Problem Transformed CSR 04 PEANO nosorts C

Tool CaT

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

stdout:

YES(?,O(n^1))

Problem:
active(and(tt(),X)) -> mark(X)
active(plus(N,0())) -> mark(N)
active(plus(N,s(M))) -> mark(s(plus(N,M)))
active(and(X1,X2)) -> and(active(X1),X2)
active(plus(X1,X2)) -> plus(active(X1),X2)
active(plus(X1,X2)) -> plus(X1,active(X2))
active(s(X)) -> s(active(X))
and(mark(X1),X2) -> mark(and(X1,X2))
plus(mark(X1),X2) -> mark(plus(X1,X2))
plus(X1,mark(X2)) -> mark(plus(X1,X2))
s(mark(X)) -> mark(s(X))
proper(and(X1,X2)) -> and(proper(X1),proper(X2))
proper(tt()) -> ok(tt())
proper(plus(X1,X2)) -> plus(proper(X1),proper(X2))
proper(0()) -> ok(0())
proper(s(X)) -> s(proper(X))
and(ok(X1),ok(X2)) -> ok(and(X1,X2))
plus(ok(X1),ok(X2)) -> ok(plus(X1,X2))
s(ok(X)) -> ok(s(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}
   transitions:
    top1(20) -> 10*
    active1(2) -> 20*
    active1(4) -> 20*
    active1(1) -> 20*
    active1(3) -> 20*
    proper1(2) -> 20*
    proper1(4) -> 20*
    proper1(1) -> 20*
    proper1(3) -> 20*
    ok1(15) -> 15,8
    ok1(17) -> 20,9
    ok1(11) -> 11,6
    ok1(13) -> 13,7
    s1(2) -> 15*
    s1(4) -> 15*
    s1(1) -> 15*
    s1(3) -> 15*
    plus1(3,1) -> 13*
    plus1(3,3) -> 13*
    plus1(4,2) -> 13*
    plus1(4,4) -> 13*
    plus1(1,2) -> 13*
    plus1(1,4) -> 13*
    plus1(2,1) -> 13*
    plus1(2,3) -> 13*
    plus1(3,2) -> 13*
    plus1(3,4) -> 13*
    plus1(4,1) -> 13*
    plus1(4,3) -> 13*
    plus1(1,1) -> 13*
    plus1(1,3) -> 13*
    plus1(2,2) -> 13*
    plus1(2,4) -> 13*
    and1(3,1) -> 11*
    and1(3,3) -> 11*
    and1(4,2) -> 11*
    and1(4,4) -> 11*
    and1(1,2) -> 11*
    and1(1,4) -> 11*
    and1(2,1) -> 11*
    and1(2,3) -> 11*
    and1(3,2) -> 11*
    and1(3,4) -> 11*
    and1(4,1) -> 11*
    and1(4,3) -> 11*
    and1(1,1) -> 11*
    and1(1,3) -> 11*
    and1(2,2) -> 11*
    and1(2,4) -> 11*
    01() -> 17*
    tt1() -> 17*
    mark1(15) -> 15,8
    mark1(11) -> 11,6
    mark1(13) -> 13,7
    top2(21) -> 10*
    active0(2) -> 5*
    active0(4) -> 5*
    active0(1) -> 5*
    active0(3) -> 5*
    active2(17) -> 21*
    and0(3,1) -> 6*
    and0(3,3) -> 6*
    and0(4,2) -> 6*
    and0(4,4) -> 6*
    and0(1,2) -> 6*
    and0(1,4) -> 6*
    and0(2,1) -> 6*
    and0(2,3) -> 6*
    and0(3,2) -> 6*
    and0(3,4) -> 6*
    and0(4,1) -> 6*
    and0(4,3) -> 6*
    and0(1,1) -> 6*
    and0(1,3) -> 6*
    and0(2,2) -> 6*
    and0(2,4) -> 6*
    tt0() -> 1*
    mark0(2) -> 2*
    mark0(4) -> 2*
    mark0(1) -> 2*
    mark0(3) -> 2*
    plus0(3,1) -> 7*
    plus0(3,3) -> 7*
    plus0(4,2) -> 7*
    plus0(4,4) -> 7*
    plus0(1,2) -> 7*
    plus0(1,4) -> 7*
    plus0(2,1) -> 7*
    plus0(2,3) -> 7*
    plus0(3,2) -> 7*
    plus0(3,4) -> 7*
    plus0(4,1) -> 7*
    plus0(4,3) -> 7*
    plus0(1,1) -> 7*
    plus0(1,3) -> 7*
    plus0(2,2) -> 7*
    plus0(2,4) -> 7*
    00() -> 3*
    s0(2) -> 8*
    s0(4) -> 8*
    s0(1) -> 8*
    s0(3) -> 8*
    proper0(2) -> 9*
    proper0(4) -> 9*
    proper0(1) -> 9*
    proper0(3) -> 9*
    ok0(2) -> 4*
    ok0(4) -> 4*
    ok0(1) -> 4*
    ok0(3) -> 4*
    top0(2) -> 10*
    top0(4) -> 10*
    top0(1) -> 10*
    top0(3) -> 10*
  problem:

  Qed

Tool IRC1

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

stdout:

YES(?,O(n^1))

