MAYBE

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict Trs:
  { a__U11(X1, X2) -> U11(X1, X2)
  , a__U11(tt(), N) -> mark(N)
  , mark(tt()) -> tt()
  , mark(s(X)) -> s(mark(X))
  , mark(0()) -> 0()
  , mark(plus(X1, X2)) -> a__plus(mark(X1), mark(X2))
  , mark(isNat(X)) -> a__isNat(X)
  , mark(U11(X1, X2)) -> a__U11(mark(X1), X2)
  , mark(U21(X1, X2, X3)) -> a__U21(mark(X1), X2, X3)
  , mark(and(X1, X2)) -> a__and(mark(X1), X2)
  , a__U21(X1, X2, X3) -> U21(X1, X2, X3)
  , a__U21(tt(), M, N) -> s(a__plus(mark(N), mark(M)))
  , a__plus(X1, X2) -> plus(X1, X2)
  , a__plus(N, s(M)) -> a__U21(a__and(a__isNat(M), isNat(N)), M, N)
  , a__plus(N, 0()) -> a__U11(a__isNat(N), N)
  , a__and(X1, X2) -> and(X1, X2)
  , a__and(tt(), X) -> mark(X)
  , a__isNat(X) -> isNat(X)
  , a__isNat(s(V1)) -> a__isNat(V1)
  , a__isNat(0()) -> tt()
  , a__isNat(plus(V1, V2)) -> a__and(a__isNat(V1), isNat(V2)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We add following dependency tuples:

Strict DPs:
  { a__U11^#(X1, X2) -> c_1()
  , a__U11^#(tt(), N) -> c_2(mark^#(N))
  , mark^#(tt()) -> c_3()
  , mark^#(s(X)) -> c_4(mark^#(X))
  , mark^#(0()) -> c_5()
  , mark^#(plus(X1, X2)) ->
    c_6(a__plus^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(isNat(X)) -> c_7(a__isNat^#(X))
  , mark^#(U11(X1, X2)) -> c_8(a__U11^#(mark(X1), X2), mark^#(X1))
  , mark^#(U21(X1, X2, X3)) ->
    c_9(a__U21^#(mark(X1), X2, X3), mark^#(X1))
  , mark^#(and(X1, X2)) -> c_10(a__and^#(mark(X1), X2), mark^#(X1))
  , a__plus^#(X1, X2) -> c_13()
  , a__plus^#(N, s(M)) ->
    c_14(a__U21^#(a__and(a__isNat(M), isNat(N)), M, N),
         a__and^#(a__isNat(M), isNat(N)),
         a__isNat^#(M))
  , a__plus^#(N, 0()) ->
    c_15(a__U11^#(a__isNat(N), N), a__isNat^#(N))
  , a__isNat^#(X) -> c_18()
  , a__isNat^#(s(V1)) -> c_19(a__isNat^#(V1))
  , a__isNat^#(0()) -> c_20()
  , a__isNat^#(plus(V1, V2)) ->
    c_21(a__and^#(a__isNat(V1), isNat(V2)), a__isNat^#(V1))
  , a__U21^#(X1, X2, X3) -> c_11()
  , a__U21^#(tt(), M, N) ->
    c_12(a__plus^#(mark(N), mark(M)), mark^#(N), mark^#(M))
  , a__and^#(X1, X2) -> c_16()
  , a__and^#(tt(), X) -> c_17(mark^#(X)) }

and mark the set of starting terms.

