MAYBE 'epo* (timeout of 60.0 seconds)' -------------------------------- Answer: MAYBE Input Problem: innermost runtime-complexity with respect to Rules: { a__U11(tt(), V1, V2) -> a__U12(a__isNat(V1), V2) , a__U12(tt(), V2) -> a__U13(a__isNat(V2)) , a__U13(tt()) -> tt() , a__U21(tt(), V1) -> a__U22(a__isNat(V1)) , a__U22(tt()) -> tt() , a__U31(tt(), N) -> mark(N) , a__U41(tt(), M, N) -> s(a__plus(mark(N), mark(M))) , a__and(tt(), X) -> mark(X) , a__isNat(0()) -> tt() , a__isNat(plus(V1, V2)) -> a__U11(a__and(a__isNatKind(V1), isNatKind(V2)), V1, V2) , a__isNat(s(V1)) -> a__U21(a__isNatKind(V1), V1) , a__isNatKind(0()) -> tt() , a__isNatKind(plus(V1, V2)) -> a__and(a__isNatKind(V1), isNatKind(V2)) , a__isNatKind(s(V1)) -> a__isNatKind(V1) , a__plus(N, 0()) -> a__U31(a__and(a__isNat(N), isNatKind(N)), N) , a__plus(N, s(M)) -> a__U41(a__and(a__and(a__isNat(M), isNatKind(M)), and(isNat(N), isNatKind(N))), M, N) , mark(U11(X1, X2, X3)) -> a__U11(mark(X1), X2, X3) , mark(U12(X1, X2)) -> a__U12(mark(X1), X2) , mark(isNat(X)) -> a__isNat(X) , mark(U13(X)) -> a__U13(mark(X)) , mark(U21(X1, X2)) -> a__U21(mark(X1), X2) , mark(U22(X)) -> a__U22(mark(X)) , mark(U31(X1, X2)) -> a__U31(mark(X1), X2) , mark(U41(X1, X2, X3)) -> a__U41(mark(X1), X2, X3) , mark(plus(X1, X2)) -> a__plus(mark(X1), mark(X2)) , mark(and(X1, X2)) -> a__and(mark(X1), X2) , mark(isNatKind(X)) -> a__isNatKind(X) , mark(tt()) -> tt() , mark(s(X)) -> s(mark(X)) , mark(0()) -> 0() , a__U11(X1, X2, X3) -> U11(X1, X2, X3) , a__U12(X1, X2) -> U12(X1, X2) , a__isNat(X) -> isNat(X) , a__U13(X) -> U13(X) , a__U21(X1, X2) -> U21(X1, X2) , a__U22(X) -> U22(X) , a__U31(X1, X2) -> U31(X1, X2) , a__U41(X1, X2, X3) -> U41(X1, X2, X3) , a__plus(X1, X2) -> plus(X1, X2) , a__and(X1, X2) -> and(X1, X2) , a__isNatKind(X) -> isNatKind(X)} Proof Output: The input cannot be shown compatible