Problem Transformed CSR 04 Ex15 Luc06 C

Tool CaT

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex15 Luc06 C

stdout:

YES(?,O(n^1))

Problem:
active(f(f(a()))) -> mark(f(g(f(a()))))
active(g(X)) -> g(active(X))
g(mark(X)) -> mark(g(X))
proper(f(X)) -> f(proper(X))
proper(a()) -> ok(a())
proper(g(X)) -> g(proper(X))
f(ok(X)) -> ok(f(X))
g(ok(X)) -> ok(g(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Proof:
Bounds Processor:
  bound: 2
  enrichment: match
  automaton:
   final states: {8,7,6,5,4}
   transitions:
    top1(46) -> 47*
    active1(69) -> 70*
    active1(61) -> 62*
    active1(63) -> 64*
    proper1(55) -> 56*
    proper1(45) -> 46*
    proper1(53) -> 54*
    ok1(25) -> 26*
    ok1(43) -> 44*
    ok1(28) -> 29*
    g1(17) -> 18*
    g1(19) -> 20*
    g1(9) -> 10*
    f1(35) -> 36*
    f1(37) -> 38*
    f1(27) -> 28*
    a1() -> 25*
    mark1(10) -> 11*
    top2(75) -> 76*
    active0(2) -> 4*
    active0(1) -> 4*
    active0(3) -> 4*
    active2(74) -> 75*
    f0(2) -> 7*
    f0(1) -> 7*
    f0(3) -> 7*
    a0() -> 1*
    mark0(2) -> 2*
    mark0(1) -> 2*
    mark0(3) -> 2*
    g0(2) -> 5*
    g0(1) -> 5*
    g0(3) -> 5*
    proper0(2) -> 6*
    proper0(1) -> 6*
    proper0(3) -> 6*
    ok0(2) -> 3*
    ok0(1) -> 3*
    ok0(3) -> 3*
    top0(2) -> 8*
    top0(1) -> 8*
    top0(3) -> 8*
    1 -> 63,53,35,17
    2 -> 61,45,27,9
    3 -> 69,55,37,19
    10 -> 43*
    11 -> 10,43,5
    18 -> 10*
    20 -> 10*
    25 -> 74*
    26 -> 54,46,6
    29 -> 38,28,7
    36 -> 28*
    38 -> 28*
    44 -> 20,10,43,5
    47 -> 8*
    54 -> 46*
    56 -> 46*
    62 -> 46*
    64 -> 46*
    70 -> 46*
    76 -> 47,8
  problem:

  Qed

Tool IRC1

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex15 Luc06 C

stdout:

YES(?,O(n^1))

Tool IRC2

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex15 Luc06 C

stdout:

YES(?,O(n^1))

'Fastest (timeout of 60.0 seconds)'
-----------------------------------
Answer:           YES(?,O(n^1))
Input Problem:    innermost runtime-complexity with respect to
  Rules:
    {  active(f(f(a()))) -> mark(f(g(f(a()))))
     , active(g(X)) -> g(active(X))
     , g(mark(X)) -> mark(g(X))
     , proper(f(X)) -> f(proper(X))
     , proper(a()) -> ok(a())
     , proper(g(X)) -> g(proper(X))
     , f(ok(X)) -> ok(f(X))
     , g(ok(X)) -> ok(g(X))
     , top(mark(X)) -> top(proper(X))
     , top(ok(X)) -> top(active(X))}

Proof Output:
  'Bounds with minimal-enrichment and initial automaton 'match'' proved the best result:

  Details:
  --------
    'Bounds with minimal-enrichment and initial automaton 'match'' succeeded with the following output:
     'Bounds with minimal-enrichment and initial automaton 'match''
     --------------------------------------------------------------
     Answer:           YES(?,O(n^1))
     Input Problem:    innermost runtime-complexity with respect to
       Rules:
         {  active(f(f(a()))) -> mark(f(g(f(a()))))
          , active(g(X)) -> g(active(X))
          , g(mark(X)) -> mark(g(X))
          , proper(f(X)) -> f(proper(X))
          , proper(a()) -> ok(a())
          , proper(g(X)) -> g(proper(X))
          , f(ok(X)) -> ok(f(X))
          , g(ok(X)) -> ok(g(X))
          , top(mark(X)) -> top(proper(X))
          , top(ok(X)) -> top(active(X))}