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict DPs:
  { a__U11^#(X1, X2) -> c_1()
  , a__U11^#(tt(), N) -> c_2(mark^#(N))
  , mark^#(tt()) -> c_3()
  , mark^#(s(X)) -> c_4(mark^#(X))
  , mark^#(0()) -> c_5()
  , mark^#(plus(X1, X2)) ->
    c_6(a__plus^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(isNat(X)) -> c_7(a__isNat^#(X))
  , mark^#(U11(X1, X2)) -> c_8(a__U11^#(mark(X1), X2), mark^#(X1))
  , mark^#(U21(X1, X2, X3)) ->
    c_9(a__U21^#(mark(X1), X2, X3), mark^#(X1))
  , mark^#(and(X1, X2)) -> c_10(a__and^#(mark(X1), X2), mark^#(X1))
  , a__plus^#(X1, X2) -> c_13()
  , a__plus^#(N, s(M)) ->
    c_14(a__U21^#(a__and(a__isNat(M), isNat(N)), M, N),
         a__and^#(a__isNat(M), isNat(N)),
         a__isNat^#(M))
  , a__plus^#(N, 0()) ->
    c_15(a__U11^#(a__isNat(N), N), a__isNat^#(N))
  , a__isNat^#(X) -> c_18()
  , a__isNat^#(s(V1)) -> c_19(a__isNat^#(V1))
  , a__isNat^#(0()) -> c_20()
  , a__isNat^#(plus(V1, V2)) ->
    c_21(a__and^#(a__isNat(V1), isNat(V2)), a__isNat^#(V1))
  , a__U21^#(X1, X2, X3) -> c_11()
  , a__U21^#(tt(), M, N) ->
    c_12(a__plus^#(mark(N), mark(M)), mark^#(N), mark^#(M))
  , a__and^#(X1, X2) -> c_16()
  , a__and^#(tt(), X) -> c_17(mark^#(X)) }
Weak Trs:
  { a__U11(X1, X2) -> U11(X1, X2)
  , a__U11(tt(), N) -> mark(N)
  , mark(tt()) -> tt()
  , mark(s(X)) -> s(mark(X))
  , mark(0()) -> 0()
  , mark(plus(X1, X2)) -> a__plus(mark(X1), mark(X2))
  , mark(isNat(X)) -> a__isNat(X)
  , mark(U11(X1, X2)) -> a__U11(mark(X1), X2)
  , mark(U21(X1, X2, X3)) -> a__U21(mark(X1), X2, X3)
  , mark(and(X1, X2)) -> a__and(mark(X1), X2)
  , a__U21(X1, X2, X3) -> U21(X1, X2, X3)
  , a__U21(tt(), M, N) -> s(a__plus(mark(N), mark(M)))
  , a__plus(X1, X2) -> plus(X1, X2)
  , a__plus(N, s(M)) -> a__U21(a__and(a__isNat(M), isNat(N)), M, N)
  , a__plus(N, 0()) -> a__U11(a__isNat(N), N)
  , a__and(X1, X2) -> and(X1, X2)
  , a__and(tt(), X) -> mark(X)
  , a__isNat(X) -> isNat(X)
  , a__isNat(s(V1)) -> a__isNat(V1)
  , a__isNat(0()) -> tt()
  , a__isNat(plus(V1, V2)) -> a__and(a__isNat(V1), isNat(V2)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

We estimate the number of application of {1,3,5,11,14,16,18,20} by
applications of Pre({1,3,5,11,14,16,18,20}) =
{2,4,6,7,8,9,10,12,13,15,17,19,21}. Here rules are labeled as
follows:

