Problem Transformed CSR 04 ExProp7 Luc06 L

Tool CaT

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 ExProp7 Luc06 L

stdout:

YES(?,O(n^1))

Problem:
f(0()) -> cons(0())
f(s(0())) -> f(p(s(0())))
p(s(X)) -> X

Proof:
Bounds Processor:
  bound: 2
  enrichment: match
  automaton:
   final states: {5,4}
   transitions:
    f1(12) -> 13*
    p1(11) -> 12*
    s1(10) -> 11*
    01() -> 8*
    cons1(8) -> 9*
    cons2(20) -> 21*
    f0(2) -> 4*
    f0(1) -> 4*
    f0(3) -> 4*
    02() -> 20*
    00() -> 1*
    cons0(2) -> 2*
    cons0(1) -> 2*
    cons0(3) -> 2*
    s0(2) -> 3*
    s0(1) -> 3*
    s0(3) -> 3*
    p0(2) -> 5*
    p0(1) -> 5*
    p0(3) -> 5*
    1 -> 5*
    2 -> 5*
    3 -> 5*
    8 -> 10*
    9 -> 4*
    10 -> 12*
    13 -> 4*
    21 -> 13,4
  problem:

  Qed

Tool IRC1

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 ExProp7 Luc06 L

stdout:

YES(?,O(n^1))

Tool IRC2

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 ExProp7 Luc06 L

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(0()) -> cons(0())
     , f(s(0())) -> f(p(s(0())))
     , p(s(X)) -> 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(0()) -> cons(0())
          , f(s(0())) -> f(p(s(0())))
          , p(s(X)) -> X}

     Proof Output:
       The problem is match-bounded by 2.
       The enriched problem is compatible with the following automaton:
       {  f_0(2) -> 1
        , f_1(4) -> 1
        , 0_0() -> 1
        , 0_0() -> 2
        , 0_1() -> 3
        , 0_1() -> 4
        , 0_2() -> 6
        , cons_0(2) -> 1
        , cons_0(2) -> 2
        , cons_1(3) -> 1
        , cons_2(6) -> 1
        , s_0(2) -> 1
        , s_0(2) -> 2
        , s_1(3) -> 5
        , p_0(2) -> 1
        , p_1(5) -> 4}

Tool RC1

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 ExProp7 Luc06 L

stdout:

YES(?,O(n^1))

Tool RC2

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 ExProp7 Luc06 L

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(0()) -> cons(0())
     , f(s(0())) -> f(p(s(0())))
     , p(s(X)) -> 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(0()) -> cons(0())
          , f(s(0())) -> f(p(s(0())))
          , p(s(X)) -> X}

     Proof Output:
       The problem is match-bounded by 2.
       The enriched problem is compatible with the following automaton:
       {  f_0(2) -> 1
        , f_1(4) -> 1
        , 0_0() -> 1
        , 0_0() -> 2
        , 0_1() -> 3
        , 0_1() -> 4
        , 0_2() -> 6
        , cons_0(2) -> 1
        , cons_0(2) -> 2
        , cons_1(3) -> 1
        , cons_2(6) -> 1
        , s_0(2) -> 1
        , s_0(2) -> 2
        , s_1(3) -> 5
        , p_0(2) -> 1
        , p_1(5) -> 4}