MAYBE

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

Strict Trs:
  { U11(tt(), N) -> activate(N)
  , activate(X) -> X
  , activate(n__0()) -> 0()
  , activate(n__plus(X1, X2)) -> plus(activate(X1), activate(X2))
  , activate(n__isNat(X)) -> isNat(X)
  , activate(n__s(X)) -> s(activate(X))
  , activate(n__x(X1, X2)) -> x(activate(X1), activate(X2))
  , U21(tt(), M, N) -> s(plus(activate(N), activate(M)))
  , s(X) -> n__s(X)
  , plus(X1, X2) -> n__plus(X1, X2)
  , plus(N, s(M)) -> U21(and(isNat(M), n__isNat(N)), M, N)
  , plus(N, 0()) -> U11(isNat(N), N)
  , U31(tt()) -> 0()
  , 0() -> n__0()
  , U41(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N))
  , x(X1, X2) -> n__x(X1, X2)
  , x(N, s(M)) -> U41(and(isNat(M), n__isNat(N)), M, N)
  , x(N, 0()) -> U31(isNat(N))
  , and(tt(), X) -> activate(X)
  , isNat(X) -> n__isNat(X)
  , isNat(n__0()) -> tt()
  , isNat(n__plus(V1, V2)) ->
    and(isNat(activate(V1)), n__isNat(activate(V2)))
  , isNat(n__s(V1)) -> isNat(activate(V1))
  , isNat(n__x(V1, V2)) ->
    and(isNat(activate(V1)), n__isNat(activate(V2))) }
Obligation:
  runtime complexity
Answer:
  MAYBE

None of the processors succeeded.

Details of failed attempt(s):
-----------------------------
1) 'WithProblem (timeout of 60 seconds)' failed due to the
   following reason:
   
   Computation stopped due to timeout after 60.0 seconds.

2) 'Best' failed due to the following reason:
   
   None of the processors succeeded.
   
   Details of failed attempt(s):
   -----------------------------
   1) 'WithProblem (timeout of 30 seconds) (timeout of 60 seconds)'
      failed due to the following reason:
      
      Computation stopped due to timeout after 30.0 seconds.
   
   2) 'Best' failed due to the following reason:
      
      None of the processors succeeded.
      
      Details of failed attempt(s):
      -----------------------------
      1) 'bsearch-popstar (timeout of 60 seconds)' failed due to the
         following reason:
         
         The processor is inapplicable, reason:
           Processor only applicable for innermost runtime complexity analysis
      
      2) 'Polynomial Path Order (PS) (timeout of 60 seconds)' failed due
         to the following reason:
         
         The processor is inapplicable, reason:
           Processor only applicable for innermost runtime complexity analysis
      
   
   3) 'Fastest (timeout of 5 seconds) (timeout of 60 seconds)' failed
      due to the following reason:
      
      None of the processors succeeded.
      
      Details of failed attempt(s):
      -----------------------------
      1) 'Bounds with perSymbol-enrichment and initial automaton 'match''
         failed due to the following reason:
         
         match-boundness of the problem could not be verified.
      
      2) 'Bounds with minimal-enrichment and initial automaton 'match''
         failed due to the following reason:
         
         match-boundness of the problem could not be verified.
      
   

3) 'Innermost Weak Dependency Pairs (timeout of 60 seconds)' failed
   due to the following reason:
   
   We add the following weak dependency pairs:
   
