Problem Transformed CSR 04 PEANO nosorts noand C

Tool CaT

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

stdout:

YES(?,O(n^1))

Problem:
active(U11(tt(),M,N)) -> mark(U12(tt(),M,N))
active(U12(tt(),M,N)) -> mark(s(plus(N,M)))
active(plus(N,0())) -> mark(N)
active(plus(N,s(M))) -> mark(U11(tt(),M,N))
active(U11(X1,X2,X3)) -> U11(active(X1),X2,X3)
active(U12(X1,X2,X3)) -> U12(active(X1),X2,X3)
active(s(X)) -> s(active(X))
active(plus(X1,X2)) -> plus(active(X1),X2)
active(plus(X1,X2)) -> plus(X1,active(X2))
U11(mark(X1),X2,X3) -> mark(U11(X1,X2,X3))
U12(mark(X1),X2,X3) -> mark(U12(X1,X2,X3))
s(mark(X)) -> mark(s(X))
plus(mark(X1),X2) -> mark(plus(X1,X2))
plus(X1,mark(X2)) -> mark(plus(X1,X2))
proper(U11(X1,X2,X3)) -> U11(proper(X1),proper(X2),proper(X3))
proper(tt()) -> ok(tt())
proper(U12(X1,X2,X3)) -> U12(proper(X1),proper(X2),proper(X3))
proper(s(X)) -> s(proper(X))
proper(plus(X1,X2)) -> plus(proper(X1),proper(X2))
proper(0()) -> ok(0())
U11(ok(X1),ok(X2),ok(X3)) -> ok(U11(X1,X2,X3))
U12(ok(X1),ok(X2),ok(X3)) -> ok(U12(X1,X2,X3))
s(ok(X)) -> ok(s(X))
plus(ok(X1),ok(X2)) -> ok(plus(X1,X2))
top(mark(X)) -> top(proper(X))
top(ok(X)) -> top(active(X))

