Problem CSR 04 Ex23 Luc06

Tool IRC1

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

stdout:

YES(?,O(n^1))

Tool IRC2

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

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: {f(f(a())) -> c(f(g(f(a()))))}

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: {f(f(a())) -> c(f(g(f(a()))))}

     Proof Output:
       The problem is match-bounded by 0.
       The enriched problem is compatible with the following automaton:
       {  f_0(2) -> 1
        , a_0() -> 2
        , c_0(2) -> 2
        , g_0(2) -> 2}

Tool RC1

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

stdout:

YES(?,O(n^1))

Tool RC2

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

stdout:

YES(?,O(n^1))

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

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: {f(f(a())) -> c(f(g(f(a()))))}

     Proof Output:
       The problem is match-bounded by 0.
       The enriched problem is compatible with the following automaton:
       {  f_0(2) -> 1
        , a_0() -> 2
        , c_0(2) -> 2
        , g_0(2) -> 2}