Problem SK90 4.16

Tool CaT

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	SK90 4.16

stdout:

YES(?,O(n^1))

Problem:
f(0()) -> s(0())
f(s(0())) -> s(s(0()))
f(s(0())) -> *(s(s(0())),f(0()))
f(+(x,s(0()))) -> +(s(s(0())),f(x))
f(+(x,y)) -> *(f(x),f(y))

Proof:
Bounds Processor:
  bound: 2
  enrichment: match
  automaton:
   final states: {5}
   transitions:
    *1(5,9) -> 10,5
    *1(10,10) -> 10,5
    f1(7) -> 9*
    f1(2) -> 10*
    f1(4) -> 10*
    f1(1) -> 10*
    f1(3) -> 10*
    +1(5,10) -> 10,5
    s1(5) -> 10,5
    s1(7) -> 10,5
    01() -> 7*
    s2(12) -> 9*
    f0(2) -> 5*
    f0(4) -> 5*
    f0(1) -> 5*
    f0(3) -> 5*
    02() -> 12*
    00() -> 1*
    s0(2) -> 2*
    s0(4) -> 2*
    s0(1) -> 2*
    s0(3) -> 2*
    *0(3,1) -> 3*
    *0(3,3) -> 3*
    *0(4,2) -> 3*
    *0(4,4) -> 3*
    *0(1,2) -> 3*
    *0(1,4) -> 3*
    *0(2,1) -> 3*
    *0(2,3) -> 3*
    *0(3,2) -> 3*
    *0(3,4) -> 3*
    *0(4,1) -> 3*
    *0(4,3) -> 3*
    *0(1,1) -> 3*
    *0(1,3) -> 3*
    *0(2,2) -> 3*
    *0(2,4) -> 3*
    +0(3,1) -> 4*
    +0(3,3) -> 4*
    +0(4,2) -> 4*
    +0(4,4) -> 4*
    +0(1,2) -> 4*
    +0(1,4) -> 4*
    +0(2,1) -> 4*
    +0(2,3) -> 4*
    +0(3,2) -> 4*
    +0(3,4) -> 4*
    +0(4,1) -> 4*
    +0(4,3) -> 4*
    +0(1,1) -> 4*
    +0(1,3) -> 4*
    +0(2,2) -> 4*
    +0(2,4) -> 4*
  problem:

  Qed

Tool IRC1

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	SK90 4.16

stdout:

YES(?,O(n^1))

Tool IRC2

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	SK90 4.16

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()) -> s(0())
     , f(s(0())) -> s(s(0()))
     , f(s(0())) -> *(s(s(0())), f(0()))
     , f(+(x, s(0()))) -> +(s(s(0())), f(x))
     , f(+(x, y)) -> *(f(x), f(y))}

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()) -> s(0())
          , f(s(0())) -> s(s(0()))
          , f(s(0())) -> *(s(s(0())), f(0()))
          , f(+(x, s(0()))) -> +(s(s(0())), f(x))
          , f(+(x, y)) -> *(f(x), f(y))}

     Proof Output:
       The problem is match-bounded by 2.
       The enriched problem is compatible with the following automaton:
       {  f_0(2) -> 1
        , f_1(2) -> 5
        , f_1(3) -> 4
        , 0_0() -> 2
        , 0_1() -> 3
        , 0_2() -> 6
        , s_0(2) -> 2
        , s_1(1) -> 1
        , s_1(3) -> 1
        , s_1(3) -> 5
        , s_1(5) -> 5
        , s_2(6) -> 4
        , *_0(2, 2) -> 2
        , *_1(1, 4) -> 1
        , *_1(5, 4) -> 5
        , *_1(5, 5) -> 1
        , *_1(5, 5) -> 5
        , +_0(2, 2) -> 2
        , +_1(1, 5) -> 1
        , +_1(5, 5) -> 5}

Tool RC1

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	SK90 4.16

stdout:

YES(?,O(n^1))

Tool RC2

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	SK90 4.16

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()) -> s(0())
     , f(s(0())) -> s(s(0()))
     , f(s(0())) -> *(s(s(0())), f(0()))
     , f(+(x, s(0()))) -> +(s(s(0())), f(x))
     , f(+(x, y)) -> *(f(x), f(y))}

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()) -> s(0())
          , f(s(0())) -> s(s(0()))
          , f(s(0())) -> *(s(s(0())), f(0()))
          , f(+(x, s(0()))) -> +(s(s(0())), f(x))
          , f(+(x, y)) -> *(f(x), f(y))}

