Problem Transformed CSR 04 Ex23 Luc06 C

Tool CaT

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex23 Luc06 C

stdout:

YES(?,O(n^1))

Problem:
active(f(f(a()))) -> mark(c(f(g(f(a())))))
active(f(X)) -> f(active(X))
active(g(X)) -> g(active(X))
f(mark(X)) -> mark(f(X))
g(mark(X)) -> mark(g(X))
proper(f(X)) -> f(proper(X))
proper(a()) -> ok(a())
proper(c(X)) -> c(proper(X))
proper(g(X)) -> g(proper(X))
f(ok(X)) -> ok(f(X))
c(ok(X)) -> ok(c(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: {9,8,7,6,5,4}
   transitions:
    top1(68) -> 69*
    active1(84) -> 85*
    active1(86) -> 87*
    active1(78) -> 79*
    proper1(70) -> 71*
    proper1(67) -> 68*
    proper1(76) -> 77*
    ok1(60) -> 61*
    ok1(50) -> 51*
    ok1(40) -> 41*
    ok1(42) -> 43*
    g1(32) -> 33*
    g1(34) -> 35*
    g1(24) -> 25*
    c1(52) -> 53*
    c1(49) -> 50*
    c1(58) -> 59*
    f1(22) -> 23*
    f1(16) -> 17*
    f1(13) -> 14*
    a1() -> 40*
    mark1(25) -> 26*
    mark1(14) -> 15*
    top2(94) -> 95*
    active0(2) -> 4*
    active0(1) -> 4*
    active0(3) -> 4*
    active2(93) -> 94*
    f0(2) -> 5*
    f0(1) -> 5*
    f0(3) -> 5*
    a0() -> 1*
    mark0(2) -> 2*
    mark0(1) -> 2*
    mark0(3) -> 2*
    c0(2) -> 8*
    c0(1) -> 8*
    c0(3) -> 8*
    g0(2) -> 6*
    g0(1) -> 6*
    g0(3) -> 6*
    proper0(2) -> 7*
    proper0(1) -> 7*
    proper0(3) -> 7*
    ok0(2) -> 3*
    ok0(1) -> 3*
    ok0(3) -> 3*
    top0(2) -> 9*
    top0(1) -> 9*
    top0(3) -> 9*
    1 -> 84,70,52,32,16
    2 -> 78,67,49,24,13
    3 -> 86,76,58,34,22
    14 -> 42*
    15 -> 14,42,5
    17 -> 14*
    23 -> 14*
    25 -> 60*
    26 -> 25,60,6
    33 -> 25*
    35 -> 25*
    40 -> 93*
    41 -> 71,68,7
    43 -> 23,14,42,5
    51 -> 59,50,8
    53 -> 50*
    59 -> 50*
    61 -> 35,25,60,6
    69 -> 9*
    71 -> 68*
    77 -> 68*
    79 -> 68*
    85 -> 68*
    87 -> 68*
    95 -> 69,9
  problem:

  Qed

Tool IRC1

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex23 Luc06 C

stdout:

YES(?,O(n^1))

Tool IRC2

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex23 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(c(f(g(f(a())))))
     , active(f(X)) -> f(active(X))
     , active(g(X)) -> g(active(X))
     , f(mark(X)) -> mark(f(X))
     , g(mark(X)) -> mark(g(X))
     , proper(f(X)) -> f(proper(X))
     , proper(a()) -> ok(a())
     , proper(c(X)) -> c(proper(X))
     , proper(g(X)) -> g(proper(X))
     , f(ok(X)) -> ok(f(X))
     , c(ok(X)) -> ok(c(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(c(f(g(f(a())))))
          , active(f(X)) -> f(active(X))
          , active(g(X)) -> g(active(X))
          , f(mark(X)) -> mark(f(X))
          , g(mark(X)) -> mark(g(X))
          , proper(f(X)) -> f(proper(X))
          , proper(a()) -> ok(a())
          , proper(c(X)) -> c(proper(X))
          , proper(g(X)) -> g(proper(X))
          , f(ok(X)) -> ok(f(X))
          , c(ok(X)) -> ok(c(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) -> 3
        , a_0() -> 2
        , a_1() -> 4
        , mark_0(2) -> 2
        , mark_1(3) -> 1
        , mark_1(3) -> 3
        , c_0(2) -> 1
        , c_1(2) -> 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) -> 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 Ex23 Luc06 C

stdout:

YES(?,O(n^1))

Tool RC2

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex23 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(c(f(g(f(a())))))
     , active(f(X)) -> f(active(X))
     , active(g(X)) -> g(active(X))
     , f(mark(X)) -> mark(f(X))
     , g(mark(X)) -> mark(g(X))
     , proper(f(X)) -> f(proper(X))
     , proper(a()) -> ok(a())
     , proper(c(X)) -> c(proper(X))
     , proper(g(X)) -> g(proper(X))
     , f(ok(X)) -> ok(f(X))
     , c(ok(X)) -> ok(c(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(c(f(g(f(a())))))
          , active(f(X)) -> f(active(X))
          , active(g(X)) -> g(active(X))
          , f(mark(X)) -> mark(f(X))
          , g(mark(X)) -> mark(g(X))
          , proper(f(X)) -> f(proper(X))
          , proper(a()) -> ok(a())
          , proper(c(X)) -> c(proper(X))
          , proper(g(X)) -> g(proper(X))
          , f(ok(X)) -> ok(f(X))
          , c(ok(X)) -> ok(c(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) -> 3
        , a_0() -> 2
        , a_1() -> 4
        , mark_0(2) -> 2
        , mark_1(3) -> 1
        , mark_1(3) -> 3
        , c_0(2) -> 1
        , c_1(2) -> 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) -> 5
        , top_0(2) -> 1
        , top_1(5) -> 1
        , top_2(6) -> 1}