Problem Transformed CSR 04 PALINDROME nosorts noand C

Tool CaT

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

stdout:

YES(?,O(n^1))

Problem:
active(__(__(X,Y),Z)) -> mark(__(X,__(Y,Z)))
active(__(X,nil())) -> mark(X)
active(__(nil(),X)) -> mark(X)
active(U11(tt())) -> mark(U12(tt()))
active(U12(tt())) -> mark(tt())
active(isNePal(__(I,__(P,I)))) -> mark(U11(tt()))
active(__(X1,X2)) -> __(active(X1),X2)
active(__(X1,X2)) -> __(X1,active(X2))
active(U11(X)) -> U11(active(X))
active(U12(X)) -> U12(active(X))
active(isNePal(X)) -> isNePal(active(X))
__(mark(X1),X2) -> mark(__(X1,X2))
__(X1,mark(X2)) -> mark(__(X1,X2))
U11(mark(X)) -> mark(U11(X))
U12(mark(X)) -> mark(U12(X))
isNePal(mark(X)) -> mark(isNePal(X))
proper(__(X1,X2)) -> __(proper(X1),proper(X2))
proper(nil()) -> ok(nil())
proper(U11(X)) -> U11(proper(X))
proper(tt()) -> ok(tt())
proper(U12(X)) -> U12(proper(X))
proper(isNePal(X)) -> isNePal(proper(X))
__(ok(X1),ok(X2)) -> ok(__(X1,X2))
U11(ok(X)) -> ok(U11(X))
U12(ok(X)) -> ok(U12(X))
isNePal(ok(X)) -> ok(isNePal(X))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Proof:
Bounds Processor:
  bound: 2
  enrichment: match
  automaton:
   final states: {19,18,11,10,9,8,7,6,5}
   transitions:
    active0(20) -> 5*
    active0(2) -> 5*
    active0(19) -> 5*
    active0(4) -> 5*
    active0(1) -> 5*
    active0(3) -> 5*
    __0(2,20) -> 6*
    __0(3,1) -> 6*
    __0(3,3) -> 6*
    __0(19,2) -> 6*
    __0(3,19) -> 6*
    __0(19,4) -> 6*
    __0(4,2) -> 6*
    __0(4,4) -> 6*
    __0(19,20) -> 6*
    __0(20,1) -> 6*
    __0(20,3) -> 6*
    __0(4,20) -> 6*
    __0(20,19) -> 6*
    __0(1,2) -> 6*
    __0(1,4) -> 6*
    __0(1,20) -> 6*
    __0(2,1) -> 6*
    __0(2,3) -> 6*
    __0(2,19) -> 6*
    __0(3,2) -> 6*
    __0(3,4) -> 6*
    __0(19,1) -> 6*
    __0(19,3) -> 6*
    __0(3,20) -> 6*
    __0(4,1) -> 6*
    __0(4,3) -> 6*
    __0(19,19) -> 6*
    __0(20,2) -> 6*
    __0(4,19) -> 6*
    __0(20,4) -> 6*
    __0(20,20) -> 6*
    __0(1,1) -> 6*
    __0(1,3) -> 6*
    __0(1,19) -> 6*
    __0(2,2) -> 6*
    __0(2,4) -> 6*
    mark0(20) -> 1*
    mark0(2) -> 1*
    mark0(19) -> 1*
    mark0(4) -> 1*
    mark0(1) -> 1*
    mark0(3) -> 1*
    nil0() -> 2*
    U110(20) -> 7*
    U110(2) -> 7*
    U110(19) -> 7*
    U110(4) -> 7*
    U110(1) -> 7*
    U110(3) -> 7*
    tt0() -> 3*
    U120(20) -> 8*
    U120(2) -> 8*
    U120(19) -> 8*
    U120(4) -> 8*
    U120(1) -> 8*
    U120(3) -> 8*
    isNePal0(20) -> 9*
    isNePal0(2) -> 9*
    isNePal0(19) -> 9*