   Strict DPs:
     { U11^#(tt(), N) -> c_1(activate^#(N))
     , activate^#(X) -> c_2(X)
     , activate^#(n__0()) -> c_3(0^#())
     , activate^#(n__plus(X1, X2)) ->
       c_4(plus^#(activate(X1), activate(X2)))
     , activate^#(n__isNat(X)) -> c_5(isNat^#(X))
     , activate^#(n__s(X)) -> c_6(s^#(activate(X)))
     , activate^#(n__x(X1, X2)) -> c_7(x^#(activate(X1), activate(X2)))
     , 0^#() -> c_14()
     , plus^#(X1, X2) -> c_10(X1, X2)
     , plus^#(N, s(M)) -> c_11(U21^#(and(isNat(M), n__isNat(N)), M, N))
     , plus^#(N, 0()) -> c_12(U11^#(isNat(N), N))
     , isNat^#(X) -> c_20(X)
     , isNat^#(n__0()) -> c_21()
     , isNat^#(n__plus(V1, V2)) ->
       c_22(and^#(isNat(activate(V1)), n__isNat(activate(V2))))
     , isNat^#(n__s(V1)) -> c_23(isNat^#(activate(V1)))
     , isNat^#(n__x(V1, V2)) ->
       c_24(and^#(isNat(activate(V1)), n__isNat(activate(V2))))
     , s^#(X) -> c_9(X)
     , x^#(X1, X2) -> c_16(X1, X2)
     , x^#(N, s(M)) -> c_17(U41^#(and(isNat(M), n__isNat(N)), M, N))
     , x^#(N, 0()) -> c_18(U31^#(isNat(N)))
     , U21^#(tt(), M, N) -> c_8(s^#(plus(activate(N), activate(M))))
     , U31^#(tt()) -> c_13(0^#())
     , U41^#(tt(), M, N) ->
       c_15(plus^#(x(activate(N), activate(M)), activate(N)))
     , and^#(tt(), X) -> c_19(activate^#(X)) }
   
   and mark the set of starting terms.
   
   We are left with following problem, upon which TcT provides the
   certificate MAYBE.
   
   Strict DPs:
     { U11^#(tt(), N) -> c_1(activate^#(N))
     , activate^#(X) -> c_2(X)
     , activate^#(n__0()) -> c_3(0^#())
     , activate^#(n__plus(X1, X2)) ->
       c_4(plus^#(activate(X1), activate(X2)))
     , activate^#(n__isNat(X)) -> c_5(isNat^#(X))
     , activate^#(n__s(X)) -> c_6(s^#(activate(X)))
     , activate^#(n__x(X1, X2)) -> c_7(x^#(activate(X1), activate(X2)))
     , 0^#() -> c_14()
     , plus^#(X1, X2) -> c_10(X1, X2)
     , plus^#(N, s(M)) -> c_11(U21^#(and(isNat(M), n__isNat(N)), M, N))
     , plus^#(N, 0()) -> c_12(U11^#(isNat(N), N))
     , isNat^#(X) -> c_20(X)
     , isNat^#(n__0()) -> c_21()
     , isNat^#(n__plus(V1, V2)) ->
       c_22(and^#(isNat(activate(V1)), n__isNat(activate(V2))))
     , isNat^#(n__s(V1)) -> c_23(isNat^#(activate(V1)))
     , isNat^#(n__x(V1, V2)) ->
       c_24(and^#(isNat(activate(V1)), n__isNat(activate(V2))))
     , s^#(X) -> c_9(X)
     , x^#(X1, X2) -> c_16(X1, X2)
     , x^#(N, s(M)) -> c_17(U41^#(and(isNat(M), n__isNat(N)), M, N))
     , x^#(N, 0()) -> c_18(U31^#(isNat(N)))
     , U21^#(tt(), M, N) -> c_8(s^#(plus(activate(N), activate(M))))
     , U31^#(tt()) -> c_13(0^#())
     , U41^#(tt(), M, N) ->
       c_15(plus^#(x(activate(N), activate(M)), activate(N)))
     , and^#(tt(), X) -> c_19(activate^#(X)) }
   Strict Trs:
     { U11(tt(), N) -> activate(N)
     , activate(X) -> X
     , activate(n__0()) -> 0()
     , activate(n__plus(X1, X2)) -> plus(activate(X1), activate(X2))
     , activate(n__isNat(X)) -> isNat(X)
     , activate(n__s(X)) -> s(activate(X))
     , activate(n__x(X1, X2)) -> x(activate(X1), activate(X2))
     , U21(tt(), M, N) -> s(plus(activate(N), activate(M)))
     , s(X) -> n__s(X)
     , plus(X1, X2) -> n__plus(X1, X2)
     , plus(N, s(M)) -> U21(and(isNat(M), n__isNat(N)), M, N)
     , plus(N, 0()) -> U11(isNat(N), N)
     , U31(tt()) -> 0()
     , 0() -> n__0()
     , U41(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N))
     , x(X1, X2) -> n__x(X1, X2)
     , x(N, s(M)) -> U41(and(isNat(M), n__isNat(N)), M, N)
     , x(N, 0()) -> U31(isNat(N))
     , and(tt(), X) -> activate(X)
     , isNat(X) -> n__isNat(X)
     , isNat(n__0()) -> tt()
     , isNat(n__plus(V1, V2)) ->
       and(isNat(activate(V1)), n__isNat(activate(V2)))
     , isNat(n__s(V1)) -> isNat(activate(V1))
     , isNat(n__x(V1, V2)) ->
       and(isNat(activate(V1)), n__isNat(activate(V2))) }
   Obligation:
     runtime complexity
   Answer:
     MAYBE
   