     Proof Output:
       The problem is match-bounded by 2.
       The enriched problem is compatible with the following automaton:
       {  active_0(2) -> 1
        , active_1(2) -> 5
        , active_2(4) -> 6
        , f_0(2) -> 1
        , f_1(2) -> 4
        , a_0() -> 2
        , a_1() -> 4
        , mark_0(2) -> 2
        , mark_1(3) -> 1
        , mark_1(3) -> 3
        , g_0(2) -> 1
        , g_1(2) -> 3
        , proper_0(2) -> 1
        , proper_1(2) -> 5
        , ok_0(2) -> 2
        , ok_1(3) -> 1
        , ok_1(3) -> 3
        , ok_1(4) -> 1
        , ok_1(4) -> 4
        , ok_1(4) -> 5
        , top_0(2) -> 1
        , top_1(5) -> 1
        , top_2(6) -> 1}

Tool RC1

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex15 Luc06 C

stdout:

YES(?,O(n^1))

Tool RC2

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex15 Luc06 C

stdout:

YES(?,O(n^1))

'Fastest (timeout of 60.0 seconds)'
-----------------------------------
Answer:           YES(?,O(n^1))
Input Problem:    runtime-complexity with respect to
  Rules:
    {  active(f(f(a()))) -> mark(f(g(f(a()))))
     , active(g(X)) -> g(active(X))
     , g(mark(X)) -> mark(g(X))
     , proper(f(X)) -> f(proper(X))
     , proper(a()) -> ok(a())
     , proper(g(X)) -> g(proper(X))
     , f(ok(X)) -> ok(f(X))
     , g(ok(X)) -> ok(g(X))
     , top(mark(X)) -> top(proper(X))
     , top(ok(X)) -> top(active(X))}

Proof Output:
  'Bounds with minimal-enrichment and initial automaton 'match'' proved the best result:

  Details:
  --------
    'Bounds with minimal-enrichment and initial automaton 'match'' succeeded with the following output:
     'Bounds with minimal-enrichment and initial automaton 'match''
     --------------------------------------------------------------
     Answer:           YES(?,O(n^1))
     Input Problem:    runtime-complexity with respect to
       Rules:
         {  active(f(f(a()))) -> mark(f(g(f(a()))))
          , active(g(X)) -> g(active(X))
          , g(mark(X)) -> mark(g(X))
          , proper(f(X)) -> f(proper(X))
          , proper(a()) -> ok(a())
          , proper(g(X)) -> g(proper(X))
          , f(ok(X)) -> ok(f(X))
          , g(ok(X)) -> ok(g(X))
          , top(mark(X)) -> top(proper(X))
          , top(ok(X)) -> top(active(X))}

     Proof Output:
       The problem is match-bounded by 2.
       The enriched problem is compatible with the following automaton:
       {  active_0(2) -> 1
        , active_1(2) -> 5
        , active_2(4) -> 6
        , f_0(2) -> 1
        , f_1(2) -> 4
        , a_0() -> 2
        , a_1() -> 4
        , mark_0(2) -> 2
        , mark_1(3) -> 1
        , mark_1(3) -> 3
        , g_0(2) -> 1
        , g_1(2) -> 3
        , proper_0(2) -> 1
        , proper_1(2) -> 5
        , ok_0(2) -> 2
        , ok_1(3) -> 1
        , ok_1(3) -> 3
        , ok_1(4) -> 1
        , ok_1(4) -> 4
        , ok_1(4) -> 5
        , top_0(2) -> 1
        , top_1(5) -> 1
        , top_2(6) -> 1}