Proof:
Bounds Processor:
  bound: 2
  enrichment: match
  automaton:
   final states: {11,10,9,8,7,6,5}
   transitions:
    top1(23) -> 11*
    active1(2) -> 23*
    active1(4) -> 23*
    active1(1) -> 23*
    active1(3) -> 23*
    proper1(2) -> 23*
    proper1(4) -> 23*
    proper1(1) -> 23*
    proper1(3) -> 23*
    ok1(15) -> 15,7
    ok1(17) -> 17,8
    ok1(19) -> 19,9
    ok1(21) -> 23,10
    ok1(13) -> 13,6
    plus1(3,1) -> 19*
    plus1(3,3) -> 19*
    plus1(4,2) -> 19*
    plus1(4,4) -> 19*
    plus1(1,2) -> 19*
    plus1(1,4) -> 19*
    plus1(2,1) -> 19*
    plus1(2,3) -> 19*
    plus1(3,2) -> 19*
    plus1(3,4) -> 19*
    plus1(4,1) -> 19*
    plus1(4,3) -> 19*
    plus1(1,1) -> 19*
    plus1(1,3) -> 19*
    plus1(2,2) -> 19*
    plus1(2,4) -> 19*
    s1(2) -> 17*
    s1(4) -> 17*
    s1(1) -> 17*
    s1(3) -> 17*
    U121(3,2,2) -> 15*
    U121(3,4,2) -> 15*
    U121(1,1,4) -> 15*
    U121(1,3,4) -> 15*
    U121(4,1,2) -> 15*
    U121(1,1,1) -> 15*
    U121(4,3,2) -> 15*
    U121(1,3,1) -> 15*
    U121(2,2,4) -> 15*
    U121(2,4,4) -> 15*
    U121(1,2,3) -> 15*
    U121(1,4,3) -> 15*
    U121(2,2,1) -> 15*
    U121(2,4,1) -> 15*
    U121(3,1,4) -> 15*
    U121(3,3,4) -> 15*
    U121(2,1,3) -> 15*
    U121(2,3,3) -> 15*
    U121(3,1,1) -> 15*
    U121(1,1,2) -> 15*
    U121(3,3,1) -> 15*
    U121(1,3,2) -> 15*
    U121(4,2,4) -> 15*
    U121(4,4,4) -> 15*
    U121(3,2,3) -> 15*
    U121(3,4,3) -> 15*
    U121(4,2,1) -> 15*
    U121(2,2,2) -> 15*
    U121(4,4,1) -> 15*
    U121(2,4,2) -> 15*
    U121(4,1,3) -> 15*
    U121(4,3,3) -> 15*
    U121(3,1,2) -> 15*
    U121(3,3,2) -> 15*
    U121(1,2,4) -> 15*
    U121(1,4,4) -> 15*
    U121(4,2,2) -> 15*
    U121(1,2,1) -> 15*
    U121(4,4,2) -> 15*
    U121(1,4,1) -> 15*
    U121(2,1,4) -> 15*
    U121(2,3,4) -> 15*
    U121(1,1,3) -> 15*
    U121(1,3,3) -> 15*
    U121(2,1,1) -> 15*
    U121(2,3,1) -> 15*
    U121(3,2,4) -> 15*
    U121(3,4,4) -> 15*
    U121(2,2,3) -> 15*
    U121(2,4,3) -> 15*
    U121(3,2,1) -> 15*
    U121(1,2,2) -> 15*
    U121(3,4,1) -> 15*
    U121(1,4,2) -> 15*
    U121(4,1,4) -> 15*
    U121(4,3,4) -> 15*
    U121(3,1,3) -> 15*
    U121(3,3,3) -> 15*
    U121(4,1,1) -> 15*
    U121(2,1,2) -> 15*
    U121(4,3,1) -> 15*
    U121(2,3,2) -> 15*
    U121(4,2,3) -> 15*
    U121(4,4,3) -> 15*
    U111(3,2,2) -> 13*
    U111(3,4,2) -> 13*
    U111(1,1,4) -> 13*
    U111(1,3,4) -> 13*
    U111(4,1,2) -> 13*
    U111(1,1,1) -> 13*
    U111(4,3,2) -> 13*
    U111(1,3,1) -> 13*
    U111(2,2,4) -> 13*
    U111(2,4,4) -> 13*
