Problem Transformed CSR 04 Ex4 7 15 Bor03 GM

Tool CaT

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

stdout:

YES(?,O(n^1))

Problem:
a__f(0()) -> cons(0(),f(s(0())))
a__f(s(0())) -> a__f(a__p(s(0())))
a__p(s(0())) -> 0()
mark(f(X)) -> a__f(mark(X))
mark(p(X)) -> a__p(mark(X))
mark(0()) -> 0()
mark(cons(X1,X2)) -> cons(mark(X1),X2)
mark(s(X)) -> s(mark(X))
a__f(X) -> f(X)
a__p(X) -> p(X)

Proof:
Bounds Processor:
  bound: 3
  enrichment: match
  automaton:
   final states: {8,7,6}
   transitions:
    f3(27) -> 28*
    f3(24) -> 18,8
    p1(5) -> 7*
    p1(2) -> 7*
    p1(4) -> 7*
    p1(1) -> 7*
    p1(3) -> 7*
    03() -> 24*
    f1(10) -> 11*
    f1(5) -> 6*
    f1(2) -> 6*
    f1(4) -> 6*
    f1(1) -> 6*
    f1(3) -> 6*
    cons3(24,28) -> 18,8
    s1(9) -> 10*
    s1(18) -> 18,8
    s3(24) -> 27*
    mark1(5) -> 18*
    mark1(2) -> 18*
    mark1(4) -> 18*
    mark1(1) -> 18*
    mark1(3) -> 18*
    cons1(18,1) -> 18,8
    cons1(18,3) -> 18,8
    cons1(18,5) -> 18,8
    cons1(18,2) -> 18,8
    cons1(18,4) -> 18,8
    cons1(9,11) -> 6*
    01() -> 18,8,7,9
    a__p1(10) -> 16*
    a__p1(18) -> 18,8
    a__f1(16) -> 6*
    a__f1(18) -> 18,8
    p2(10) -> 16*
    p2(18) -> 18,8
    a__f0(5) -> 6*
    a__f0(2) -> 6*
    a__f0(4) -> 6*
    a__f0(1) -> 6*
    a__f0(3) -> 6*
    f2(19) -> 20*
    f2(16) -> 6*
    f2(18) -> 18,8
    00() -> 1*
    02() -> 8,18,16
    cons0(3,1) -> 2*
    cons0(3,3) -> 2*
    cons0(3,5) -> 2*
    cons0(4,2) -> 2*
    cons0(4,4) -> 2*
    cons0(5,1) -> 2*
    cons0(5,3) -> 2*
    cons0(5,5) -> 2*
    cons0(1,2) -> 2*
    cons0(1,4) -> 2*
    cons0(2,1) -> 2*
    cons0(2,3) -> 2*
    cons0(2,5) -> 2*
    cons0(3,2) -> 2*
    cons0(3,4) -> 2*
    cons0(4,1) -> 2*
    cons0(4,3) -> 2*
    cons0(4,5) -> 2*
    cons0(5,2) -> 2*
    cons0(5,4) -> 2*
    cons0(1,1) -> 2*
    cons0(1,3) -> 2*
    cons0(1,5) -> 2*
    cons0(2,2) -> 2*
    cons0(2,4) -> 2*
    a__f2(24) -> 8,18
    f0(5) -> 3*
    f0(2) -> 3*
    f0(4) -> 3*
    f0(1) -> 3*
    f0(3) -> 3*
    a__p2(19) -> 24*
    s0(5) -> 4*
    s0(2) -> 4*
    s0(4) -> 4*
    s0(1) -> 4*
    s0(3) -> 4*
    s2(16) -> 19*
    a__p0(5) -> 7*
    a__p0(2) -> 7*
    a__p0(4) -> 7*
    a__p0(1) -> 7*
    a__p0(3) -> 7*
    cons2(16,20) -> 6,18,8
    mark0(5) -> 8*
    mark0(2) -> 8*
    mark0(4) -> 8*
    mark0(1) -> 8*
    mark0(3) -> 8*
    p3(19) -> 24*
    p0(5) -> 5*
    p0(2) -> 5*
    p0(4) -> 5*
    p0(1) -> 5*
    p0(3) -> 5*
  problem:

  Qed

Tool IRC1

Execution Time	Unknown
Answer	YES(?,O(n^3))
Input	Transformed CSR 04 Ex4 7 15 Bor03 GM

stdout:

YES(?,O(n^3))

Tool IRC2

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

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:
    {  a__f(0()) -> cons(0(), f(s(0())))
     , a__f(s(0())) -> a__f(a__p(s(0())))
     , a__p(s(0())) -> 0()
     , mark(f(X)) -> a__f(mark(X))
     , mark(p(X)) -> a__p(mark(X))
     , mark(0()) -> 0()
     , mark(cons(X1, X2)) -> cons(mark(X1), X2)
     , mark(s(X)) -> s(mark(X))
     , a__f(X) -> f(X)
     , a__p(X) -> p(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:
         {  a__f(0()) -> cons(0(), f(s(0())))
          , a__f(s(0())) -> a__f(a__p(s(0())))
          , a__p(s(0())) -> 0()
          , mark(f(X)) -> a__f(mark(X))
          , mark(p(X)) -> a__p(mark(X))
          , mark(0()) -> 0()
          , mark(cons(X1, X2)) -> cons(mark(X1), X2)
          , mark(s(X)) -> s(mark(X))
          , a__f(X) -> f(X)
          , a__p(X) -> p(X)}

