MAYBE We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { U11(tt(), V1, V2) -> U12(isNat(activate(V1)), activate(V2)) , U12(tt(), V2) -> U13(isNat(activate(V2))) , isNat(X) -> n__isNat(X) , isNat(n__0()) -> tt() , isNat(n__plus(V1, V2)) -> U11(and(isNatKind(activate(V1)), n__isNatKind(activate(V2))), activate(V1), activate(V2)) , isNat(n__s(V1)) -> U21(isNatKind(activate(V1)), activate(V1)) , isNat(n__x(V1, V2)) -> U31(and(isNatKind(activate(V1)), n__isNatKind(activate(V2))), activate(V1), activate(V2)) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(activate(X1), activate(X2)) , activate(n__isNatKind(X)) -> isNatKind(X) , activate(n__s(X)) -> s(activate(X)) , activate(n__x(X1, X2)) -> x(activate(X1), activate(X2)) , activate(n__and(X1, X2)) -> and(activate(X1), X2) , activate(n__isNat(X)) -> isNat(X) , U13(tt()) -> tt() , U21(tt(), V1) -> U22(isNat(activate(V1))) , U22(tt()) -> tt() , U31(tt(), V1, V2) -> U32(isNat(activate(V1)), activate(V2)) , U32(tt(), V2) -> U33(isNat(activate(V2))) , U33(tt()) -> tt() , U41(tt(), N) -> activate(N) , U51(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)) -> U51(and(and(isNat(M), n__isNatKind(M)), n__and(n__isNat(N), n__isNatKind(N))), M, N) , plus(N, 0()) -> U41(and(isNat(N), n__isNatKind(N)), N) , U61(tt()) -> 0() , 0() -> n__0() , U71(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N)) , x(X1, X2) -> n__x(X1, X2) , x(N, s(M)) -> U71(and(and(isNat(M), n__isNatKind(M)), n__and(n__isNat(N), n__isNatKind(N))), M, N) , x(N, 0()) -> U61(and(isNat(N), n__isNatKind(N))) , and(X1, X2) -> n__and(X1, X2) , and(tt(), X) -> activate(X) , isNatKind(X) -> n__isNatKind(X) , isNatKind(n__0()) -> tt() , isNatKind(n__plus(V1, V2)) -> and(isNatKind(activate(V1)), n__isNatKind(activate(V2))) , isNatKind(n__s(V1)) -> isNatKind(activate(V1)) , isNatKind(n__x(V1, V2)) -> and(isNatKind(activate(V1)), n__isNatKind(activate(V2))) } Obligation: innermost runtime complexity Answer: MAYBE Arguments of following rules are not normal-forms: { plus(N, s(M)) -> U51(and(and(isNat(M), n__isNatKind(M)), n__and(n__isNat(N), n__isNatKind(N))), M, N) , plus(N, 0()) -> U41(and(isNat(N), n__isNatKind(N)), N) , x(N, s(M)) -> U71(and(and(isNat(M), n__isNatKind(M)), n__and(n__isNat(N), n__isNatKind(N))), M, N) , x(N, 0()) -> U61(and(isNat(N), n__isNatKind(N))) } All above mentioned rules can be savely removed. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { U11(tt(), V1, V2) -> U12(isNat(activate(V1)), activate(V2)) , U12(tt(), V2) -> U13(isNat(activate(V2))) , isNat(X) -> n__isNat(X) , isNat(n__0()) -> tt() , isNat(n__plus(V1, V2)) -> U11(and(isNatKind(activate(V1)), n__isNatKind(activate(V2))), activate(V1), activate(V2)) , isNat(n__s(V1)) -> U21(isNatKind(activate(V1)), activate(V1)) , isNat(n__x(V1, V2)) -> U31(and(isNatKind(activate(V1)), n__isNatKind(activate(V2))), activate(V1), activate(V2)) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(activate(X1), activate(X2)) , activate(n__isNatKind(X)) -> isNatKind(X) , activate(n__s(X)) -> s(activate(X)) , activate(n__x(X1, X2)) -> x(activate(X1), activate(X2)) , activate(n__and(X1, X2)) -> and(activate(X1), X2) , activate(n__isNat(X)) -> isNat(X) , U13(tt()) -> tt() , U21(tt(), V1) -> U22(isNat(activate(V1))) , U22(tt()) -> tt() , U31(tt(), V1, V2) -> U32(isNat(activate(V1)), activate(V2)) , U32(tt(), V2) -> U33(isNat(activate(V2))) , U33(tt()) -> tt() , U41(tt(), N) -> activate(N) , U51(tt(), M, N) -> s(plus(activate(N), activate(M))) , s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) , U61(tt()) -> 0() , 0() -> n__0() , U71(tt(), M, N) -> plus(x(activate(N), activate(M)), activate(N)) , x(X1, X2) -> n__x(X1, X2) , and(X1, X2) -> n__and(X1, X2) , and(tt(), X) -> activate(X) , isNatKind(X) -> n__isNatKind(X) , isNatKind(n__0()) -> tt() , isNatKind(n__plus(V1, V2)) -> and(isNatKind(activate(V1)), n__isNatKind(activate(V2))) , isNatKind(n__s(V1)) -> isNatKind(activate(V1)) , isNatKind(n__x(V1, V2)) -> and(isNatKind(activate(V1)), n__isNatKind(activate(V2))) } Obligation: innermost runtime complexity Answer: MAYBE None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'empty' failed due to the following reason: Empty strict component of the problem is NOT empty. 2) 'Best' failed due to the following reason: 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) 'Polynomial Path Order (PS) (timeout of 60 seconds)' failed due to the following reason: The input cannot be shown compatible 2) 'bsearch-popstar (timeout of 60 seconds)' failed due to the following reason: The input cannot be shown compatible 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 minimal-enrichment and initial automaton 'match'' failed due to the following reason: match-boundness of the problem could not be verified. 2) 'Bounds with perSymbol-enrichment and initial automaton 'match'' failed due to the following reason: match-boundness of the problem could not be verified. Arrrr..