Problem Transformed CSR 04 ExProp7 Luc06 C

Tool CaT

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

stdout:

YES(?,O(n^1))

Problem:
active(f(0())) -> mark(cons(0(),f(s(0()))))
active(f(s(0()))) -> mark(f(p(s(0()))))
active(p(s(X))) -> mark(X)
active(f(X)) -> f(active(X))
active(cons(X1,X2)) -> cons(active(X1),X2)
active(s(X)) -> s(active(X))
active(p(X)) -> p(active(X))
f(mark(X)) -> mark(f(X))
cons(mark(X1),X2) -> mark(cons(X1,X2))
s(mark(X)) -> mark(s(X))
p(mark(X)) -> mark(p(X))
proper(f(X)) -> f(proper(X))
proper(0()) -> ok(0())
proper(cons(X1,X2)) -> cons(proper(X1),proper(X2))
proper(s(X)) -> s(proper(X))
proper(p(X)) -> p(proper(X))
f(ok(X)) -> ok(f(X))
cons(ok(X1),ok(X2)) -> ok(cons(X1,X2))
s(ok(X)) -> ok(s(X))
p(ok(X)) -> ok(p(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Proof:
Bounds Processor:
  bound: 2
  enrichment: match
  automaton:
   final states: {10,9,8,7,6,5,4}
   transitions:
    top1(22) -> 10*
    active1(2) -> 22*
    active1(1) -> 22*
    active1(3) -> 22*
    proper1(2) -> 22*
    proper1(1) -> 22*
    proper1(3) -> 22*
    ok1(15) -> 15,7
    ok1(19) -> 22,9
    ok1(14) -> 14,6
    ok1(11) -> 11,5
    ok1(18) -> 18,8
    p1(2) -> 18*
    p1(1) -> 18*
    p1(3) -> 18*
    s1(2) -> 15*
    s1(1) -> 15*
    s1(3) -> 15*
    cons1(3,1) -> 14*
    cons1(3,3) -> 14*
    cons1(1,2) -> 14*
    cons1(2,1) -> 14*
    cons1(2,3) -> 14*
    cons1(3,2) -> 14*
    cons1(1,1) -> 14*
    cons1(1,3) -> 14*
    cons1(2,2) -> 14*
    f1(2) -> 11*
    f1(1) -> 11*
    f1(3) -> 11*
    01() -> 19*
    mark1(15) -> 15,7
    mark1(14) -> 14,6
    mark1(11) -> 11,5
    mark1(18) -> 18,8
    top2(23) -> 10*
    active0(2) -> 4*
    active0(1) -> 4*
    active0(3) -> 4*
    active2(19) -> 23*
    f0(2) -> 5*
    f0(1) -> 5*
    f0(3) -> 5*
    00() -> 1*
    mark0(2) -> 2*
    mark0(1) -> 2*
    mark0(3) -> 2*
    cons0(3,1) -> 6*
    cons0(3,3) -> 6*
    cons0(1,2) -> 6*
    cons0(2,1) -> 6*
    cons0(2,3) -> 6*
    cons0(3,2) -> 6*
    cons0(1,1) -> 6*
    cons0(1,3) -> 6*
    cons0(2,2) -> 6*
    s0(2) -> 7*
    s0(1) -> 7*
    s0(3) -> 7*
    p0(2) -> 8*
    p0(1) -> 8*
    p0(3) -> 8*
    proper0(2) -> 9*
    proper0(1) -> 9*
    proper0(3) -> 9*
    ok0(2) -> 3*
    ok0(1) -> 3*
    ok0(3) -> 3*
    top0(2) -> 10*
    top0(1) -> 10*
    top0(3) -> 10*
  problem:

  Qed

Tool IRC1

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

stdout:

YES(?,O(n^1))