     Proof Output:
       The problem is match-bounded by 3.
       The enriched problem is compatible with the following automaton:
       {  a__f_0(2) -> 1
        , a__f_1(7) -> 1
        , a__f_1(7) -> 7
        , a__f_2(10) -> 1
        , a__f_2(10) -> 7
        , 0_0() -> 2
        , 0_1() -> 1
        , 0_1() -> 3
        , 0_1() -> 6
        , 0_1() -> 7
        , 0_2() -> 1
        , 0_2() -> 7
        , 0_2() -> 10
        , 0_3() -> 10
        , cons_0(2, 2) -> 2
        , cons_1(3, 4) -> 1
        , cons_1(7, 2) -> 1
        , cons_1(7, 2) -> 7
        , cons_2(7, 8) -> 1
        , cons_2(10, 8) -> 7
        , cons_3(10, 11) -> 1
        , cons_3(10, 11) -> 7
        , f_0(2) -> 2
        , f_1(2) -> 1
        , f_1(5) -> 4
        , f_2(7) -> 1
        , f_2(7) -> 7
        , f_2(9) -> 8
        , f_3(10) -> 1
        , f_3(10) -> 7
        , f_3(12) -> 11
        , s_0(2) -> 2
        , s_1(6) -> 5
        , s_1(7) -> 1
        , s_1(7) -> 7
        , s_2(7) -> 9
        , s_3(10) -> 12
        , a__p_0(2) -> 1
        , a__p_1(5) -> 7
        , a__p_1(7) -> 1
        , a__p_1(7) -> 7
        , a__p_2(9) -> 10
        , mark_0(2) -> 1
        , mark_1(2) -> 7
        , p_0(2) -> 2
        , p_1(2) -> 1
        , p_2(5) -> 7
        , p_2(7) -> 1
        , p_2(7) -> 7
        , p_3(9) -> 10}

Tool RC1

Execution Time	Unknown
Answer	YES(?,O(n^2))
Input	Transformed CSR 04 Ex4 7 15 Bor03 GM

stdout:

YES(?,O(n^2))

Tool RC2

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

stdout:

YES(?,O(n^1))

'Fastest (timeout of 60.0 seconds)'
-----------------------------------
Answer:           YES(?,O(n^1))
Input Problem:    runtime-complexity with respect to
  Rules:
    {  a__f(0()) -> cons(0(), f(s(0())))
     , a__f(s(0())) -> a__f(a__p(s(0())))
     , a__p(s(0())) -> 0()
     , mark(f(X)) -> a__f(mark(X))
     , mark(p(X)) -> a__p(mark(X))
     , mark(0()) -> 0()
     , mark(cons(X1, X2)) -> cons(mark(X1), X2)
     , mark(s(X)) -> s(mark(X))
     , a__f(X) -> f(X)
     , a__p(X) -> p(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:
         {  a__f(0()) -> cons(0(), f(s(0())))
          , a__f(s(0())) -> a__f(a__p(s(0())))
          , a__p(s(0())) -> 0()
          , mark(f(X)) -> a__f(mark(X))
          , mark(p(X)) -> a__p(mark(X))
          , mark(0()) -> 0()
          , mark(cons(X1, X2)) -> cons(mark(X1), X2)
          , mark(s(X)) -> s(mark(X))
          , a__f(X) -> f(X)
          , a__p(X) -> p(X)}

     Proof Output:
       The problem is match-bounded by 3.
       The enriched problem is compatible with the following automaton:
       {  a__f_0(2) -> 1
        , a__f_1(7) -> 1
        , a__f_1(7) -> 7
        , a__f_2(10) -> 1
        , a__f_2(10) -> 7
        , 0_0() -> 2
        , 0_1() -> 1
        , 0_1() -> 3
        , 0_1() -> 6
        , 0_1() -> 7
        , 0_2() -> 1
        , 0_2() -> 7
        , 0_2() -> 10
        , 0_3() -> 10
        , cons_0(2, 2) -> 2
        , cons_1(3, 4) -> 1
        , cons_1(7, 2) -> 1
        , cons_1(7, 2) -> 7
        , cons_2(7, 8) -> 1
        , cons_2(10, 8) -> 7
        , cons_3(10, 11) -> 1
        , cons_3(10, 11) -> 7
        , f_0(2) -> 2
        , f_1(2) -> 1
        , f_1(5) -> 4
        , f_2(7) -> 1
        , f_2(7) -> 7
        , f_2(9) -> 8
        , f_3(10) -> 1
        , f_3(10) -> 7
        , f_3(12) -> 11
        , s_0(2) -> 2
        , s_1(6) -> 5
        , s_1(7) -> 1
        , s_1(7) -> 7
        , s_2(7) -> 9
        , s_3(10) -> 12
        , a__p_0(2) -> 1
        , a__p_1(5) -> 7
        , a__p_1(7) -> 1
        , a__p_1(7) -> 7
        , a__p_2(9) -> 10
        , mark_0(2) -> 1
        , mark_1(2) -> 7
        , p_0(2) -> 2
        , p_1(2) -> 1
        , p_2(5) -> 7
        , p_2(7) -> 1
        , p_2(7) -> 7
        , p_3(9) -> 10}