    U111(1,2,3) -> 13*
    U111(1,4,3) -> 13*
    U111(2,2,1) -> 13*
    U111(2,4,1) -> 13*
    U111(3,1,4) -> 13*
    U111(3,3,4) -> 13*
    U111(2,1,3) -> 13*
    U111(2,3,3) -> 13*
    U111(3,1,1) -> 13*
    U111(1,1,2) -> 13*
    U111(3,3,1) -> 13*
    U111(1,3,2) -> 13*
    U111(4,2,4) -> 13*
    U111(4,4,4) -> 13*
    U111(3,2,3) -> 13*
    U111(3,4,3) -> 13*
    U111(4,2,1) -> 13*
    U111(2,2,2) -> 13*
    U111(4,4,1) -> 13*
    U111(2,4,2) -> 13*
    U111(4,1,3) -> 13*
    U111(4,3,3) -> 13*
    U111(3,1,2) -> 13*
    U111(3,3,2) -> 13*
    U111(1,2,4) -> 13*
    U111(1,4,4) -> 13*
    U111(4,2,2) -> 13*
    U111(1,2,1) -> 13*
    U111(4,4,2) -> 13*
    U111(1,4,1) -> 13*
    U111(2,1,4) -> 13*
    U111(2,3,4) -> 13*
    U111(1,1,3) -> 13*
    U111(1,3,3) -> 13*
    U111(2,1,1) -> 13*
    U111(2,3,1) -> 13*
    U111(3,2,4) -> 13*
    U111(3,4,4) -> 13*
    U111(2,2,3) -> 13*
    U111(2,4,3) -> 13*
    U111(3,2,1) -> 13*
    U111(1,2,2) -> 13*
    U111(3,4,1) -> 13*
    U111(1,4,2) -> 13*
    U111(4,1,4) -> 13*
    U111(4,3,4) -> 13*
    U111(3,1,3) -> 13*
    U111(3,3,3) -> 13*
    U111(4,1,1) -> 13*
    U111(2,1,2) -> 13*
    U111(4,3,1) -> 13*
    U111(2,3,2) -> 13*
    U111(4,2,3) -> 13*
    U111(4,4,3) -> 13*
    01() -> 21*
    tt1() -> 21*
    mark1(15) -> 15,7
    mark1(17) -> 17,8
    mark1(19) -> 19,9
    mark1(13) -> 13,6
    top2(24) -> 11*
    active0(2) -> 5*
    active0(4) -> 5*
    active0(1) -> 5*
    active0(3) -> 5*
    active2(21) -> 24*
    U110(3,4,2) -> 6*
    U110(1,1,4) -> 6*
    U110(1,3,4) -> 6*
    U110(4,1,2) -> 6*
    U110(1,1,1) -> 6*
    U110(4,3,2) -> 6*
    U110(1,3,1) -> 6*
    U110(2,2,4) -> 6*
    U110(2,4,4) -> 6*
    U110(1,2,3) -> 6*
    U110(1,4,3) -> 6*
    U110(2,2,1) -> 6*
    U110(2,4,1) -> 6*
    U110(3,1,4) -> 6*
    U110(3,3,4) -> 6*
    U110(2,1,3) -> 6*
    U110(2,3,3) -> 6*
    U110(3,1,1) -> 6*
    U110(1,1,2) -> 6*
    U110(3,3,1) -> 6*
    U110(1,3,2) -> 6*
    U110(4,2,4) -> 6*
    U110(4,4,4) -> 6*
    U110(3,2,3) -> 6*
    U110(3,4,3) -> 6*
    U110(4,2,1) -> 6*
    U110(2,2,2) -> 6*
    U110(4,4,1) -> 6*
    U110(2,4,2) -> 6*
    U110(4,1,3) -> 6*
    U110(4,3,3) -> 6*
    U110(3,1,2) -> 6*
    U110(3,3,2) -> 6*
    U110(1,2,4) -> 6*
    U110(1,4,4) -> 6*
    U110(4,2,2) -> 6*
    U110(1,2,1) -> 6*
    U110(4,4,2) -> 6*
    U110(1,4,1) -> 6*
    U110(2,1,4) -> 6*
    U110(2,3,4) -> 6*
    U110(1,1,3) -> 6*
    U110(1,3,3) -> 6*
    U110(2,1,1) -> 6*
    U110(2,3,1) -> 6*
    U110(3,2,4) -> 6*
    U110(3,4,4) -> 6*
