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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        top#_A(x1) = max{1, x1 - 79}
        active_A(x1) = max{68, x1 - 11}
        c_A = 92
        mark_A(x1) = 3
        check_A(x1) = 53
        found_A(x1) = max{19, x1 - 10}
        proper_A(x1) = 89
        ok_A(x1) = max{20, x1 - 4}
        f_A(x1) = 25
        match_A(x1,x2) = max{58, x1 + 29}
        X_A = 61
        start_A(x1) = max{53, x1 - 6}
  
    precedence:
    
      c = check = start > active > mark = f > found > match > proper = ok > top# = X
    
    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) = []

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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        top#_A(x1) = max{40, x1 + 4}
        mark_A(x1) = x1 + 68
        check_A(x1) = x1 + 55
        found_A(x1) = max{39, x1 + 37}
        active_A(x1) = x1 + 37
        proper_A(x1) = x1
        c_A = 15
        ok_A(x1) = max{14, x1 - 16}
        f_A(x1) = x1 + 31
        match_A(x1,x2) = max{x1 - 82, x2}
        X_A = 24
        start_A(x1) = x1 + 54
  
    precedence:
    
      c > active > check = ok = f = match = start > top# = mark = found = proper = X
    
    partial status:
  
      pi(top#) = []
      pi(mark) = [1]
      pi(check) = []
      pi(found) = []
      pi(active) = []
      pi(proper) = [1]
      pi(c) = []
      pi(ok) = []
      pi(f) = []
      pi(match) = [2]
      pi(X) = []
      pi(start) = []

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#(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}


-- Reduction pair.

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

p1: 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, r10, r12, r14, r15

Take the reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        top#_A(x1) = max{5, x1 + 4}
        found_A(x1) = x1 + 4
        active_A(x1) = x1 + 2
        f_A(x1) = x1 + 3
        ok_A(x1) = x1 + 4
        mark_A(x1) = x1 + 4
  
    precedence:
    
      top# = found = active = f = ok = mark
    
    partial status:
  
      pi(top#) = []
      pi(found) = [1]
      pi(active) = [1]
      pi(f) = [1]
      pi(ok) = [1]
      pi(mark) = []

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: 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 monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        check#_A(x1) = x1 + 2
        f_A(x1) = x1 + 2
  
    precedence:
    
      check# = f
    
    partial status:
  
      pi(check#) = [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: 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 monotone reduction pair:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        active#_A(x1) = x1 + 2
        f_A(x1) = x1 + 2
  
    precedence:
    
      active# = f
    
    partial status:
  
      pi(active#) = [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: 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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        match#_A(x1,x2) = max{0, x1 - 1, x2 - 1}
        f_A(x1) = x1 + 2
  
    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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        proper#_A(x1) = x1 + 1
        f_A(x1) = x1
  
    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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        f#_A(x1) = x1 + 3
        ok_A(x1) = x1
        mark_A(x1) = x1 + 1
        found_A(x1) = x1 + 2
  
    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

We remove them from the problem.

-- SCC decomposition.

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

p1: f#(mark(x)) -> f#(x)
p2: 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 estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

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

p1: f#(mark(x)) -> f#(x)
p2: 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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        f#_A(x1) = max{3, x1 + 2}
        mark_A(x1) = x1 + 3
        found_A(x1) = max{1, x1}
  
    precedence:
    
      f# = mark = found
    
    partial status:
  
      pi(f#) = [1]
      pi(mark) = [1]
      pi(found) = [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: 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 estimated dependency graph contains the following SCCs:

  {p1}


-- Reduction pair.

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

p1: 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:

  weighted path order
  
    base order:
  
      max/plus interpretations on natural numbers:
      
        f#_A(x1) = x1 + 2
        found_A(x1) = x1 + 2
  
    precedence:
    
      f# = found
    
    partial status:
  
      pi(f#) = [1]
      pi(found) = [1]

The next rules are strictly ordered:

  p1

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