  DPs:
    { 1: a__U11^#(X1, X2) -> c_1()
    , 2: a__U11^#(tt(), N) -> c_2(mark^#(N))
    , 3: mark^#(tt()) -> c_3()
    , 4: mark^#(s(X)) -> c_4(mark^#(X))
    , 5: mark^#(0()) -> c_5()
    , 6: mark^#(plus(X1, X2)) ->
         c_6(a__plus^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
    , 7: mark^#(isNat(X)) -> c_7(a__isNat^#(X))
    , 8: mark^#(U11(X1, X2)) -> c_8(a__U11^#(mark(X1), X2), mark^#(X1))
    , 9: mark^#(U21(X1, X2, X3)) ->
         c_9(a__U21^#(mark(X1), X2, X3), mark^#(X1))
    , 10: mark^#(and(X1, X2)) ->
          c_10(a__and^#(mark(X1), X2), mark^#(X1))
    , 11: a__plus^#(X1, X2) -> c_13()
    , 12: a__plus^#(N, s(M)) ->
          c_14(a__U21^#(a__and(a__isNat(M), isNat(N)), M, N),
               a__and^#(a__isNat(M), isNat(N)),
               a__isNat^#(M))
    , 13: a__plus^#(N, 0()) ->
          c_15(a__U11^#(a__isNat(N), N), a__isNat^#(N))
    , 14: a__isNat^#(X) -> c_18()
    , 15: a__isNat^#(s(V1)) -> c_19(a__isNat^#(V1))
    , 16: a__isNat^#(0()) -> c_20()
    , 17: a__isNat^#(plus(V1, V2)) ->
          c_21(a__and^#(a__isNat(V1), isNat(V2)), a__isNat^#(V1))
    , 18: a__U21^#(X1, X2, X3) -> c_11()
    , 19: a__U21^#(tt(), M, N) ->
          c_12(a__plus^#(mark(N), mark(M)), mark^#(N), mark^#(M))
    , 20: a__and^#(X1, X2) -> c_16()
    , 21: a__and^#(tt(), X) -> c_17(mark^#(X)) }

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict DPs:
  { a__U11^#(tt(), N) -> c_2(mark^#(N))
  , mark^#(s(X)) -> c_4(mark^#(X))
  , mark^#(plus(X1, X2)) ->
    c_6(a__plus^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(isNat(X)) -> c_7(a__isNat^#(X))
  , mark^#(U11(X1, X2)) -> c_8(a__U11^#(mark(X1), X2), mark^#(X1))
  , mark^#(U21(X1, X2, X3)) ->
    c_9(a__U21^#(mark(X1), X2, X3), mark^#(X1))
  , mark^#(and(X1, X2)) -> c_10(a__and^#(mark(X1), X2), mark^#(X1))
  , a__plus^#(N, s(M)) ->
    c_14(a__U21^#(a__and(a__isNat(M), isNat(N)), M, N),
         a__and^#(a__isNat(M), isNat(N)),
         a__isNat^#(M))
  , a__plus^#(N, 0()) ->
    c_15(a__U11^#(a__isNat(N), N), a__isNat^#(N))
  , a__isNat^#(s(V1)) -> c_19(a__isNat^#(V1))
  , a__isNat^#(plus(V1, V2)) ->
    c_21(a__and^#(a__isNat(V1), isNat(V2)), a__isNat^#(V1))
  , a__U21^#(tt(), M, N) ->
    c_12(a__plus^#(mark(N), mark(M)), mark^#(N), mark^#(M))
  , a__and^#(tt(), X) -> c_17(mark^#(X)) }
Weak DPs:
  { a__U11^#(X1, X2) -> c_1()
  , mark^#(tt()) -> c_3()
  , mark^#(0()) -> c_5()
  , a__plus^#(X1, X2) -> c_13()
  , a__isNat^#(X) -> c_18()
  , a__isNat^#(0()) -> c_20()
  , a__U21^#(X1, X2, X3) -> c_11()
  , a__and^#(X1, X2) -> c_16() }
Weak Trs:
  { a__U11(X1, X2) -> U11(X1, X2)
  , a__U11(tt(), N) -> mark(N)
  , mark(tt()) -> tt()
  , mark(s(X)) -> s(mark(X))
  , mark(0()) -> 0()
  , mark(plus(X1, X2)) -> a__plus(mark(X1), mark(X2))
  , mark(isNat(X)) -> a__isNat(X)
  , mark(U11(X1, X2)) -> a__U11(mark(X1), X2)
  , mark(U21(X1, X2, X3)) -> a__U21(mark(X1), X2, X3)
  , mark(and(X1, X2)) -> a__and(mark(X1), X2)
  , a__U21(X1, X2, X3) -> U21(X1, X2, X3)
  , a__U21(tt(), M, N) -> s(a__plus(mark(N), mark(M)))
  , a__plus(X1, X2) -> plus(X1, X2)
  , a__plus(N, s(M)) -> a__U21(a__and(a__isNat(M), isNat(N)), M, N)
  , a__plus(N, 0()) -> a__U11(a__isNat(N), N)
  , a__and(X1, X2) -> and(X1, X2)
  , a__and(tt(), X) -> mark(X)
  , a__isNat(X) -> isNat(X)
  , a__isNat(s(V1)) -> a__isNat(V1)
  , a__isNat(0()) -> tt()
  , a__isNat(plus(V1, V2)) -> a__and(a__isNat(V1), isNat(V2)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.

