Problem Transformed CSR 04 Ex5 Zan97 FR

Tool CaT

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex5 Zan97 FR

stdout:

YES(?,O(n^1))

Problem:
f(X) -> if(X,c(),n__f(n__true()))
if(true(),X,Y) -> X
if(false(),X,Y) -> activate(Y)
f(X) -> n__f(X)
true() -> n__true()
activate(n__f(X)) -> f(activate(X))
activate(n__true()) -> true()
activate(X) -> X

Proof:
Bounds Processor:
  bound: 4
  enrichment: match
  automaton:
   final states: {8,7,6,5}
   transitions:
    c3() -> 25*
    true1() -> 6,8
    n__true3() -> 21,23
    f1(6) -> 6,8
    true3() -> 29*
    activate1(10) -> 5*
    activate1(2) -> 6*
    activate1(19) -> 6,8
    activate1(4) -> 6*
    activate1(1) -> 6*
    activate1(3) -> 6*
    n__true4() -> 29*
    n__true1() -> 7,9
    n__f1(2) -> 5*
    n__f1(9) -> 10*
    n__f1(4) -> 5*
    n__f1(1) -> 5*
    n__f1(3) -> 5*
    if1(1,11,10) -> 5*
    if1(3,11,10) -> 5*
    if1(2,11,10) -> 5*
    if1(4,11,10) -> 5*
    c1() -> 11*
    n__true2() -> 6,8,18
    n__f2(6) -> 6,8
    n__f2(18) -> 19*
    f0(2) -> 5*
    f0(4) -> 5*
    f0(1) -> 5*
    f0(3) -> 5*
    if2(6,20,19) -> 6,8
    if0(3,4,2) -> 6*
    if0(1,1,4) -> 6*
    if0(1,3,4) -> 6*
    if0(4,1,2) -> 6*
    if0(1,1,1) -> 6*
    if0(4,3,2) -> 6*
    if0(1,3,1) -> 6*
    if0(2,2,4) -> 6*
    if0(2,4,4) -> 6*
    if0(1,2,3) -> 6*
    if0(1,4,3) -> 6*
    if0(2,2,1) -> 6*
    if0(2,4,1) -> 6*
    if0(3,1,4) -> 6*
    if0(3,3,4) -> 6*
    if0(2,1,3) -> 6*
    if0(2,3,3) -> 6*
    if0(3,1,1) -> 6*
    if0(1,1,2) -> 6*
    if0(3,3,1) -> 6*
    if0(1,3,2) -> 6*
    if0(4,2,4) -> 6*
    if0(4,4,4) -> 6*
    if0(3,2,3) -> 6*
    if0(3,4,3) -> 6*
    if0(4,2,1) -> 6*
    if0(2,2,2) -> 6*
    if0(4,4,1) -> 6*
    if0(2,4,2) -> 6*
    if0(4,1,3) -> 6*
    if0(4,3,3) -> 6*
    if0(3,1,2) -> 6*
    if0(3,3,2) -> 6*
    if0(1,2,4) -> 6*
    if0(1,4,4) -> 6*
    if0(4,2,2) -> 6*
    if0(1,2,1) -> 6*
    if0(4,4,2) -> 6*
    if0(1,4,1) -> 6*
    if0(2,1,4) -> 6*
    if0(2,3,4) -> 6*
    if0(1,1,3) -> 6*
    if0(1,3,3) -> 6*
    if0(2,1,1) -> 6*
    if0(2,3,1) -> 6*
    if0(3,2,4) -> 6*
    if0(3,4,4) -> 6*
    if0(2,2,3) -> 6*
    if0(2,4,3) -> 6*
    if0(3,2,1) -> 6*
    if0(1,2,2) -> 6*
    if0(3,4,1) -> 6*
    if0(1,4,2) -> 6*
    if0(4,1,4) -> 6*
    if0(4,3,4) -> 6*
    if0(3,1,3) -> 6*
    if0(3,3,3) -> 6*
    if0(4,1,1) -> 6*
    if0(2,1,2) -> 6*
    if0(4,3,1) -> 6*
    if0(2,3,2) -> 6*
    if0(4,2,3) -> 6*
    if0(4,4,3) -> 6*
    if0(3,2,2) -> 6*
    c2() -> 20*
    c0() -> 1*
    f2(29) -> 6,8
    f2(21) -> 5*
    n__f0(2) -> 2*
    n__f0(4) -> 2*
    n__f0(1) -> 2*
    n__f0(3) -> 2*
    activate2(9) -> 21*
    activate2(18) -> 29*
    n__true0() -> 3*
    true2() -> 21*
    true0() -> 7*
    n__f3(29) -> 6,8
    n__f3(21) -> 5*
    n__f3(23) -> 24*
    false0() -> 4*
    if3(29,25,5) -> 6*
    if3(29,25,24) -> 8*
    if3(21,25,24) -> 5*
    activate0(2) -> 8*
    activate0(4) -> 8*
    activate0(1) -> 8*
    activate0(3) -> 8*
    1 -> 6,8
    2 -> 6,8
    3 -> 6,8
    4 -> 6,8
    9 -> 21*
    10 -> 5*
    18 -> 29*
    19 -> 6,8
    20 -> 6,8
    25 -> 6,8,5
  problem:

  Qed

Tool IRC1

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex5 Zan97 FR

stdout:

YES(?,O(n^1))

Tool IRC2

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex5 Zan97 FR

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(X) -> if(X, c(), n__f(n__true()))
     , if(true(), X, Y) -> X
     , if(false(), X, Y) -> activate(Y)
     , f(X) -> n__f(X)
     , true() -> n__true()
     , activate(n__f(X)) -> f(activate(X))
     , activate(n__true()) -> true()
     , activate(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(X) -> if(X, c(), n__f(n__true()))
          , if(true(), X, Y) -> X
          , if(false(), X, Y) -> activate(Y)
          , f(X) -> n__f(X)
          , true() -> n__true()
          , activate(n__f(X)) -> f(activate(X))
          , activate(n__true()) -> true()
          , activate(X) -> X}

     Proof Output:
       The problem is match-bounded by 4.
       The enriched problem is compatible with the following automaton:
       {  f_0(2) -> 1
        , f_1(1) -> 1
        , f_2(6) -> 1
        , if_0(2, 2, 2) -> 1
        , if_1(2, 3, 4) -> 1
        , if_2(1, 7, 8) -> 1
        , if_3(6, 10, 11) -> 1
        , c_0() -> 1
        , c_0() -> 2
        , c_1() -> 3
        , c_2() -> 1
        , c_2() -> 7
        , c_3() -> 1
        , c_3() -> 10
        , n__f_0(2) -> 1
        , n__f_0(2) -> 2
        , n__f_1(2) -> 1
        , n__f_1(5) -> 1
        , n__f_1(5) -> 4
        , n__f_2(1) -> 1
        , n__f_2(9) -> 1
        , n__f_2(9) -> 8
        , n__f_3(6) -> 1
        , n__f_3(12) -> 11
        , n__true_0() -> 1
        , n__true_0() -> 2
        , n__true_1() -> 1
        , n__true_1() -> 5
        , n__true_1() -> 6
        , n__true_2() -> 1
        , n__true_2() -> 6
        , n__true_2() -> 9
        , n__true_3() -> 6
        , n__true_3() -> 12
        , n__true_4() -> 6
        , true_0() -> 1
        , true_1() -> 1
        , true_2() -> 6
        , true_3() -> 6
        , false_0() -> 1
        , false_0() -> 2
        , activate_0(2) -> 1
        , activate_1(2) -> 1
        , activate_1(4) -> 1
        , activate_1(8) -> 1
        , activate_2(5) -> 6
        , activate_2(9) -> 6}

Tool RC1

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex5 Zan97 FR

stdout:

YES(?,O(n^1))

Tool RC2

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex5 Zan97 FR

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(X) -> if(X, c(), n__f(n__true()))
     , if(true(), X, Y) -> X
     , if(false(), X, Y) -> activate(Y)
     , f(X) -> n__f(X)
     , true() -> n__true()
     , activate(n__f(X)) -> f(activate(X))
     , activate(n__true()) -> true()
     , activate(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(X) -> if(X, c(), n__f(n__true()))
          , if(true(), X, Y) -> X
          , if(false(), X, Y) -> activate(Y)
          , f(X) -> n__f(X)
          , true() -> n__true()
          , activate(n__f(X)) -> f(activate(X))
          , activate(n__true()) -> true()
          , activate(X) -> X}

     Proof Output:
       The problem is match-bounded by 4.
       The enriched problem is compatible with the following automaton:
       {  f_0(2) -> 1
        , f_1(1) -> 1
        , f_2(6) -> 1
        , if_0(2, 2, 2) -> 1
        , if_1(2, 3, 4) -> 1
        , if_2(1, 7, 8) -> 1
        , if_3(6, 10, 11) -> 1
        , c_0() -> 1
        , c_0() -> 2
        , c_1() -> 3
        , c_2() -> 1
        , c_2() -> 7
        , c_3() -> 1
        , c_3() -> 10
        , n__f_0(2) -> 1
        , n__f_0(2) -> 2
        , n__f_1(2) -> 1
        , n__f_1(5) -> 1
        , n__f_1(5) -> 4
        , n__f_2(1) -> 1
        , n__f_2(9) -> 1
        , n__f_2(9) -> 8
        , n__f_3(6) -> 1
        , n__f_3(12) -> 11
        , n__true_0() -> 1
        , n__true_0() -> 2
        , n__true_1() -> 1
        , n__true_1() -> 5
        , n__true_1() -> 6
        , n__true_2() -> 1
        , n__true_2() -> 6
        , n__true_2() -> 9
        , n__true_3() -> 6
        , n__true_3() -> 12
        , n__true_4() -> 6
        , true_0() -> 1
        , true_1() -> 1
        , true_2() -> 6
        , true_3() -> 6
        , false_0() -> 1
        , false_0() -> 2
        , activate_0(2) -> 1
        , activate_1(2) -> 1
        , activate_1(4) -> 1
        , activate_1(8) -> 1
        , activate_2(5) -> 6
        , activate_2(9) -> 6}