Problem Transformed CSR 04 Ex4 7 15 Bor03 FR

Tool CaT

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex4 7 15 Bor03 FR

stdout:

YES(?,O(n^1))

Problem:
f(0()) -> cons(0(),n__f(n__s(n__0())))
f(s(0())) -> f(p(s(0())))
p(s(0())) -> 0()
f(X) -> n__f(X)
s(X) -> n__s(X)
0() -> n__0()
activate(n__f(X)) -> f(activate(X))
activate(n__s(X)) -> s(activate(X))
activate(n__0()) -> 0()
activate(X) -> X

Proof:
Bounds Processor:
  bound: 4
  enrichment: match
  automaton:
   final states: {9,8,7,6,5}
   transitions:
    03() -> 19*
    01() -> 10,9
    n__04() -> 19*
    s1(10) -> 10,9
    cons3(19,22) -> 10,9
    activate1(2) -> 10*
    activate1(4) -> 10*
    activate1(1) -> 10*
    activate1(3) -> 10*
    f1(10) -> 10,9
    n__01() -> 8*
    n__s1(2) -> 7*
    n__s1(4) -> 7*
    n__s1(1) -> 7*
    n__s1(3) -> 7*
    n__f1(2) -> 5*
    n__f1(4) -> 5*
    n__f1(1) -> 5*
    n__f1(3) -> 5*
    n__02() -> 10,9
    n__s2(10) -> 10,9
    n__s2(9) -> 12*
    n__f2(10) -> 10,9
    n__f2(12) -> 13*
    f0(2) -> 5*
    f0(4) -> 5*
    f0(1) -> 5*
    f0(3) -> 5*
    f2(19) -> 9,10
    00() -> 8*
    p2(18) -> 19*
    cons0(3,1) -> 1*
    cons0(3,3) -> 1*
    cons0(4,2) -> 1*
    cons0(4,4) -> 1*
    cons0(1,2) -> 1*
    cons0(1,4) -> 1*
    cons0(2,1) -> 1*
    cons0(2,3) -> 1*
    cons0(3,2) -> 1*
    cons0(3,4) -> 1*
    cons0(4,1) -> 1*
    cons0(4,3) -> 1*
    cons0(1,1) -> 1*
    cons0(1,3) -> 1*
    cons0(2,2) -> 1*
    cons0(2,4) -> 1*
    s2(14) -> 18*
    n__f0(2) -> 2*
    n__f0(4) -> 2*
    n__f0(1) -> 2*
    n__f0(3) -> 2*
    02() -> 14*
    n__s0(2) -> 3*
    n__s0(4) -> 3*
    n__s0(1) -> 3*
    n__s0(3) -> 3*
    cons2(14,13) -> 9,10
    n__00() -> 4*
    n__03() -> 14*
    s0(2) -> 7*
    s0(4) -> 7*
    s0(1) -> 7*
    s0(3) -> 7*
    n__s3(14) -> 18*
    p0(2) -> 6*
    p0(4) -> 6*
    p0(1) -> 6*
    p0(3) -> 6*
    n__f3(19) -> 10,9
    n__f3(18) -> 22*
    activate0(2) -> 9*
    activate0(4) -> 9*
    activate0(1) -> 9*
    activate0(3) -> 9*
    1 -> 10,9
    2 -> 10,9
    3 -> 10,9
    4 -> 10,9
  problem:

  Qed

Tool IRC1

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex4 7 15 Bor03 FR

stdout:

YES(?,O(n^1))

