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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      top#_A(x1) = ((1,0,0,0),(0,0,0,0),(1,0,0,0),(0,0,0,0)) x1
      active_A(x1) = ((1,0,0,0),(1,0,0,0),(0,0,0,0),(0,0,0,0)) x1 + (0,5,2,3)
      c_A() = (1,0,1,1)
      mark_A(x1) = (0,4,1,2)
      check_A(x1) = (0,5,6,2)
      found_A(x1) = x1 + (0,4,1,1)
      proper_A(x1) = ((0,0,0,0),(1,0,0,0),(1,0,0,0),(0,0,0,0)) x1 + (2,2,4,0)
      ok_A(x1) = x1 + (0,3,6,1)
      f_A(x1) = (0,4,7,2)
      match_A(x1,x2) = x1 + (0,1,7,1)
      X_A() = (3,0,1,0)
      start_A(x1) = ((1,0,0,0),(0,0,0,0),(0,1,0,0),(0,0,0,0)) x1 + (0,5,0,2)

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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      top#_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),(0,0,0,0),(0,0,0,0),(1,0,0,0)) x1 + (7,8,9,23)
      check_A(x1) = ((1,0,0,0),(0,1,0,0),(1,0,0,0),(1,0,0,0)) x1 + (6,1,17,21)
      found_A(x1) = ((1,0,0,0),(0,0,0,0),(0,1,0,0),(1,1,1,0)) x1 + (2,0,2,13)
      active_A(x1) = ((1,0,0,0),(0,1,0,0),(1,1,0,0),(1,1,0,0)) x1 + (1,0,1,13)
      proper_A(x1) = ((1,0,0,0),(0,0,0,0),(0,0,0,0),(0,1,0,0)) x1 + (2,9,11,2)
      c_A() = (0,11,1,1)
      ok_A(x1) = ((1,0,0,0),(0,1,0,0),(0,1,0,0),(0,0,0,0)) x1 + (1,0,1,0)
      f_A(x1) = x1 + (7,8,10,27)
      match_A(x1,x2) = ((1,0,0,0),(1,0,0,0),(1,1,0,0),(1,1,1,0)) x2 + (3,1,12,1)
      X_A() = (1,0,1,1)
      start_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,0)) x1 + (2,0,2,6)

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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      check#_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,1,1,1)) x1
      f_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,0)) x1 + (1,0,0,1)

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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      active#_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,1,1,1)) x1
      f_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,0)) x1 + (1,0,0,1)

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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      match#_A(x1,x2) = ((1,0,0,0),(0,0,0,0),(1,1,0,0),(1,0,0,0)) x1 + ((1,0,0,0),(0,0,0,0),(1,0,0,0),(1,0,0,0)) x2
      f_A(x1) = ((1,0,0,0),(0,0,0,0),(0,1,0,0),(1,1,1,0)) x1 + (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: 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 reduction pair:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      proper#_A(x1) = ((1,0,0,0),(0,1,0,0),(0,0,1,0),(0,1,1,1)) x1
      f_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,0)) x1 + (1,0,0,1)

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:

  matrix interpretations:
  
    carrier: N^4
    order: lexicographic order
    interpretations:
      f#_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,0,1)) x1
      ok_A(x1) = ((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + (1,1,1,1)
      mark_A(x1) = ((1,0,0,0),(0,1,0,0),(1,1,1,0),(1,1,1,1)) x1 + (1,1,1,1)
      found_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
  r1, r2, r3, r4, r5, r6, r7, r8, r9, r10, r11, r12, r13, r14, r15

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