Tool IRC2

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 PEANO nosorts 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(and(tt(), X)) -> mark(X)
     , active(plus(N, 0())) -> mark(N)
     , active(plus(N, s(M))) -> mark(s(plus(N, M)))
     , active(and(X1, X2)) -> and(active(X1), X2)
     , active(plus(X1, X2)) -> plus(active(X1), X2)
     , active(plus(X1, X2)) -> plus(X1, active(X2))
     , active(s(X)) -> s(active(X))
     , and(mark(X1), X2) -> mark(and(X1, X2))
     , plus(mark(X1), X2) -> mark(plus(X1, X2))
     , plus(X1, mark(X2)) -> mark(plus(X1, X2))
     , s(mark(X)) -> mark(s(X))
     , proper(and(X1, X2)) -> and(proper(X1), proper(X2))
     , proper(tt()) -> ok(tt())
     , proper(plus(X1, X2)) -> plus(proper(X1), proper(X2))
     , proper(0()) -> ok(0())
     , proper(s(X)) -> s(proper(X))
     , and(ok(X1), ok(X2)) -> ok(and(X1, X2))
     , plus(ok(X1), ok(X2)) -> ok(plus(X1, X2))
     , s(ok(X)) -> ok(s(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(and(tt(), X)) -> mark(X)
          , active(plus(N, 0())) -> mark(N)
          , active(plus(N, s(M))) -> mark(s(plus(N, M)))
          , active(and(X1, X2)) -> and(active(X1), X2)
          , active(plus(X1, X2)) -> plus(active(X1), X2)
          , active(plus(X1, X2)) -> plus(X1, active(X2))
          , active(s(X)) -> s(active(X))
          , and(mark(X1), X2) -> mark(and(X1, X2))
          , plus(mark(X1), X2) -> mark(plus(X1, X2))
          , plus(X1, mark(X2)) -> mark(plus(X1, X2))
          , s(mark(X)) -> mark(s(X))
          , proper(and(X1, X2)) -> and(proper(X1), proper(X2))
          , proper(tt()) -> ok(tt())
          , proper(plus(X1, X2)) -> plus(proper(X1), proper(X2))
          , proper(0()) -> ok(0())
          , proper(s(X)) -> s(proper(X))
          , and(ok(X1), ok(X2)) -> ok(and(X1, X2))
          , plus(ok(X1), ok(X2)) -> ok(plus(X1, X2))
          , s(ok(X)) -> ok(s(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
        , and_0(2, 2) -> 1
        , and_1(2, 2) -> 3
        , tt_0() -> 2
        , tt_1() -> 4
        , mark_0(2) -> 2
        , mark_1(3) -> 1
        , mark_1(3) -> 3
        , plus_0(2, 2) -> 1
        , plus_1(2, 2) -> 3
        , 0_0() -> 2
        , 0_1() -> 4
        , s_0(2) -> 1
        , s_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 PEANO nosorts C

stdout:

YES(?,O(n^1))

Tool RC2

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 PEANO nosorts 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(and(tt(), X)) -> mark(X)
     , active(plus(N, 0())) -> mark(N)
     , active(plus(N, s(M))) -> mark(s(plus(N, M)))
     , active(and(X1, X2)) -> and(active(X1), X2)
     , active(plus(X1, X2)) -> plus(active(X1), X2)
     , active(plus(X1, X2)) -> plus(X1, active(X2))
     , active(s(X)) -> s(active(X))
     , and(mark(X1), X2) -> mark(and(X1, X2))
     , plus(mark(X1), X2) -> mark(plus(X1, X2))
     , plus(X1, mark(X2)) -> mark(plus(X1, X2))
     , s(mark(X)) -> mark(s(X))
     , proper(and(X1, X2)) -> and(proper(X1), proper(X2))
     , proper(tt()) -> ok(tt())
     , proper(plus(X1, X2)) -> plus(proper(X1), proper(X2))
     , proper(0()) -> ok(0())
     , proper(s(X)) -> s(proper(X))
     , and(ok(X1), ok(X2)) -> ok(and(X1, X2))
     , plus(ok(X1), ok(X2)) -> ok(plus(X1, X2))
     , s(ok(X)) -> ok(s(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(and(tt(), X)) -> mark(X)
          , active(plus(N, 0())) -> mark(N)
          , active(plus(N, s(M))) -> mark(s(plus(N, M)))
          , active(and(X1, X2)) -> and(active(X1), X2)
          , active(plus(X1, X2)) -> plus(active(X1), X2)
          , active(plus(X1, X2)) -> plus(X1, active(X2))
          , active(s(X)) -> s(active(X))
          , and(mark(X1), X2) -> mark(and(X1, X2))
          , plus(mark(X1), X2) -> mark(plus(X1, X2))
          , plus(X1, mark(X2)) -> mark(plus(X1, X2))
          , s(mark(X)) -> mark(s(X))
          , proper(and(X1, X2)) -> and(proper(X1), proper(X2))
          , proper(tt()) -> ok(tt())
          , proper(plus(X1, X2)) -> plus(proper(X1), proper(X2))
          , proper(0()) -> ok(0())
          , proper(s(X)) -> s(proper(X))
          , and(ok(X1), ok(X2)) -> ok(and(X1, X2))
          , plus(ok(X1), ok(X2)) -> ok(plus(X1, X2))
          , s(ok(X)) -> ok(s(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
        , and_0(2, 2) -> 1
        , and_1(2, 2) -> 3
        , tt_0() -> 2
        , tt_1() -> 4
        , mark_0(2) -> 2
        , mark_1(3) -> 1
        , mark_1(3) -> 3
        , plus_0(2, 2) -> 1
        , plus_1(2, 2) -> 3
        , 0_0() -> 2
        , 0_1() -> 4
        , s_0(2) -> 1
        , s_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}