   We estimate the number of application of {8,13} by applications of
   Pre({8,13}) = {2,3,5,9,12,15,17,18,22}. Here rules are labeled as
   follows:
   
     DPs:
       { 1: U11^#(tt(), N) -> c_1(activate^#(N))
       , 2: activate^#(X) -> c_2(X)
       , 3: activate^#(n__0()) -> c_3(0^#())
       , 4: activate^#(n__plus(X1, X2)) ->
            c_4(plus^#(activate(X1), activate(X2)))
       , 5: activate^#(n__isNat(X)) -> c_5(isNat^#(X))
       , 6: activate^#(n__s(X)) -> c_6(s^#(activate(X)))
       , 7: activate^#(n__x(X1, X2)) ->
            c_7(x^#(activate(X1), activate(X2)))
       , 8: 0^#() -> c_14()
       , 9: plus^#(X1, X2) -> c_10(X1, X2)
       , 10: plus^#(N, s(M)) ->
             c_11(U21^#(and(isNat(M), n__isNat(N)), M, N))
       , 11: plus^#(N, 0()) -> c_12(U11^#(isNat(N), N))
       , 12: isNat^#(X) -> c_20(X)
       , 13: isNat^#(n__0()) -> c_21()
       , 14: isNat^#(n__plus(V1, V2)) ->
             c_22(and^#(isNat(activate(V1)), n__isNat(activate(V2))))
       , 15: isNat^#(n__s(V1)) -> c_23(isNat^#(activate(V1)))
       , 16: isNat^#(n__x(V1, V2)) ->
             c_24(and^#(isNat(activate(V1)), n__isNat(activate(V2))))
       , 17: s^#(X) -> c_9(X)
       , 18: x^#(X1, X2) -> c_16(X1, X2)
       , 19: x^#(N, s(M)) -> c_17(U41^#(and(isNat(M), n__isNat(N)), M, N))
       , 20: x^#(N, 0()) -> c_18(U31^#(isNat(N)))
       , 21: U21^#(tt(), M, N) -> c_8(s^#(plus(activate(N), activate(M))))
       , 22: U31^#(tt()) -> c_13(0^#())
       , 23: U41^#(tt(), M, N) ->
             c_15(plus^#(x(activate(N), activate(M)), activate(N)))
       , 24: and^#(tt(), X) -> c_19(activate^#(X)) }
   
   We are left with following problem, upon which TcT provides the
   certificate MAYBE.
   
