Problem Transformed CSR 04 ExConc Zan97 C

Tool CaT

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 ExConc Zan97 C

stdout:

YES(?,O(n^1))

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

Proof:
Bounds Processor:
  bound: 1
  enrichment: match
  automaton:
   final states: {8,7,6,5,4,3}
   transitions:
    top1(48) -> 49*
    active1(57) -> 58*
    active1(63) -> 64*
    proper1(55) -> 56*
    proper1(47) -> 48*
    ok1(45) -> 46*
    ok1(37) -> 38*
    ok1(29) -> 30*
    h1(27) -> 28*
    h1(19) -> 20*
    g1(39) -> 40*
    g1(36) -> 37*
    f1(17) -> 18*
    f1(9) -> 10*
    mark1(20) -> 21*
    mark1(10) -> 11*
    active0(2) -> 3*
    active0(1) -> 3*
    f0(2) -> 4*
    f0(1) -> 4*
    mark0(2) -> 1*
    mark0(1) -> 1*
    g0(2) -> 7*
    g0(1) -> 7*
    h0(2) -> 5*
    h0(1) -> 5*
    proper0(2) -> 6*
    proper0(1) -> 6*
    ok0(2) -> 2*
    ok0(1) -> 2*
    top0(2) -> 8*
    top0(1) -> 8*
    1 -> 63,55,39,27,17
    2 -> 57,47,36,19,9
    10 -> 29*
    11 -> 18,10,29,4
    18 -> 10*
    20 -> 45*
    21 -> 28,20,45,5
    28 -> 20*
    30 -> 10,29,4
    38 -> 37,7
    40 -> 37*
    46 -> 20,45,5
    49 -> 8*
    56 -> 48*
    58 -> 48*
    64 -> 48*
  problem:

  Qed

Tool IRC1

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 ExConc Zan97 C

stdout:

YES(?,O(n^1))

Tool IRC2

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 ExConc Zan97 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(X)) -> mark(g(h(f(X))))
     , active(f(X)) -> f(active(X))
     , active(h(X)) -> h(active(X))
     , f(mark(X)) -> mark(f(X))
     , h(mark(X)) -> mark(h(X))
     , proper(f(X)) -> f(proper(X))
     , proper(g(X)) -> g(proper(X))
     , proper(h(X)) -> h(proper(X))
     , f(ok(X)) -> ok(f(X))
     , g(ok(X)) -> ok(g(X))
     , h(ok(X)) -> ok(h(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(X)) -> mark(g(h(f(X))))
          , active(f(X)) -> f(active(X))
          , active(h(X)) -> h(active(X))
          , f(mark(X)) -> mark(f(X))
          , h(mark(X)) -> mark(h(X))
          , proper(f(X)) -> f(proper(X))
          , proper(g(X)) -> g(proper(X))
          , proper(h(X)) -> h(proper(X))
          , f(ok(X)) -> ok(f(X))
          , g(ok(X)) -> ok(g(X))
          , h(ok(X)) -> ok(h(X))
          , top(mark(X)) -> top(proper(X))
          , top(ok(X)) -> top(active(X))}

     Proof Output:
       The problem is match-bounded by 1.
       The enriched problem is compatible with the following automaton:
       {  active_0(2) -> 1
        , active_1(2) -> 4
        , f_0(2) -> 1
        , f_1(2) -> 3
        , mark_0(2) -> 2
        , mark_1(3) -> 1
        , mark_1(3) -> 3
        , g_0(2) -> 1
        , g_1(2) -> 3
        , h_0(2) -> 1
        , h_1(2) -> 3
        , proper_0(2) -> 1
        , proper_1(2) -> 4
        , ok_0(2) -> 2
        , ok_1(3) -> 1
        , ok_1(3) -> 3
        , top_0(2) -> 1
        , top_1(4) -> 1}

Tool RC1

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 ExConc Zan97 C

stdout:

YES(?,O(n^1))

Tool RC2

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 ExConc Zan97 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(X)) -> mark(g(h(f(X))))
     , active(f(X)) -> f(active(X))
     , active(h(X)) -> h(active(X))
     , f(mark(X)) -> mark(f(X))
     , h(mark(X)) -> mark(h(X))
     , proper(f(X)) -> f(proper(X))
     , proper(g(X)) -> g(proper(X))
     , proper(h(X)) -> h(proper(X))
     , f(ok(X)) -> ok(f(X))
     , g(ok(X)) -> ok(g(X))
     , h(ok(X)) -> ok(h(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(X)) -> mark(g(h(f(X))))
          , active(f(X)) -> f(active(X))
          , active(h(X)) -> h(active(X))
          , f(mark(X)) -> mark(f(X))
          , h(mark(X)) -> mark(h(X))
          , proper(f(X)) -> f(proper(X))
          , proper(g(X)) -> g(proper(X))
          , proper(h(X)) -> h(proper(X))
          , f(ok(X)) -> ok(f(X))
          , g(ok(X)) -> ok(g(X))
          , h(ok(X)) -> ok(h(X))
          , top(mark(X)) -> top(proper(X))
          , top(ok(X)) -> top(active(X))}

     Proof Output:
       The problem is match-bounded by 1.
       The enriched problem is compatible with the following automaton:
       {  active_0(2) -> 1
        , active_1(2) -> 4
        , f_0(2) -> 1
        , f_1(2) -> 3
        , mark_0(2) -> 2
        , mark_1(3) -> 1
        , mark_1(3) -> 3
        , g_0(2) -> 1
        , g_1(2) -> 3
        , h_0(2) -> 1
        , h_1(2) -> 3
        , proper_0(2) -> 1
        , proper_1(2) -> 4
        , ok_0(2) -> 2
        , ok_1(3) -> 1
        , ok_1(3) -> 3
        , top_0(2) -> 1
        , top_1(4) -> 1}