Tool IRC2

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex4 7 15 Bor03 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(0()) -> cons(0(), n__f(n__s(n__0())))
     , f(s(0())) -> f(p(s(0())))
     , p(s(0())) -> 0()
     , f(X) -> n__f(X)
     , s(X) -> n__s(X)
     , 0() -> n__0()
     , activate(n__f(X)) -> f(activate(X))
     , activate(n__s(X)) -> s(activate(X))
     , activate(n__0()) -> 0()
     , 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(0()) -> cons(0(), n__f(n__s(n__0())))
          , f(s(0())) -> f(p(s(0())))
          , p(s(0())) -> 0()
          , f(X) -> n__f(X)
          , s(X) -> n__s(X)
          , 0() -> n__0()
          , activate(n__f(X)) -> f(activate(X))
          , activate(n__s(X)) -> s(activate(X))
          , activate(n__0()) -> 0()
          , 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(3) -> 1
        , f_1(3) -> 3
        , f_2(8) -> 1
        , f_2(8) -> 3
        , 0_0() -> 1
        , 0_1() -> 1
        , 0_1() -> 3
        , 0_2() -> 4
        , 0_3() -> 8
        , cons_0(2, 2) -> 1
        , cons_0(2, 2) -> 2
        , cons_0(2, 2) -> 3
        , cons_2(4, 5) -> 1
        , cons_2(4, 5) -> 3
        , cons_3(8, 10) -> 1
        , cons_3(8, 10) -> 3
        , n__f_0(2) -> 1
        , n__f_0(2) -> 2
        , n__f_0(2) -> 3
        , n__f_1(2) -> 1
        , n__f_2(3) -> 1
        , n__f_2(3) -> 3
        , n__f_2(6) -> 5
        , n__f_3(8) -> 1
        , n__f_3(8) -> 3
        , n__f_3(9) -> 10
        , n__s_0(2) -> 1
        , n__s_0(2) -> 2
        , n__s_0(2) -> 3
        , n__s_1(2) -> 1
        , n__s_2(3) -> 1
        , n__s_2(3) -> 3
        , n__s_2(7) -> 6
        , n__s_3(4) -> 9
        , n__0_0() -> 1
        , n__0_0() -> 2
        , n__0_0() -> 3
        , n__0_1() -> 1
        , n__0_2() -> 1
        , n__0_2() -> 3
        , n__0_2() -> 7
        , n__0_3() -> 4
        , n__0_4() -> 8
        , s_0(2) -> 1
        , s_1(3) -> 1
        , s_1(3) -> 3
        , s_2(4) -> 9
        , p_0(2) -> 1
        , p_2(9) -> 8
        , activate_0(2) -> 1
        , activate_1(2) -> 3}

Tool RC1

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex4 7 15 Bor03 FR

stdout:

YES(?,O(n^1))

Tool RC2

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex4 7 15 Bor03 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(0()) -> cons(0(), n__f(n__s(n__0())))
     , f(s(0())) -> f(p(s(0())))
     , p(s(0())) -> 0()
     , f(X) -> n__f(X)
     , s(X) -> n__s(X)
     , 0() -> n__0()
     , activate(n__f(X)) -> f(activate(X))
     , activate(n__s(X)) -> s(activate(X))
     , activate(n__0()) -> 0()
     , 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(0()) -> cons(0(), n__f(n__s(n__0())))
          , f(s(0())) -> f(p(s(0())))
          , p(s(0())) -> 0()
          , f(X) -> n__f(X)
          , s(X) -> n__s(X)
          , 0() -> n__0()
          , activate(n__f(X)) -> f(activate(X))
          , activate(n__s(X)) -> s(activate(X))
          , activate(n__0()) -> 0()
          , 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(3) -> 1
        , f_1(3) -> 3
        , f_2(8) -> 1
        , f_2(8) -> 3
        , 0_0() -> 1
        , 0_1() -> 1
        , 0_1() -> 3
        , 0_2() -> 4
        , 0_3() -> 8
        , cons_0(2, 2) -> 1
        , cons_0(2, 2) -> 2
        , cons_0(2, 2) -> 3
        , cons_2(4, 5) -> 1
        , cons_2(4, 5) -> 3
        , cons_3(8, 10) -> 1
        , cons_3(8, 10) -> 3
        , n__f_0(2) -> 1
        , n__f_0(2) -> 2
        , n__f_0(2) -> 3
        , n__f_1(2) -> 1
        , n__f_2(3) -> 1
        , n__f_2(3) -> 3
        , n__f_2(6) -> 5
        , n__f_3(8) -> 1
        , n__f_3(8) -> 3
        , n__f_3(9) -> 10
        , n__s_0(2) -> 1
        , n__s_0(2) -> 2
        , n__s_0(2) -> 3
        , n__s_1(2) -> 1
        , n__s_2(3) -> 1
        , n__s_2(3) -> 3
        , n__s_2(7) -> 6
        , n__s_3(4) -> 9
        , n__0_0() -> 1
        , n__0_0() -> 2
        , n__0_0() -> 3
        , n__0_1() -> 1
        , n__0_2() -> 1
        , n__0_2() -> 3
        , n__0_2() -> 7
        , n__0_3() -> 4
        , n__0_4() -> 8
        , s_0(2) -> 1
        , s_1(3) -> 1
        , s_1(3) -> 3
        , s_2(4) -> 9
        , p_0(2) -> 1
        , p_2(9) -> 8
        , activate_0(2) -> 1
        , activate_1(2) -> 3}