   Strict DPs:
     { U11^#(tt(), N) -> c_1(activate^#(N))
     , activate^#(X) -> c_2(X)
     , activate^#(n__0()) -> c_3(0^#())
     , activate^#(n__plus(X1, X2)) ->
       c_4(plus^#(activate(X1), activate(X2)))
     , activate^#(n__isNat(X)) -> c_5(isNat^#(X))
     , activate^#(n__s(X)) -> c_6(s^#(activate(X)))
     , activate^#(n__x(X1, X2)) -> c_7(x^#(activate(X1), activate(X2)))
     , plus^#(X1, X2) -> c_10(X1, X2)
     , plus^#(N, s(M)) -> c_11(U21^#(and(isNat(M), n__isNat(N)), M, N))
     , plus^#(N, 0()) -> c_12(U11^#(isNat(N), N))
     , isNat^#(X) -> c_20(X)
     , isNat^#(n__plus(V1, V2)) ->
       c_22(and^#(isNat(activate(V1)), n__isNat(activate(V2))))
     , isNat^#(n__s(V1)) -> c_23(isNat^#(activate(V1)))
     , isNat^#(n__x(V1, V2)) ->
       c_24(and^#(isNat(activate(V1)), n__isNat(activate(V2))))
     , s^#(X) -> c_9(X)
     , x^#(X1, X2) -> c_16(X1, X2)
     , x^#(N, s(M)) -> c_17(U41^#(and(isNat(M), n__isNat(N)), M, N))
     , x^#(N, 0()) -> c_18(U31^#(isNat(N)))
     , U21^#(tt(), M, N) -> c_8(s^#(plus(activate(N), activate(M))))
     , U31^#(tt()) -> c_13(0^#())
     , U41^#(tt(), M, N) ->
       c_15(plus^#(x(activate(N), activate(M)), activate(N)))
     , and^#(tt(), X) -> c_19(activate^#(X)) }
   Strict Trs:
     { U11(tt(), N) -> activate(N)
     , activate(X) -> X
     , activate(n__0()) -> 0()
     , activate(n__plus(X1, X2)) -> plus(activate(X1), activate(X2))
     , activate(n__isNat(X)) -> isNat(X)
     , activate(n__s(X)) -> s(activate(X))
     , activate(n__x(X1, X2)) -> x(activate(X1), activate(X2))
     , U21(tt(), M, N) -> s(plus(activate(N), activate(M)))
     , s(X) -> n__s(X)
     , plus(X1, X2) -> n__plus(X1, X2)
     , plus(N, s(M)) -> U21(and(isNat(M), n__isNat(N)), M, N)
     , plus(N, 0()) -> U11(isNat(N), N)
     , U31(tt()) -> 0()
     , 0() -> n__0()
     , U41(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N))
     , x(X1, X2) -> n__x(X1, X2)
     , x(N, s(M)) -> U41(and(isNat(M), n__isNat(N)), M, N)
     , x(N, 0()) -> U31(isNat(N))
     , and(tt(), X) -> activate(X)
     , isNat(X) -> n__isNat(X)
     , isNat(n__0()) -> tt()
     , isNat(n__plus(V1, V2)) ->
       and(isNat(activate(V1)), n__isNat(activate(V2)))
     , isNat(n__s(V1)) -> isNat(activate(V1))
     , isNat(n__x(V1, V2)) ->
       and(isNat(activate(V1)), n__isNat(activate(V2))) }
   Weak DPs:
     { 0^#() -> c_14()
     , isNat^#(n__0()) -> c_21() }
   Obligation:
     runtime complexity
   Answer:
     MAYBE
   
   We estimate the number of application of {3,20} by applications of
   Pre({3,20}) = {1,2,8,11,15,16,18,22}. Here rules are labeled as
   follows:
   