Tool IRC2

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

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:
    {  active(f(0())) -> mark(cons(0(), f(s(0()))))
     , active(f(s(0()))) -> mark(f(p(s(0()))))
     , active(p(s(X))) -> mark(X)
     , active(f(X)) -> f(active(X))
     , active(cons(X1, X2)) -> cons(active(X1), X2)
     , active(s(X)) -> s(active(X))
     , active(p(X)) -> p(active(X))
     , f(mark(X)) -> mark(f(X))
     , cons(mark(X1), X2) -> mark(cons(X1, X2))
     , s(mark(X)) -> mark(s(X))
     , p(mark(X)) -> mark(p(X))
     , proper(f(X)) -> f(proper(X))
     , proper(0()) -> ok(0())
     , proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
     , proper(s(X)) -> s(proper(X))
     , proper(p(X)) -> p(proper(X))
     , f(ok(X)) -> ok(f(X))
     , cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
     , s(ok(X)) -> ok(s(X))
     , p(ok(X)) -> ok(p(X))
     , top(mark(X)) -> top(proper(X))
     , top(ok(X)) -> top(active(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:
         {  active(f(0())) -> mark(cons(0(), f(s(0()))))
          , active(f(s(0()))) -> mark(f(p(s(0()))))
          , active(p(s(X))) -> mark(X)
          , active(f(X)) -> f(active(X))
          , active(cons(X1, X2)) -> cons(active(X1), X2)
          , active(s(X)) -> s(active(X))
          , active(p(X)) -> p(active(X))
          , f(mark(X)) -> mark(f(X))
          , cons(mark(X1), X2) -> mark(cons(X1, X2))
          , s(mark(X)) -> mark(s(X))
          , p(mark(X)) -> mark(p(X))
          , proper(f(X)) -> f(proper(X))
          , proper(0()) -> ok(0())
          , proper(cons(X1, X2)) -> cons(proper(X1), proper(X2))
          , proper(s(X)) -> s(proper(X))
          , proper(p(X)) -> p(proper(X))
          , f(ok(X)) -> ok(f(X))
          , cons(ok(X1), ok(X2)) -> ok(cons(X1, X2))
          , s(ok(X)) -> ok(s(X))
          , p(ok(X)) -> ok(p(X))
          , top(mark(X)) -> top(proper(X))
          , top(ok(X)) -> top(active(X))}

     Proof Output:
       The problem is match-bounded by 2.
       The enriched problem is compatible with the following automaton:
       {  active_0(2) -> 1
        , active_1(2) -> 5
        , active_2(4) -> 6
        , f_0(2) -> 1
        , f_1(2) -> 3
        , 0_0() -> 2
        , 0_1() -> 4
        , mark_0(2) -> 2
        , mark_1(3) -> 1
        , mark_1(3) -> 3
        , cons_0(2, 2) -> 1
        , cons_1(2, 2) -> 3
        , s_0(2) -> 1
        , s_1(2) -> 3
        , p_0(2) -> 1
        , p_1(2) -> 3
        , proper_0(2) -> 1
        , proper_1(2) -> 5
        , ok_0(2) -> 2
        , ok_1(3) -> 1
        , ok_1(3) -> 3
        , ok_1(4) -> 1
        , ok_1(4) -> 5
        , top_0(2) -> 1
        , top_1(5) -> 1
        , top_2(6) -> 1}

Tool RC1

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

stdout:

YES(?,O(n^1))

Tool RC2

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

stdout:

YES(?,O(n^1))

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

     Proof Output:
       The problem is match-bounded by 2.
       The enriched problem is compatible with the following automaton:
       {  active_0(2) -> 1
        , active_1(2) -> 5
        , active_2(4) -> 6
        , f_0(2) -> 1
        , f_1(2) -> 3
        , 0_0() -> 2
        , 0_1() -> 4
        , mark_0(2) -> 2
        , mark_1(3) -> 1
        , mark_1(3) -> 3
        , cons_0(2, 2) -> 1
        , cons_1(2, 2) -> 3
        , s_0(2) -> 1
        , s_1(2) -> 3
        , p_0(2) -> 1
        , p_1(2) -> 3
        , proper_0(2) -> 1
        , proper_1(2) -> 5
        , ok_0(2) -> 2
        , ok_1(3) -> 1
        , ok_1(3) -> 3
        , ok_1(4) -> 1
        , ok_1(4) -> 5
        , top_0(2) -> 1
        , top_1(5) -> 1
        , top_2(6) -> 1}