     Proof Output:
       The problem is match-bounded by 2.
       The enriched problem is compatible with the following automaton:
       {  f_0(2) -> 1
        , f_1(2) -> 5
        , f_1(3) -> 4
        , 0_0() -> 2
        , 0_1() -> 3
        , 0_2() -> 6
        , s_0(2) -> 2
        , s_1(1) -> 1
        , s_1(3) -> 1
        , s_1(3) -> 5
        , s_1(5) -> 5
        , s_2(6) -> 4
        , *_0(2, 2) -> 2
        , *_1(1, 4) -> 1
        , *_1(5, 4) -> 5
        , *_1(5, 5) -> 1
        , *_1(5, 5) -> 5
        , +_0(2, 2) -> 2
        , +_1(1, 5) -> 1
        , +_1(5, 5) -> 5}

Tool pair1rc

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	SK90 4.16

stdout:

Tool pair2rc

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	SK90 4.16

stdout:

Tool pair3irc

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	SK90 4.16

stdout:

Tool pair3rc

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	SK90 4.16

stdout:

YES(?,O(n^1))

We consider the following Problem:

  Strict Trs:
    {  f(0()) -> s(0())
     , f(s(0())) -> s(s(0()))
     , f(s(0())) -> *(s(s(0())), f(0()))
     , f(+(x, s(0()))) -> +(s(s(0())), f(x))
     , f(+(x, y)) -> *(f(x), f(y))}
  StartTerms: basic terms
  Strategy: none

Certificate: YES(?,O(n^1))

Application of 'pair3 (timeout of 60.0 seconds)':
-------------------------------------------------
  The processor is not applicable, reason is:
   Input problem is not restricted to innermost rewriting

  We abort the transformation and continue with the subprocessor on the problem

  Strict Trs:
    {  f(0()) -> s(0())
     , f(s(0())) -> s(s(0()))
     , f(s(0())) -> *(s(s(0())), f(0()))
     , f(+(x, s(0()))) -> +(s(s(0())), f(x))
     , f(+(x, y)) -> *(f(x), f(y))}
  StartTerms: basic terms
  Strategy: none

  1) 'Fastest' proved the goal fastest:

     'Fastest' proved the goal fastest:

     'Bounds with perSymbol-enrichment and initial automaton 'match'' proved the goal fastest:

     The problem is match-bounded by 2.
     The enriched problem is compatible with the following automaton:
     {  f_0(2) -> 1
      , f_0(3) -> 1
      , f_0(4) -> 1
      , f_0(5) -> 1
      , f_1(2) -> 8
      , f_1(3) -> 8
      , f_1(4) -> 8
      , f_1(5) -> 8
      , f_1(6) -> 7
      , 0_0() -> 2
      , 0_1() -> 6
      , 0_2() -> 9
      , s_0(2) -> 3
      , s_0(3) -> 3
      , s_0(4) -> 3
      , s_0(5) -> 3
      , s_1(1) -> 1
      , s_1(6) -> 1
      , s_1(6) -> 8
      , s_1(8) -> 8
      , s_2(9) -> 7
      , *_0(2, 2) -> 4
      , *_0(2, 3) -> 4
      , *_0(2, 4) -> 4
      , *_0(2, 5) -> 4
      , *_0(3, 2) -> 4
      , *_0(3, 3) -> 4
      , *_0(3, 4) -> 4
      , *_0(3, 5) -> 4
      , *_0(4, 2) -> 4
      , *_0(4, 3) -> 4
      , *_0(4, 4) -> 4
      , *_0(4, 5) -> 4
      , *_0(5, 2) -> 4
      , *_0(5, 3) -> 4
      , *_0(5, 4) -> 4
      , *_0(5, 5) -> 4
      , *_1(1, 7) -> 1
      , *_1(8, 7) -> 8
      , *_1(8, 8) -> 1
      , *_1(8, 8) -> 8
      , +_0(2, 2) -> 5
      , +_0(2, 3) -> 5
      , +_0(2, 4) -> 5
      , +_0(2, 5) -> 5
      , +_0(3, 2) -> 5
      , +_0(3, 3) -> 5
      , +_0(3, 4) -> 5
      , +_0(3, 5) -> 5
      , +_0(4, 2) -> 5
      , +_0(4, 3) -> 5
      , +_0(4, 4) -> 5
      , +_0(4, 5) -> 5
      , +_0(5, 2) -> 5
      , +_0(5, 3) -> 5
      , +_0(5, 4) -> 5
      , +_0(5, 5) -> 5
      , +_1(1, 8) -> 1
      , +_1(8, 8) -> 8}


Hurray, we answered YES(?,O(n^1))

Tool rc

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	SK90 4.16

stdout:

Tool tup3irc

Execution Time	7.107702ms
Answer	YES(?,O(n^1))
Input	SK90 4.16