1) = ((1,1),(0,1)) x1 + (1,1)
        ok_A(x1) = ((1,1),(1,1)) x1 + (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.