    U110(2,2,3) -> 6*
    U110(2,4,3) -> 6*
    U110(3,2,1) -> 6*
    U110(1,2,2) -> 6*
    U110(3,4,1) -> 6*
    U110(1,4,2) -> 6*
    U110(4,1,4) -> 6*
    U110(4,3,4) -> 6*
    U110(3,1,3) -> 6*
    U110(3,3,3) -> 6*
    U110(4,1,1) -> 6*
    U110(2,1,2) -> 6*
    U110(4,3,1) -> 6*
    U110(2,3,2) -> 6*
    U110(4,2,3) -> 6*
    U110(4,4,3) -> 6*
    U110(3,2,2) -> 6*
    tt0() -> 1*
    mark0(2) -> 2*
    mark0(4) -> 2*
    mark0(1) -> 2*
    mark0(3) -> 2*
    U120(3,2,2) -> 7*
    U120(3,4,2) -> 7*
    U120(1,1,4) -> 7*
    U120(1,3,4) -> 7*
    U120(4,1,2) -> 7*
    U120(1,1,1) -> 7*
    U120(4,3,2) -> 7*
    U120(1,3,1) -> 7*
    U120(2,2,4) -> 7*
    U120(2,4,4) -> 7*
    U120(1,2,3) -> 7*
    U120(1,4,3) -> 7*
    U120(2,2,1) -> 7*
    U120(2,4,1) -> 7*
    U120(3,1,4) -> 7*
    U120(3,3,4) -> 7*
    U120(2,1,3) -> 7*
    U120(2,3,3) -> 7*
    U120(3,1,1) -> 7*
    U120(1,1,2) -> 7*
    U120(3,3,1) -> 7*
    U120(1,3,2) -> 7*
    U120(4,2,4) -> 7*
    U120(4,4,4) -> 7*
    U120(3,2,3) -> 7*
    U120(3,4,3) -> 7*
    U120(4,2,1) -> 7*
    U120(2,2,2) -> 7*
    U120(4,4,1) -> 7*
    U120(2,4,2) -> 7*
    U120(4,1,3) -> 7*
    U120(4,3,3) -> 7*
    U120(3,1,2) -> 7*
    U120(3,3,2) -> 7*
    U120(1,2,4) -> 7*
    U120(1,4,4) -> 7*
    U120(4,2,2) -> 7*
    U120(1,2,1) -> 7*
    U120(4,4,2) -> 7*
    U120(1,4,1) -> 7*
    U120(2,1,4) -> 7*
    U120(2,3,4) -> 7*
    U120(1,1,3) -> 7*
    U120(1,3,3) -> 7*
    U120(2,1,1) -> 7*
    U120(2,3,1) -> 7*
    U120(3,2,4) -> 7*
    U120(3,4,4) -> 7*
    U120(2,2,3) -> 7*
    U120(2,4,3) -> 7*
    U120(3,2,1) -> 7*
    U120(1,2,2) -> 7*
    U120(3,4,1) -> 7*
    U120(1,4,2) -> 7*
    U120(4,1,4) -> 7*
    U120(4,3,4) -> 7*
    U120(3,1,3) -> 7*
    U120(3,3,3) -> 7*
    U120(4,1,1) -> 7*
    U120(2,1,2) -> 7*
    U120(4,3,1) -> 7*
    U120(2,3,2) -> 7*
    U120(4,2,3) -> 7*
    U120(4,4,3) -> 7*
    s0(2) -> 8*
    s0(4) -> 8*
    s0(1) -> 8*
    s0(3) -> 8*
    plus0(3,1) -> 9*
    plus0(3,3) -> 9*
    plus0(4,2) -> 9*
    plus0(4,4) -> 9*
    plus0(1,2) -> 9*
    plus0(1,4) -> 9*
    plus0(2,1) -> 9*
    plus0(2,3) -> 9*
    plus0(3,2) -> 9*
    plus0(3,4) -> 9*
    plus0(4,1) -> 9*
    plus0(4,3) -> 9*
    plus0(1,1) -> 9*
    plus0(1,3) -> 9*
    plus0(2,2) -> 9*
    plus0(2,4) -> 9*
    00() -> 3*
    proper0(2) -> 10*
    proper0(4) -> 10*
    proper0(1) -> 10*
    proper0(3) -> 10*
    ok0(2) -> 4*
    ok0(4) -> 4*
    ok0(1) -> 4*
    ok0(3) -> 4*
    top0(2) -> 11*
    top0(4) -> 11*
    top0(1) -> 11*
    top0(3) -> 11*
  problem:

  Qed

Tool IRC1

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

stdout:

YES(?,O(n^1))

Tool IRC2

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 PEANO 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(U11(tt(), M, N)) -> mark(U12(tt(), M, N))
     , active(U12(tt(), M, N)) -> mark(s(plus(N, M)))
     , active(plus(N, 0())) -> mark(N)
     , active(plus(N, s(M))) -> mark(U11(tt(), M, N))
     , active(U11(X1, X2, X3)) -> U11(active(X1), X2, X3)
     , active(U12(X1, X2, X3)) -> U12(active(X1), X2, X3)
     , active(s(X)) -> s(active(X))
     , active(plus(X1, X2)) -> plus(active(X1), X2)
     , active(plus(X1, X2)) -> plus(X1, active(X2))
     , U11(mark(X1), X2, X3) -> mark(U11(X1, X2, X3))
     , U12(mark(X1), X2, X3) -> mark(U12(X1, X2, X3))
     , s(mark(X)) -> mark(s(X))
     , plus(mark(X1), X2) -> mark(plus(X1, X2))
     , plus(X1, mark(X2)) -> mark(plus(X1, X2))
     , proper(U11(X1, X2, X3)) ->
       U11(proper(X1), proper(X2), proper(X3))
     , proper(tt()) -> ok(tt())
     , proper(U12(X1, X2, X3)) ->
       U12(proper(X1), proper(X2), proper(X3))
     , proper(s(X)) -> s(proper(X))
     , proper(plus(X1, X2)) -> plus(proper(X1), proper(X2))
     , proper(0()) -> ok(0())
     , U11(ok(X1), ok(X2), ok(X3)) -> ok(U11(X1, X2, X3))
     , U12(ok(X1), ok(X2), ok(X3)) -> ok(U12(X1, X2, X3))
     , s(ok(X)) -> ok(s(X))
     , plus(ok(X1), ok(X2)) -> ok(plus(X1, X2))
     , 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(U11(tt(), M, N)) -> mark(U12(tt(), M, N))
          , active(U12(tt(), M, N)) -> mark(s(plus(N, M)))
          , active(plus(N, 0())) -> mark(N)
          , active(plus(N, s(M))) -> mark(U11(tt(), M, N))
          , active(U11(X1, X2, X3)) -> U11(active(X1), X2, X3)
          , active(U12(X1, X2, X3)) -> U12(active(X1), X2, X3)
          , active(s(X)) -> s(active(X))
          , active(plus(X1, X2)) -> plus(active(X1), X2)
          , active(plus(X1, X2)) -> plus(X1, active(X2))
          , U11(mark(X1), X2, X3) -> mark(U11(X1, X2, X3))
          , U12(mark(X1), X2, X3) -> mark(U12(X1, X2, X3))
          , s(mark(X)) -> mark(s(X))
          , plus(mark(X1), X2) -> mark(plus(X1, X2))
          , plus(X1, mark(X2)) -> mark(plus(X1, X2))
          , proper(U11(X1, X2, X3)) ->
            U11(proper(X1), proper(X2), proper(X3))
          , proper(tt()) -> ok(tt())
          , proper(U12(X1, X2, X3)) ->
            U12(proper(X1), proper(X2), proper(X3))
          , proper(s(X)) -> s(proper(X))
          , proper(plus(X1, X2)) -> plus(proper(X1), proper(X2))
          , proper(0()) -> ok(0())
          , U11(ok(X1), ok(X2), ok(X3)) -> ok(U11(X1, X2, X3))
          , U12(ok(X1), ok(X2), ok(X3)) -> ok(U12(X1, X2, X3))
          , s(ok(X)) -> ok(s(X))
          , plus(ok(X1), ok(X2)) -> ok(plus(X1, X2))
          , 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
        , U11_0(2, 2, 2) -> 1
        , U11_1(2, 2, 2) -> 3
        , tt_0() -> 2
        , tt_1() -> 4
        , mark_0(2) -> 2
        , mark_1(3) -> 1
        , mark_1(3) -> 3
        , U12_0(2, 2, 2) -> 1
        , U12_1(2, 2, 2) -> 3
        , s_0(2) -> 1
        , s_1(2) -> 3
        , plus_0(2, 2) -> 1
        , plus_1(2, 2) -> 3
        , 0_0() -> 2
        , 0_1() -> 4
        , 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 noand C

stdout:

YES(?,O(n^1))

