YES

We show the termination of the TRS R:

  active(f(x)) -> mark(x)
  top(active(c())) -> top(mark(c()))
  top(mark(x)) -> top(check(x))
  check(f(x)) -> f(check(x))
  check(x) -> start(match(f(X()),x))
  match(f(x),f(y)) -> f(match(x,y))
  match(X(),x) -> proper(x)
  proper(c()) -> ok(c())
  proper(f(x)) -> f(proper(x))
  f(ok(x)) -> ok(f(x))
  start(ok(x)) -> found(x)
  f(found(x)) -> found(f(x))
  top(found(x)) -> top(active(x))
  active(f(x)) -> f(active(x))
  f(mark(x)) -> mark(f(x))

-- SCC decomposition.

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

p1: top#(active(c())) -> top#(mark(c()))
p2: top#(mark(x)) -> top#(check(x))
p3: top#(mark(x)) -> check#(x)
p4: check#(f(x)) -> f#(check(x))
p5: check#(f(x)) -> check#(x)
p6: check#(x) -> start#(match(f(X()),x))
p7: check#(x) -> match#(f(X()),x)
p8: check#(x) -> f#(X())
p9: match#(f(x),f(y)) -> f#(match(x,y))
p10: match#(f(x),f(y)) -> match#(x,y)
p11: match#(X(),x) -> proper#(x)
p12: proper#(f(x)) -> f#(proper(x))
p13: proper#(f(x)) -> proper#(x)
p14: f#(ok(x)) -> f#(x)
p15: f#(found(x)) -> f#(x)
p16: top#(found(x)) -> top#(active(x))
p17: top#(found(x)) -> active#(x)
p18: active#(f(x)) -> f#(active(x))
p19: active#(f(x)) -> active#(x)
p20: f#(mark(x)) -> f#(x)

and R consists of:

r1: active(f(x)) -> mark(x)
r2: top(active(c())) -> top(mark(c()))
r3: top(mark(x)) -> top(check(x))
r4: check(f(x)) -> f(check(x))
r5: check(x) -> start(match(f(X()),x))
r6: match(f(x),f(y)) -> f(match(x,y))
r7: match(X(),x) -> proper(x)
r8: proper(c()) -> ok(c())
r9: proper(f(x)) -> f(proper(x))
r10: f(ok(x)) -> ok(f(x))
r11: start(ok(x)) -> found(x)
r12: f(found(x)) -> found(f(x))
r13: top(found(x)) -> top(active(x))
r14: active(f(x)) -> f(active(x))
r15: f(mark(x)) -> mark(f(x))

The estimated dependency graph contains the following SCCs:

  {p1, p2, p16}
  {p5}
  {p19}
  {p10}
  {p13}
  {p14, p15, p20}


-- Reduction pair.

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

p1: top#(active(c())) -> top#(mark(c()))
p2: top#(mark(x)) -> top#(check(x))
p3: top#(found(x)) -> top#(active(x))

and R consists of:

r1: active(f(x)) -> mark(x)
r2: top(active(c())) -> top(mark(c()))
r3: top(mark(x)) -> top(check(x))
r4: check(f(x)) -> f(check(x))
r5: check(x) -> start(match(f(X()),x))
r6: match(f(x),f(y)) -> f(match(x,y))
r7: match(X(),x) -> proper(x)
r8: proper(c()) -> ok(c())
r9: proper(f(x)) -> f(proper(x))
r10: f(ok(x)) -> ok(f(x))
r11: start(ok(x)) -> found(x)
r12: f(found(x)) -> found(f(x))
r13: top(found(x)) -> top(active(x))
r14: active(f(x)) -> f(active(x))
r15: f(mark(x)) -> mark(f(x))