     DPs:
       { 1: U11^#(tt(), N) -> c_1(activate^#(N))
       , 2: activate^#(X) -> c_2(X)
       , 3: activate^#(n__0()) -> c_3(0^#())
       , 4: activate^#(n__plus(X1, X2)) ->
            c_4(plus^#(activate(X1), activate(X2)))
       , 5: activate^#(n__isNat(X)) -> c_5(isNat^#(X))
       , 6: activate^#(n__s(X)) -> c_6(s^#(activate(X)))
       , 7: activate^#(n__x(X1, X2)) ->
            c_7(x^#(activate(X1), activate(X2)))
       , 8: plus^#(X1, X2) -> c_10(X1, X2)
       , 9: plus^#(N, s(M)) ->
            c_11(U21^#(and(isNat(M), n__isNat(N)), M, N))
       , 10: plus^#(N, 0()) -> c_12(U11^#(isNat(N), N))
       , 11: isNat^#(X) -> c_20(X)
       , 12: isNat^#(n__plus(V1, V2)) ->
             c_22(and^#(isNat(activate(V1)), n__isNat(activate(V2))))
       , 13: isNat^#(n__s(V1)) -> c_23(isNat^#(activate(V1)))
       , 14: isNat^#(n__x(V1, V2)) ->
             c_24(and^#(isNat(activate(V1)), n__isNat(activate(V2))))
       , 15: s^#(X) -> c_9(X)
       , 16: x^#(X1, X2) -> c_16(X1, X2)
       , 17: x^#(N, s(M)) -> c_17(U41^#(and(isNat(M), n__isNat(N)), M, N))
       , 18: x^#(N, 0()) -> c_18(U31^#(isNat(N)))
       , 19: U21^#(tt(), M, N) -> c_8(s^#(plus(activate(N), activate(M))))
       , 20: U31^#(tt()) -> c_13(0^#())
       , 21: U41^#(tt(), M, N) ->
             c_15(plus^#(x(activate(N), activate(M)), activate(N)))
       , 22: and^#(tt(), X) -> c_19(activate^#(X))
       , 23: 0^#() -> c_14()
       , 24: isNat^#(n__0()) -> c_21() }
   
   We are left with following problem, upon which TcT provides the
   certificate MAYBE.
   
   Strict DPs:
     { U11^#(tt(), N) -> c_1(activate^#(N))
     , activate^#(X) -> c_2(X)
     , activate^#(n__plus(X1, X2)) ->
       c_4(plus^#(activate(X1), activate(X2)))
     , activate^#(n__isNat(X)) -> c_5(isNat^#(X))
     , activate^#(n__s(X)) -> c_6(s^#(activate(X)))
     , activate^#(n__x(X1, X2)) -> c_7(x^#(activate(X1), activate(X2)))
     , plus^#(X1, X2) -> c_10(X1, X2)
     , plus^#(N, s(M)) -> c_11(U21^#(and(isNat(M), n__isNat(N)), M, N))
     , plus^#(N, 0()) -> c_12(U11^#(isNat(N), N))
     , isNat^#(X) -> c_20(X)
     , isNat^#(n__plus(V1, V2)) ->
       c_22(and^#(isNat(activate(V1)), n__isNat(activate(V2))))
     , isNat^#(n__s(V1)) -> c_23(isNat^#(activate(V1)))
     , isNat^#(n__x(V1, V2)) ->
       c_24(and^#(isNat(activate(V1)), n__isNat(activate(V2))))
     , s^#(X) -> c_9(X)
     , x^#(X1, X2) -> c_16(X1, X2)
     , x^#(N, s(M)) -> c_17(U41^#(and(isNat(M), n__isNat(N)), M, N))
     , x^#(N, 0()) -> c_18(U31^#(isNat(N)))
     , U21^#(tt(), M, N) -> c_8(s^#(plus(activate(N), activate(M))))
     , U41^#(tt(), M, N) ->
       c_15(plus^#(x(activate(N), activate(M)), activate(N)))
     , and^#(tt(), X) -> c_19(activate^#(X)) }
   Strict Trs:
     { U11(tt(), N) -> activate(N)
     , activate(X) -> X
     , activate(n__0()) -> 0()
     , activate(n__plus(X1, X2)) -> plus(activate(X1), activate(X2))
     , activate(n__isNat(X)) -> isNat(X)
     , activate(n__s(X)) -> s(activate(X))
     , activate(n__x(X1, X2)) -> x(activate(X1), activate(X2))
     , U21(tt(), M, N) -> s(plus(activate(N), activate(M)))
     , s(X) -> n__s(X)
     , plus(X1, X2) -> n__plus(X1, X2)
     , plus(N, s(M)) -> U21(and(isNat(M), n__isNat(N)), M, N)
     , plus(N, 0()) -> U11(isNat(N), N)
     , U31(tt()) -> 0()
     , 0() -> n__0()
     , U41(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N))
     , x(X1, X2) -> n__x(X1, X2)
     , x(N, s(M)) -> U41(and(isNat(M), n__isNat(N)), M, N)
     , x(N, 0()) -> U31(isNat(N))
     , and(tt(), X) -> activate(X)
     , isNat(X) -> n__isNat(X)
     , isNat(n__0()) -> tt()
     , isNat(n__plus(V1, V2)) ->
       and(isNat(activate(V1)), n__isNat(activate(V2)))
     , isNat(n__s(V1)) -> isNat(activate(V1))
     , isNat(n__x(V1, V2)) ->
       and(isNat(activate(V1)), n__isNat(activate(V2))) }
   Weak DPs:
     { activate^#(n__0()) -> c_3(0^#())
     , 0^#() -> c_14()
     , isNat^#(n__0()) -> c_21()
     , U31^#(tt()) -> c_13(0^#()) }
   Obligation:
     runtime complexity
   Answer:
     MAYBE
   