Tool RC2

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 PEANO 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(U11(tt(), M, N)) -> mark(U12(tt(), M, N))
     , active(U12(tt(), M, N)) -> mark(s(plus(N, M)))
     , active(plus(N, 0())) -> mark(N)
     , active(plus(N, s(M))) -> mark(U11(tt(), M, N))
     , active(U11(X1, X2, X3)) -> U11(active(X1), X2, X3)
     , active(U12(X1, X2, X3)) -> U12(active(X1), X2, X3)
     , active(s(X)) -> s(active(X))
     , active(plus(X1, X2)) -> plus(active(X1), X2)
     , active(plus(X1, X2)) -> plus(X1, active(X2))
     , U11(mark(X1), X2, X3) -> mark(U11(X1, X2, X3))
     , U12(mark(X1), X2, X3) -> mark(U12(X1, X2, X3))
     , s(mark(X)) -> mark(s(X))
     , plus(mark(X1), X2) -> mark(plus(X1, X2))
     , plus(X1, mark(X2)) -> mark(plus(X1, X2))
     , proper(U11(X1, X2, X3)) ->
       U11(proper(X1), proper(X2), proper(X3))
     , proper(tt()) -> ok(tt())
     , proper(U12(X1, X2, X3)) ->
       U12(proper(X1), proper(X2), proper(X3))
     , proper(s(X)) -> s(proper(X))
     , proper(plus(X1, X2)) -> plus(proper(X1), proper(X2))
     , proper(0()) -> ok(0())
     , U11(ok(X1), ok(X2), ok(X3)) -> ok(U11(X1, X2, X3))
     , U12(ok(X1), ok(X2), ok(X3)) -> ok(U12(X1, X2, X3))
     , s(ok(X)) -> ok(s(X))
     , plus(ok(X1), ok(X2)) -> ok(plus(X1, X2))
     , 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(U11(tt(), M, N)) -> mark(U12(tt(), M, N))
          , active(U12(tt(), M, N)) -> mark(s(plus(N, M)))
          , active(plus(N, 0())) -> mark(N)
          , active(plus(N, s(M))) -> mark(U11(tt(), M, N))
          , active(U11(X1, X2, X3)) -> U11(active(X1), X2, X3)
          , active(U12(X1, X2, X3)) -> U12(active(X1), X2, X3)
          , active(s(X)) -> s(active(X))
          , active(plus(X1, X2)) -> plus(active(X1), X2)
          , active(plus(X1, X2)) -> plus(X1, active(X2))
          , U11(mark(X1), X2, X3) -> mark(U11(X1, X2, X3))
          , U12(mark(X1), X2, X3) -> mark(U12(X1, X2, X3))
          , s(mark(X)) -> mark(s(X))
          , plus(mark(X1), X2) -> mark(plus(X1, X2))
          , plus(X1, mark(X2)) -> mark(plus(X1, X2))
          , proper(U11(X1, X2, X3)) ->
            U11(proper(X1), proper(X2), proper(X3))
          , proper(tt()) -> ok(tt())
          , proper(U12(X1, X2, X3)) ->
            U12(proper(X1), proper(X2), proper(X3))
          , proper(s(X)) -> s(proper(X))
          , proper(plus(X1, X2)) -> plus(proper(X1), proper(X2))
          , proper(0()) -> ok(0())
          , U11(ok(X1), ok(X2), ok(X3)) -> ok(U11(X1, X2, X3))
          , U12(ok(X1), ok(X2), ok(X3)) -> ok(U12(X1, X2, X3))
          , s(ok(X)) -> ok(s(X))
          , plus(ok(X1), ok(X2)) -> ok(plus(X1, X2))
          , 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
        , U11_0(2, 2, 2) -> 1
        , U11_1(2, 2, 2) -> 3
        , tt_0() -> 2
        , tt_1() -> 4
        , mark_0(2) -> 2
        , mark_1(3) -> 1
        , mark_1(3) -> 3
        , U12_0(2, 2, 2) -> 1
        , U12_1(2, 2, 2) -> 3
        , s_0(2) -> 1
        , s_1(2) -> 3
        , plus_0(2, 2) -> 1
        , plus_1(2, 2) -> 3
        , 0_0() -> 2
        , 0_1() -> 4
        , 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}