The set of usable rules consists of

  r1, r4, r5, r6, r7, r8, r9, r10, r11, r12, r14, r15

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          top#_A(x1) = x1 + 11
          active_A(x1) = x1
          c_A = 5
          mark_A(x1) = max{4, x1 - 6}
          check_A(x1) = max{4, x1 - 18}
          found_A(x1) = x1
          proper_A(x1) = 6
          ok_A(x1) = x1
          f_A(x1) = max{4, x1 - 3}
          match_A(x1,x2) = max{4, x1 - 1}
          X_A = 8
          start_A(x1) = x1
    
      precedence:
      
        top# = active = c = mark = check = found = proper = ok = f = match = X = start
      
      partial status:
    
        pi(top#) = []
        pi(active) = []
        pi(c) = []
        pi(mark) = []
        pi(check) = []
        pi(found) = []
        pi(proper) = []
        pi(ok) = []
        pi(f) = []
        pi(match) = []
        pi(X) = []
        pi(start) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          top#_A(x1) = max{0, x1 - 9}
          active_A(x1) = x1
          c_A = 0
          mark_A(x1) = 3
          check_A(x1) = 8
          found_A(x1) = max{4, x1}
          proper_A(x1) = 2
          ok_A(x1) = max{3, x1 - 3}
          f_A(x1) = 4
          match_A(x1,x2) = 4
          X_A = 9
          start_A(x1) = x1 + 3
    
      precedence:
      
        c = X > active > match > proper > start > top# = mark = check = found = ok = f
      
      partial status:
    
        pi(top#) = []
        pi(active) = []
        pi(c) = []
        pi(mark) = []
        pi(check) = []
        pi(found) = []
        pi(proper) = []
        pi(ok) = []
        pi(f) = []
        pi(match) = []
        pi(X) = []
        pi(start) = [1]
    

The next rules are strictly ordered:

  p1

We remove them from the problem.

-- SCC decomposition.

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

p1: top#(mark(x)) -> top#(check(x))
p2: top#(found(x)) -> top#(active(x))

and R consists of:

r1: active(f(x)) -> mark(x)
r2: top(active(c())) -> top(mark(c()))
r3: top(mark(x)) -> top(check(x))
r4: check(f(x)) -> f(check(x))
r5: check(x) -> start(match(f(X()),x))
r6: match(f(x),f(y)) -> f(match(x,y))
r7: match(X(),x) -> proper(x)
r8: proper(c()) -> ok(c())
r9: proper(f(x)) -> f(proper(x))
r10: f(ok(x)) -> ok(f(x))
r11: start(ok(x)) -> found(x)
r12: f(found(x)) -> found(f(x))
r13: top(found(x)) -> top(active(x))
r14: active(f(x)) -> f(active(x))
r15: f(mark(x)) -> mark(f(x))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: top#(mark(x)) -> top#(check(x))
p2: top#(found(x)) -> top#(active(x))

and R consists of:

r1: active(f(x)) -> mark(x)
r2: top(active(c())) -> top(mark(c()))
r3: top(mark(x)) -> top(check(x))
r4: check(f(x)) -> f(check(x))
r5: check(x) -> start(match(f(X()),x))
r6: match(f(x),f(y)) -> f(match(x,y))
r7: match(X(),x) -> proper(x)
r8: proper(c()) -> ok(c())
r9: proper(f(x)) -> f(proper(x))
r10: f(ok(x)) -> ok(f(x))
r11: start(ok(x)) -> found(x)
r12: f(found(x)) -> found(f(x))
r13: top(found(x)) -> top(active(x))
r14: active(f(x)) -> f(active(x))
r15: f(mark(x)) -> mark(f(x))

The set of usable rules consists of

  r1, r4, r5, r6, r7, r8, r9, r10, r11, r12, r14, r15

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          top#_A(x1) = max{11, x1 + 6}
          mark_A(x1) = max{57, x1 + 46}
          check_A(x1) = x1 + 33
          found_A(x1) = max{41, x1 + 30}
          active_A(x1) = max{35, x1 + 24}
          proper_A(x1) = x1 + 80
          c_A = 46
          ok_A(x1) = max{125, x1 + 79}
          f_A(x1) = max{45, x1 + 34}
          match_A(x1,x2) = max{x1 + 12, x2 + 80}
          X_A = 13
          start_A(x1) = max{1, x1 - 48}
    