    isNePal0(4) -> 9*
    isNePal0(1) -> 9*
    isNePal0(3) -> 9*
    proper0(20) -> 10*
    proper0(2) -> 10*
    proper0(19) -> 10*
    proper0(4) -> 10*
    proper0(1) -> 10*
    proper0(3) -> 10*
    ok0(20) -> 4*
    ok0(2) -> 4*
    ok0(19) -> 4*
    ok0(4) -> 4*
    ok0(1) -> 4*
    ok0(3) -> 4*
    top0(20) -> 11*
    top0(2) -> 11*
    top0(19) -> 11*
    top0(4) -> 11*
    top0(1) -> 11*
    top0(3) -> 11*
    top1(10) -> 11*
    top1(19) -> 11*
    top1(18) -> 11*
    active1(20) -> 5,18*
    active1(2) -> 18*,5,10
    active1(19) -> 5,18*
    active1(4) -> 18*,5,10
    active1(1) -> 18*,5,10
    active1(3) -> 18*,5,10
    proper1(20) -> 10*
    proper1(2) -> 10*
    proper1(19) -> 10*
    proper1(4) -> 10*
    proper1(1) -> 10*
    proper1(3) -> 10*
    ok1(20) -> 4,19*
    ok1(7) -> 7*
    ok1(2) -> 19*,4,10
    ok1(9) -> 9*
    ok1(6) -> 6*
    ok1(8) -> 8*
    isNePal1(20) -> 9*
    isNePal1(2) -> 9*
    isNePal1(19) -> 9*
    isNePal1(4) -> 9*
    isNePal1(1) -> 9*
    isNePal1(3) -> 9*
    U121(20) -> 8*
    U121(2) -> 8*
    U121(19) -> 8*
    U121(4) -> 8*
    U121(1) -> 8*
    U121(3) -> 8*
    U111(20) -> 7*
    U111(2) -> 7*
    U111(19) -> 7*
    U111(4) -> 7*
    U111(1) -> 7*
    U111(3) -> 7*
    __1(2,20) -> 6*
    __1(3,1) -> 6*
    __1(3,3) -> 6*
    __1(19,2) -> 6*
    __1(3,19) -> 6*
    __1(19,4) -> 6*
    __1(4,2) -> 6*
    __1(4,4) -> 6*
    __1(19,20) -> 6*
    __1(20,1) -> 6*
    __1(20,3) -> 6*
    __1(4,20) -> 6*
    __1(20,19) -> 6*
    __1(1,2) -> 6*
    __1(1,4) -> 6*
    __1(1,20) -> 6*
    __1(2,1) -> 6*
    __1(2,3) -> 6*
    __1(2,19) -> 6*
    __1(3,2) -> 6*
    __1(3,4) -> 6*
    __1(19,1) -> 6*
    __1(19,3) -> 6*
    __1(3,20) -> 6*
    __1(4,1) -> 6*
    __1(4,3) -> 6*
    __1(19,19) -> 6*
    __1(20,2) -> 6*
    __1(4,19) -> 6*
    __1(20,4) -> 6*
    __1(20,20) -> 6*
    __1(1,1) -> 6*
    __1(1,3) -> 6*
    __1(1,19) -> 6*
    __1(2,2) -> 6*
    __1(2,4) -> 6*
    tt1() -> 20*,3,2
    nil1() -> 2*
    mark1(7) -> 7*
    mark1(9) -> 9*
    mark1(6) -> 6*
    mark1(8) -> 8*
    top2(5) -> 11*
    top2(18) -> 11*
    active2(20) -> 18*,5
    active2(2) -> 10,18*,5
    ok2(20) -> 4,19*
    ok2(7) -> 7*
    ok2(2) -> 4,19*,10
    ok2(9) -> 9*
    ok2(6) -> 6*
    ok2(8) -> 8*
    isNePal2(20) -> 9*
    isNePal2(2) -> 9*
    U122(20) -> 8*
    U122(2) -> 8*
    U112(20) -> 7*
    U112(2) -> 7*
    __2(2,20) -> 6*
    __2(20,2) -> 6*
    __2(20,20) -> 6*
    __2(2,2) -> 6*
    tt2() -> 3,20*,2
    nil2() -> 2*
  problem:

  Qed

Tool IRC1

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

stdout:

YES(?,O(n^1))

Tool IRC2

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 PALINDROME nosorts noand 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(__(__(X, Y), Z)) -> mark(__(X, __(Y, Z)))
     , active(__(X, nil())) -> mark(X)
     , active(__(nil(), X)) -> mark(X)
     , active(U11(tt())) -> mark(U12(tt()))
     , active(U12(tt())) -> mark(tt())
     , active(isNePal(__(I, __(P, I)))) -> mark(U11(tt()))
     , active(__(X1, X2)) -> __(active(X1), X2)
     , active(__(X1, X2)) -> __(X1, active(X2))
     , active(U11(X)) -> U11(active(X))
     , active(U12(X)) -> U12(active(X))
     , active(isNePal(X)) -> isNePal(active(X))
     , __(mark(X1), X2) -> mark(__(X1, X2))
     , __(X1, mark(X2)) -> mark(__(X1, X2))
     , U11(mark(X)) -> mark(U11(X))
     , U12(mark(X)) -> mark(U12(X))
     , isNePal(mark(X)) -> mark(isNePal(X))
     , proper(__(X1, X2)) -> __(proper(X1), proper(X2))
     , proper(nil()) -> ok(nil())
     , proper(U11(X)) -> U11(proper(X))
     , proper(tt()) -> ok(tt())
     , proper(U12(X)) -> U12(proper(X))
     , proper(isNePal(X)) -> isNePal(proper(X))
     , __(ok(X1), ok(X2)) -> ok(__(X1, X2))
     , U11(ok(X)) -> ok(U11(X))
     , U12(ok(X)) -> ok(U12(X))
     , isNePal(ok(X)) -> ok(isNePal(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(__(__(X, Y), Z)) -> mark(__(X, __(Y, Z)))
          , active(__(X, nil())) -> mark(X)
          , active(__(nil(), X)) -> mark(X)
          , active(U11(tt())) -> mark(U12(tt()))
          , active(U12(tt())) -> mark(tt())
          , active(isNePal(__(I, __(P, I)))) -> mark(U11(tt()))
          , active(__(X1, X2)) -> __(active(X1), X2)
          , active(__(X1, X2)) -> __(X1, active(X2))
          , active(U11(X)) -> U11(active(X))
          , active(U12(X)) -> U12(active(X))
          , active(isNePal(X)) -> isNePal(active(X))
          , __(mark(X1), X2) -> mark(__(X1, X2))
          , __(X1, mark(X2)) -> mark(__(X1, X2))
          , U11(mark(X)) -> mark(U11(X))
          , U12(mark(X)) -> mark(U12(X))
          , isNePal(mark(X)) -> mark(isNePal(X))
          , proper(__(X1, X2)) -> __(proper(X1), proper(X2))
          , proper(nil()) -> ok(nil())
          , proper(U11(X)) -> U11(proper(X))
          , proper(tt()) -> ok(tt())
          , proper(U12(X)) -> U12(proper(X))
          , proper(isNePal(X)) -> isNePal(proper(X))
          , __(ok(X1), ok(X2)) -> ok(__(X1, X2))
          , U11(ok(X)) -> ok(U11(X))
          , U12(ok(X)) -> ok(U12(X))
          , isNePal(ok(X)) -> ok(isNePal(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
        , ___0(2, 2) -> 1
        , ___1(2, 2) -> 3
        , mark_0(2) -> 2
        , mark_1(3) -> 1
        , mark_1(3) -> 3
        , nil_0() -> 2
        , nil_1() -> 4
        , U11_0(2) -> 1
        , U11_1(2) -> 3
        , tt_0() -> 2
        , tt_1() -> 4
        , U12_0(2) -> 1
        , U12_1(2) -> 3
        , isNePal_0(2) -> 1
        , isNePal_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 PALINDROME nosorts noand C

stdout:

YES(?,O(n^1))

Tool RC2

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 PALINDROME nosorts noand 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(__(__(X, Y), Z)) -> mark(__(X, __(Y, Z)))
     , active(__(X, nil())) -> mark(X)
     , active(__(nil(), X)) -> mark(X)
     , active(U11(tt())) -> mark(U12(tt()))
     , active(U12(tt())) -> mark(tt())
     , active(isNePal(__(I, __(P, I)))) -> mark(U11(tt()))
     , active(__(X1, X2)) -> __(active(X1), X2)
     , active(__(X1, X2)) -> __(X1, active(X2))
     , active(U11(X)) -> U11(active(X))
     , active(U12(X)) -> U12(active(X))
     , active(isNePal(X)) -> isNePal(active(X))
     , __(mark(X1), X2) -> mark(__(X1, X2))
     , __(X1, mark(X2)) -> mark(__(X1, X2))
     , U11(mark(X)) -> mark(U11(X))
     , U12(mark(X)) -> mark(U12(X))
     , isNePal(mark(X)) -> mark(isNePal(X))
     , proper(__(X1, X2)) -> __(proper(X1), proper(X2))
     , proper(nil()) -> ok(nil())
     , proper(U11(X)) -> U11(proper(X))
     , proper(tt()) -> ok(tt())
     , proper(U12(X)) -> U12(proper(X))
     , proper(isNePal(X)) -> isNePal(proper(X))
     , __(ok(X1), ok(X2)) -> ok(__(X1, X2))
     , U11(ok(X)) -> ok(U11(X))
     , U12(ok(X)) -> ok(U12(X))
     , isNePal(ok(X)) -> ok(isNePal(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(__(__(X, Y), Z)) -> mark(__(X, __(Y, Z)))
          , active(__(X, nil())) -> mark(X)
          , active(__(nil(), X)) -> mark(X)
          , active(U11(tt())) -> mark(U12(tt()))
          , active(U12(tt())) -> mark(tt())
          , active(isNePal(__(I, __(P, I)))) -> mark(U11(tt()))
          , active(__(X1, X2)) -> __(active(X1), X2)
          , active(__(X1, X2)) -> __(X1, active(X2))
          , active(U11(X)) -> U11(active(X))
          , active(U12(X)) -> U12(active(X))
          , active(isNePal(X)) -> isNePal(active(X))
          , __(mark(X1), X2) -> mark(__(X1, X2))
          , __(X1, mark(X2)) -> mark(__(X1, X2))
          , U11(mark(X)) -> mark(U11(X))
          , U12(mark(X)) -> mark(U12(X))
          , isNePal(mark(X)) -> mark(isNePal(X))
          , proper(__(X1, X2)) -> __(proper(X1), proper(X2))
          , proper(nil()) -> ok(nil())
          , proper(U11(X)) -> U11(proper(X))
          , proper(tt()) -> ok(tt())
          , proper(U12(X)) -> U12(proper(X))
          , proper(isNePal(X)) -> isNePal(proper(X))
          , __(ok(X1), ok(X2)) -> ok(__(X1, X2))
          , U11(ok(X)) -> ok(U11(X))
          , U12(ok(X)) -> ok(U12(X))
          , isNePal(ok(X)) -> ok(isNePal(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
        , ___0(2, 2) -> 1
        , ___1(2, 2) -> 3
        , mark_0(2) -> 2
        , mark_1(3) -> 1
        , mark_1(3) -> 3
        , nil_0() -> 2
        , nil_1() -> 4
        , U11_0(2) -> 1
        , U11_1(2) -> 3
        , tt_0() -> 2
        , tt_1() -> 4
        , U12_0(2) -> 1
        , U12_1(2) -> 3
        , isNePal_0(2) -> 1
        , isNePal_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}