   We estimate the number of application of {17} by applications of
   Pre({17}) = {2,6,7,10,14,15}. Here rules are labeled as follows:
   
     DPs:
       { 1: U11^#(tt(), N) -> c_1(activate^#(N))
       , 2: activate^#(X) -> c_2(X)
       , 3: activate^#(n__plus(X1, X2)) ->
            c_4(plus^#(activate(X1), activate(X2)))
       , 4: activate^#(n__isNat(X)) -> c_5(isNat^#(X))
       , 5: activate^#(n__s(X)) -> c_6(s^#(activate(X)))
       , 6: activate^#(n__x(X1, X2)) ->
            c_7(x^#(activate(X1), activate(X2)))
       , 7: plus^#(X1, X2) -> c_10(X1, X2)
       , 8: plus^#(N, s(M)) ->
            c_11(U21^#(and(isNat(M), n__isNat(N)), M, N))
       , 9: plus^#(N, 0()) -> c_12(U11^#(isNat(N), N))
       , 10: isNat^#(X) -> c_20(X)
       , 11: isNat^#(n__plus(V1, V2)) ->
             c_22(and^#(isNat(activate(V1)), n__isNat(activate(V2))))
       , 12: isNat^#(n__s(V1)) -> c_23(isNat^#(activate(V1)))
       , 13: isNat^#(n__x(V1, V2)) ->
             c_24(and^#(isNat(activate(V1)), n__isNat(activate(V2))))
       , 14: s^#(X) -> c_9(X)
       , 15: x^#(X1, X2) -> c_16(X1, X2)
       , 16: x^#(N, s(M)) -> c_17(U41^#(and(isNat(M), n__isNat(N)), M, N))
       , 17: x^#(N, 0()) -> c_18(U31^#(isNat(N)))
       , 18: U21^#(tt(), M, N) -> c_8(s^#(plus(activate(N), activate(M))))
       , 19: U41^#(tt(), M, N) ->
             c_15(plus^#(x(activate(N), activate(M)), activate(N)))
       , 20: and^#(tt(), X) -> c_19(activate^#(X))
       , 21: activate^#(n__0()) -> c_3(0^#())
       , 22: 0^#() -> c_14()
       , 23: isNat^#(n__0()) -> c_21()
       , 24: U31^#(tt()) -> c_13(0^#()) }
   
   We are left with following problem, upon which TcT provides the
   certificate MAYBE.
   