{ a__U11^#(X1, X2) -> c_1()
, mark^#(tt()) -> c_3()
, mark^#(0()) -> c_5()
, a__plus^#(X1, X2) -> c_13()
, a__isNat^#(X) -> c_18()
, a__isNat^#(0()) -> c_20()
, a__U21^#(X1, X2, X3) -> c_11()
, a__and^#(X1, X2) -> c_16() }

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict DPs:
  { a__U11^#(tt(), N) -> c_2(mark^#(N))
  , mark^#(s(X)) -> c_4(mark^#(X))
  , mark^#(plus(X1, X2)) ->
    c_6(a__plus^#(mark(X1), mark(X2)), mark^#(X1), mark^#(X2))
  , mark^#(isNat(X)) -> c_7(a__isNat^#(X))
  , mark^#(U11(X1, X2)) -> c_8(a__U11^#(mark(X1), X2), mark^#(X1))
  , mark^#(U21(X1, X2, X3)) ->
    c_9(a__U21^#(mark(X1), X2, X3), mark^#(X1))
  , mark^#(and(X1, X2)) -> c_10(a__and^#(mark(X1), X2), mark^#(X1))
  , a__plus^#(N, s(M)) ->
    c_14(a__U21^#(a__and(a__isNat(M), isNat(N)), M, N),
         a__and^#(a__isNat(M), isNat(N)),
         a__isNat^#(M))
  , a__plus^#(N, 0()) ->
    c_15(a__U11^#(a__isNat(N), N), a__isNat^#(N))
  , a__isNat^#(s(V1)) -> c_19(a__isNat^#(V1))
  , a__isNat^#(plus(V1, V2)) ->
    c_21(a__and^#(a__isNat(V1), isNat(V2)), a__isNat^#(V1))
  , a__U21^#(tt(), M, N) ->
    c_12(a__plus^#(mark(N), mark(M)), mark^#(N), mark^#(M))
  , a__and^#(tt(), X) -> c_17(mark^#(X)) }
Weak Trs:
  { a__U11(X1, X2) -> U11(X1, X2)
  , a__U11(tt(), N) -> mark(N)
  , mark(tt()) -> tt()
  , mark(s(X)) -> s(mark(X))
  , mark(0()) -> 0()
  , mark(plus(X1, X2)) -> a__plus(mark(X1), mark(X2))
  , mark(isNat(X)) -> a__isNat(X)
  , mark(U11(X1, X2)) -> a__U11(mark(X1), X2)
  , mark(U21(X1, X2, X3)) -> a__U21(mark(X1), X2, X3)
  , mark(and(X1, X2)) -> a__and(mark(X1), X2)
  , a__U21(X1, X2, X3) -> U21(X1, X2, X3)
  , a__U21(tt(), M, N) -> s(a__plus(mark(N), mark(M)))
  , a__plus(X1, X2) -> plus(X1, X2)
  , a__plus(N, s(M)) -> a__U21(a__and(a__isNat(M), isNat(N)), M, N)
  , a__plus(N, 0()) -> a__U11(a__isNat(N), N)
  , a__and(X1, X2) -> and(X1, X2)
  , a__and(tt(), X) -> mark(X)
  , a__isNat(X) -> isNat(X)
  , a__isNat(s(V1)) -> a__isNat(V1)
  , a__isNat(0()) -> tt()
  , a__isNat(plus(V1, V2)) -> a__and(a__isNat(V1), isNat(V2)) }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

The input cannot be shown compatible

Arrrr..