Problem Rubio 04 mfp95

Tool CaT

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Rubio 04 mfp95

stdout:

YES(?,O(n^1))

Problem:
f(s(X),Y) -> h(s(f(h(Y),X)))

Proof:
Bounds Processor:
  bound: 1
  enrichment: match
  automaton:
   final states: {3}
   transitions:
    h1(2) -> 4*
    h1(6) -> 3*
    h1(1) -> 4*
    s1(5) -> 6*
    f1(4,2) -> 5*
    f1(4,1) -> 5*
    f0(1,2) -> 3*
    f0(2,1) -> 3*
    f0(1,1) -> 3*
    f0(2,2) -> 3*
    s0(2) -> 1*
    s0(1) -> 1*
    h0(2) -> 2*
    h0(1) -> 2*
  problem:

  Qed

Tool IRC1

Execution Time	Unknown
Answer	YES(?,O(1))
Input	Rubio 04 mfp95

stdout:

YES(?,O(1))

Tool IRC2

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Rubio 04 mfp95

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(s(X), Y) -> h(s(f(h(Y), 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: {f(s(X), Y) -> h(s(f(h(Y), X)))}

     Proof Output:
       The problem is match-bounded by 1.
       The enriched problem is compatible with the following automaton:
       {  f_0(2, 2) -> 1
        , f_1(5, 2) -> 4
        , s_0(2) -> 2
        , s_1(4) -> 3
        , h_0(2) -> 2
        , h_1(2) -> 5
        , h_1(3) -> 1}

Tool RC1

Execution Time	Unknown
Answer	YES(?,O(1))
Input	Rubio 04 mfp95

stdout:

YES(?,O(1))

Tool RC2

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Rubio 04 mfp95

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(s(X), Y) -> h(s(f(h(Y), 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: {f(s(X), Y) -> h(s(f(h(Y), X)))}

     Proof Output:
       The problem is match-bounded by 1.
       The enriched problem is compatible with the following automaton:
       {  f_0(2, 2) -> 1
        , f_1(5, 2) -> 4
        , s_0(2) -> 2
        , s_1(4) -> 3
        , h_0(2) -> 2
        , h_1(2) -> 5
        , h_1(3) -> 1}