      precedence:
      
        c = X = start > active = match > proper > mark = check = found = ok = f > top#
      
      partial status:
    
        pi(top#) = [1]
        pi(mark) = [1]
        pi(check) = [1]
        pi(found) = []
        pi(active) = [1]
        pi(proper) = [1]
        pi(c) = []
        pi(ok) = [1]
        pi(f) = [1]
        pi(match) = [1, 2]
        pi(X) = []
        pi(start) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          top#_A(x1) = max{28, x1 + 6}
          mark_A(x1) = x1 + 29
          check_A(x1) = x1 + 12
          found_A(x1) = 30
          active_A(x1) = 23
          proper_A(x1) = max{53, x1 + 43}
          c_A = 26
          ok_A(x1) = x1 + 28
          f_A(x1) = max{22, x1 + 13}
          match_A(x1,x2) = max{52, x1 + 40, x2 + 42}
          X_A = 11
          start_A(x1) = 13
    
      precedence:
      
        top# = mark = check = found = active = proper = c = ok = f = match = X = start
      
      partial status:
    
        pi(top#) = [1]
        pi(mark) = [1]
        pi(check) = [1]
        pi(found) = []
        pi(active) = []
        pi(proper) = []
        pi(c) = []
        pi(ok) = [1]
        pi(f) = [1]
        pi(match) = [2]
        pi(X) = []
        pi(start) = []
    

The next rules are strictly ordered:

  p1, p2

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: check#(f(x)) -> check#(x)

and R consists of:

r1: active(f(x)) -> mark(x)
r2: top(active(c())) -> top(mark(c()))
r3: top(mark(x)) -> top(check(x))
r4: check(f(x)) -> f(check(x))
r5: check(x) -> start(match(f(X()),x))
r6: match(f(x),f(y)) -> f(match(x,y))
r7: match(X(),x) -> proper(x)
r8: proper(c()) -> ok(c())
r9: proper(f(x)) -> f(proper(x))
r10: f(ok(x)) -> ok(f(x))
r11: start(ok(x)) -> found(x)
r12: f(found(x)) -> found(f(x))
r13: top(found(x)) -> top(active(x))
r14: active(f(x)) -> f(active(x))
r15: f(mark(x)) -> mark(f(x))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          check#_A(x1) = max{4, x1 + 3}
          f_A(x1) = max{3, x1 + 2}
    
      precedence:
      
        check# = f
      
      partial status:
    
        pi(check#) = [1]
        pi(f) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          check#_A(x1) = max{1, x1 - 1}
          f_A(x1) = x1
    
      precedence:
      
        check# = f
      
      partial status:
    
        pi(check#) = []
        pi(f) = [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#(f(x)) -> active#(x)

and R consists of:

r1: active(f(x)) -> mark(x)
r2: top(active(c())) -> top(mark(c()))
r3: top(mark(x)) -> top(check(x))
r4: check(f(x)) -> f(check(x))
r5: check(x) -> start(match(f(X()),x))
r6: match(f(x),f(y)) -> f(match(x,y))
r7: match(X(),x) -> proper(x)
r8: proper(c()) -> ok(c())
r9: proper(f(x)) -> f(proper(x))
r10: f(ok(x)) -> ok(f(x))
r11: start(ok(x)) -> found(x)
r12: f(found(x)) -> found(f(x))
r13: top(found(x)) -> top(active(x))
r14: active(f(x)) -> f(active(x))
r15: f(mark(x)) -> mark(f(x))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = max{4, x1 + 3}
          f_A(x1) = max{3, x1 + 2}
    
      precedence:
      
        active# = f
      
      partial status:
    
        pi(active#) = [1]
        pi(f) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          active#_A(x1) = max{1, x1 - 1}
          f_A(x1) = x1
    
      precedence:
      
        active# = f
      
      partial status:
    
        pi(active#) = []
        pi(f) = [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: match#(f(x),f(y)) -> match#(x,y)

and R consists of:

r1: active(f(x)) -> mark(x)
r2: top(active(c())) -> top(mark(c()))
r3: top(mark(x)) -> top(check(x))
r4: check(f(x)) -> f(check(x))
r5: check(x) -> start(match(f(X()),x))
r6: match(f(x),f(y)) -> f(match(x,y))
r7: match(X(),x) -> proper(x)
r8: proper(c()) -> ok(c())
r9: proper(f(x)) -> f(proper(x))
r10: f(ok(x)) -> ok(f(x))
r11: start(ok(x)) -> found(x)
r12: f(found(x)) -> found(f(x))
r13: top(found(x)) -> top(active(x))
r14: active(f(x)) -> f(active(x))
r15: f(mark(x)) -> mark(f(x))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          match#_A(x1,x2) = max{2, x1 - 1, x2 + 1}
          f_A(x1) = max{1, x1}
    
      precedence:
      
        match# = f
      
      partial status:
    
        pi(match#) = [2]
        pi(f) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          match#_A(x1,x2) = 0
          f_A(x1) = max{2, x1}
    
      precedence:
      
        match# = f
      
      partial status:
    
        pi(match#) = []
        pi(f) = [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: proper#(f(x)) -> proper#(x)

and R consists of:

r1: active(f(x)) -> mark(x)
r2: top(active(c())) -> top(mark(c()))
r3: top(mark(x)) -> top(check(x))
r4: check(f(x)) -> f(check(x))
r5: check(x) -> start(match(f(X()),x))
r6: match(f(x),f(y)) -> f(match(x,y))
r7: match(X(),x) -> proper(x)
r8: proper(c()) -> ok(c())
r9: proper(f(x)) -> f(proper(x))
r10: f(ok(x)) -> ok(f(x))
r11: start(ok(x)) -> found(x)
r12: f(found(x)) -> found(f(x))
r13: top(found(x)) -> top(active(x))
r14: active(f(x)) -> f(active(x))
r15: f(mark(x)) -> mark(f(x))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          proper#_A(x1) = max{2, x1 + 1}
          f_A(x1) = max{1, x1}
    
      precedence:
      
        proper# = f
      
      partial status:
    
        pi(proper#) = [1]
        pi(f) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          proper#_A(x1) = x1 + 2
          f_A(x1) = x1 + 1
    
      precedence:
      
        proper# = f
      
      partial status:
    
        pi(proper#) = [1]
        pi(f) = [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: f#(ok(x)) -> f#(x)
p2: f#(mark(x)) -> f#(x)
p3: f#(found(x)) -> f#(x)

and R consists of:

r1: active(f(x)) -> mark(x)
r2: top(active(c())) -> top(mark(c()))
r3: top(mark(x)) -> top(check(x))
r4: check(f(x)) -> f(check(x))
r5: check(x) -> start(match(f(X()),x))
r6: match(f(x),f(y)) -> f(match(x,y))
r7: match(X(),x) -> proper(x)
r8: proper(c()) -> ok(c())
r9: proper(f(x)) -> f(proper(x))
r10: f(ok(x)) -> ok(f(x))
r11: start(ok(x)) -> found(x)
r12: f(found(x)) -> found(f(x))
r13: top(found(x)) -> top(active(x))
r14: active(f(x)) -> f(active(x))
r15: f(mark(x)) -> mark(f(x))

The set of usable rules consists of

  (no rules)

Take the monotone reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = max{2, x1 + 1}
          ok_A(x1) = max{1, x1}
          mark_A(x1) = max{1, x1}
          found_A(x1) = max{1, x1}
    
      precedence:
      
        f# = ok = mark = found
      
      partial status:
    
        pi(f#) = [1]
        pi(ok) = [1]
        pi(mark) = [1]
        pi(found) = [1]
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1) = x1 + 1
          ok_A(x1) = x1
          mark_A(x1) = x1
          found_A(x1) = x1
    
      precedence:
      
        f# = ok = mark = found
      
      partial status:
    
        pi(f#) = [1]
        pi(ok) = [1]
        pi(mark) = [1]
        pi(found) = [1]
    

The next rules are strictly ordered:

  p1, p2, p3

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