   Strict DPs:
     { U11^#(tt(), N) -> c_1(activate^#(N))
     , activate^#(X) -> c_2(X)
     , activate^#(n__plus(X1, X2)) ->
       c_4(plus^#(activate(X1), activate(X2)))
     , activate^#(n__isNat(X)) -> c_5(isNat^#(X))
     , activate^#(n__s(X)) -> c_6(s^#(activate(X)))
     , activate^#(n__x(X1, X2)) -> c_7(x^#(activate(X1), activate(X2)))
     , plus^#(X1, X2) -> c_10(X1, X2)
     , plus^#(N, s(M)) -> c_11(U21^#(and(isNat(M), n__isNat(N)), M, N))
     , plus^#(N, 0()) -> c_12(U11^#(isNat(N), N))
     , isNat^#(X) -> c_20(X)
     , isNat^#(n__plus(V1, V2)) ->
       c_22(and^#(isNat(activate(V1)), n__isNat(activate(V2))))
     , isNat^#(n__s(V1)) -> c_23(isNat^#(activate(V1)))
     , isNat^#(n__x(V1, V2)) ->
       c_24(and^#(isNat(activate(V1)), n__isNat(activate(V2))))
     , s^#(X) -> c_9(X)
     , x^#(X1, X2) -> c_16(X1, X2)
     , x^#(N, s(M)) -> c_17(U41^#(and(isNat(M), n__isNat(N)), M, N))
     , U21^#(tt(), M, N) -> c_8(s^#(plus(activate(N), activate(M))))
     , U41^#(tt(), M, N) ->
       c_15(plus^#(x(activate(N), activate(M)), activate(N)))
     , and^#(tt(), X) -> c_19(activate^#(X)) }
   Strict Trs:
     { U11(tt(), N) -> activate(N)
     , activate(X) -> X
     , activate(n__0()) -> 0()
     , activate(n__plus(X1, X2)) -> plus(activate(X1), activate(X2))
     , activate(n__isNat(X)) -> isNat(X)
     , activate(n__s(X)) -> s(activate(X))
     , activate(n__x(X1, X2)) -> x(activate(X1), activate(X2))
     , U21(tt(), M, N) -> s(plus(activate(N), activate(M)))
     , s(X) -> n__s(X)
     , plus(X1, X2) -> n__plus(X1, X2)
     , plus(N, s(M)) -> U21(and(isNat(M), n__isNat(N)), M, N)
     , plus(N, 0()) -> U11(isNat(N), N)
     , U31(tt()) -> 0()
     , 0() -> n__0()
     , U41(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N))
     , x(X1, X2) -> n__x(X1, X2)
     , x(N, s(M)) -> U41(and(isNat(M), n__isNat(N)), M, N)
     , x(N, 0()) -> U31(isNat(N))
     , and(tt(), X) -> activate(X)
     , isNat(X) -> n__isNat(X)
     , isNat(n__0()) -> tt()
     , isNat(n__plus(V1, V2)) ->
       and(isNat(activate(V1)), n__isNat(activate(V2)))
     , isNat(n__s(V1)) -> isNat(activate(V1))
     , isNat(n__x(V1, V2)) ->
       and(isNat(activate(V1)), n__isNat(activate(V2))) }
   Weak DPs:
     { activate^#(n__0()) -> c_3(0^#())
     , 0^#() -> c_14()
     , isNat^#(n__0()) -> c_21()
     , x^#(N, 0()) -> c_18(U31^#(isNat(N)))
     , U31^#(tt()) -> c_13(0^#()) }
   Obligation:
     runtime complexity
   Answer:
     MAYBE
   
   Empty strict component of the problem is NOT empty.


Arrrr..