YES(O(1),O(n^1)) We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { U11(tt(), N) -> activate(N) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(X1, X2) , activate(n__isNat(X)) -> isNat(X) , activate(n__s(X)) -> s(X) , activate(n__x(X1, X2)) -> x(X1, 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: innermost runtime complexity Answer: YES(O(1),O(n^1)) Arguments of following rules are not normal-forms: { plus(N, s(M)) -> U21(and(isNat(M), n__isNat(N)), M, N) , plus(N, 0()) -> U11(isNat(N), N) , x(N, s(M)) -> U41(and(isNat(M), n__isNat(N)), M, N) , x(N, 0()) -> U31(isNat(N)) } All above mentioned rules can be savely removed. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { U11(tt(), N) -> activate(N) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(X1, X2) , activate(n__isNat(X)) -> isNat(X) , activate(n__s(X)) -> s(X) , activate(n__x(X1, X2)) -> x(X1, X2) , U21(tt(), M, N) -> s(plus(activate(N), activate(M))) , s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) , 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) , 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: innermost runtime complexity Answer: YES(O(1),O(n^1)) We add the following dependency tuples: Strict DPs: { U11^#(tt(), N) -> c_1(activate^#(N)) , activate^#(X) -> c_2() , activate^#(n__0()) -> c_3(0^#()) , activate^#(n__plus(X1, X2)) -> c_4(plus^#(X1, X2)) , activate^#(n__isNat(X)) -> c_5(isNat^#(X)) , activate^#(n__s(X)) -> c_6(s^#(X)) , activate^#(n__x(X1, X2)) -> c_7(x^#(X1, X2)) , 0^#() -> c_12() , plus^#(X1, X2) -> c_10() , isNat^#(X) -> c_16() , isNat^#(n__0()) -> c_17() , isNat^#(n__plus(V1, V2)) -> c_18(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) , isNat^#(n__s(V1)) -> c_19(isNat^#(activate(V1)), activate^#(V1)) , isNat^#(n__x(V1, V2)) -> c_20(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) , s^#(X) -> c_9() , x^#(X1, X2) -> c_14() , U21^#(tt(), M, N) -> c_8(s^#(plus(activate(N), activate(M))), plus^#(activate(N), activate(M)), activate^#(N), activate^#(M)) , U31^#(tt()) -> c_11(0^#()) , U41^#(tt(), M, N) -> c_13(plus^#(x(activate(N), activate(M)), activate(N)), x^#(activate(N), activate(M)), activate^#(N), activate^#(M), activate^#(N)) , and^#(tt(), X) -> c_15(activate^#(X)) } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { U11^#(tt(), N) -> c_1(activate^#(N)) , activate^#(X) -> c_2() , activate^#(n__0()) -> c_3(0^#()) , activate^#(n__plus(X1, X2)) -> c_4(plus^#(X1, X2)) , activate^#(n__isNat(X)) -> c_5(isNat^#(X)) , activate^#(n__s(X)) -> c_6(s^#(X)) , activate^#(n__x(X1, X2)) -> c_7(x^#(X1, X2)) , 0^#() -> c_12() , plus^#(X1, X2) -> c_10() , isNat^#(X) -> c_16() , isNat^#(n__0()) -> c_17() , isNat^#(n__plus(V1, V2)) -> c_18(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) , isNat^#(n__s(V1)) -> c_19(isNat^#(activate(V1)), activate^#(V1)) , isNat^#(n__x(V1, V2)) -> c_20(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) , s^#(X) -> c_9() , x^#(X1, X2) -> c_14() , U21^#(tt(), M, N) -> c_8(s^#(plus(activate(N), activate(M))), plus^#(activate(N), activate(M)), activate^#(N), activate^#(M)) , U31^#(tt()) -> c_11(0^#()) , U41^#(tt(), M, N) -> c_13(plus^#(x(activate(N), activate(M)), activate(N)), x^#(activate(N), activate(M)), activate^#(N), activate^#(M), activate^#(N)) , and^#(tt(), X) -> c_15(activate^#(X)) } Weak Trs: { U11(tt(), N) -> activate(N) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(X1, X2) , activate(n__isNat(X)) -> isNat(X) , activate(n__s(X)) -> s(X) , activate(n__x(X1, X2)) -> x(X1, X2) , U21(tt(), M, N) -> s(plus(activate(N), activate(M))) , s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) , 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) , 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: innermost runtime complexity Answer: YES(O(1),O(n^1)) We estimate the number of application of {2,8,9,10,11,15,16} by applications of Pre({2,8,9,10,11,15,16}) = {1,3,4,5,6,7,12,13,14,17,18,19,20}. Here rules are labeled as follows: DPs: { 1: U11^#(tt(), N) -> c_1(activate^#(N)) , 2: activate^#(X) -> c_2() , 3: activate^#(n__0()) -> c_3(0^#()) , 4: activate^#(n__plus(X1, X2)) -> c_4(plus^#(X1, X2)) , 5: activate^#(n__isNat(X)) -> c_5(isNat^#(X)) , 6: activate^#(n__s(X)) -> c_6(s^#(X)) , 7: activate^#(n__x(X1, X2)) -> c_7(x^#(X1, X2)) , 8: 0^#() -> c_12() , 9: plus^#(X1, X2) -> c_10() , 10: isNat^#(X) -> c_16() , 11: isNat^#(n__0()) -> c_17() , 12: isNat^#(n__plus(V1, V2)) -> c_18(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) , 13: isNat^#(n__s(V1)) -> c_19(isNat^#(activate(V1)), activate^#(V1)) , 14: isNat^#(n__x(V1, V2)) -> c_20(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) , 15: s^#(X) -> c_9() , 16: x^#(X1, X2) -> c_14() , 17: U21^#(tt(), M, N) -> c_8(s^#(plus(activate(N), activate(M))), plus^#(activate(N), activate(M)), activate^#(N), activate^#(M)) , 18: U31^#(tt()) -> c_11(0^#()) , 19: U41^#(tt(), M, N) -> c_13(plus^#(x(activate(N), activate(M)), activate(N)), x^#(activate(N), activate(M)), activate^#(N), activate^#(M), activate^#(N)) , 20: and^#(tt(), X) -> c_15(activate^#(X)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { U11^#(tt(), N) -> c_1(activate^#(N)) , activate^#(n__0()) -> c_3(0^#()) , activate^#(n__plus(X1, X2)) -> c_4(plus^#(X1, X2)) , activate^#(n__isNat(X)) -> c_5(isNat^#(X)) , activate^#(n__s(X)) -> c_6(s^#(X)) , activate^#(n__x(X1, X2)) -> c_7(x^#(X1, X2)) , isNat^#(n__plus(V1, V2)) -> c_18(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) , isNat^#(n__s(V1)) -> c_19(isNat^#(activate(V1)), activate^#(V1)) , isNat^#(n__x(V1, V2)) -> c_20(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) , U21^#(tt(), M, N) -> c_8(s^#(plus(activate(N), activate(M))), plus^#(activate(N), activate(M)), activate^#(N), activate^#(M)) , U31^#(tt()) -> c_11(0^#()) , U41^#(tt(), M, N) -> c_13(plus^#(x(activate(N), activate(M)), activate(N)), x^#(activate(N), activate(M)), activate^#(N), activate^#(M), activate^#(N)) , and^#(tt(), X) -> c_15(activate^#(X)) } Weak DPs: { activate^#(X) -> c_2() , 0^#() -> c_12() , plus^#(X1, X2) -> c_10() , isNat^#(X) -> c_16() , isNat^#(n__0()) -> c_17() , s^#(X) -> c_9() , x^#(X1, X2) -> c_14() } Weak Trs: { U11(tt(), N) -> activate(N) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(X1, X2) , activate(n__isNat(X)) -> isNat(X) , activate(n__s(X)) -> s(X) , activate(n__x(X1, X2)) -> x(X1, X2) , U21(tt(), M, N) -> s(plus(activate(N), activate(M))) , s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) , 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) , 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: innermost runtime complexity Answer: YES(O(1),O(n^1)) We estimate the number of application of {2,3,5,6,11} by applications of Pre({2,3,5,6,11}) = {1,7,8,9,10,12,13}. Here rules are labeled as follows: DPs: { 1: U11^#(tt(), N) -> c_1(activate^#(N)) , 2: activate^#(n__0()) -> c_3(0^#()) , 3: activate^#(n__plus(X1, X2)) -> c_4(plus^#(X1, X2)) , 4: activate^#(n__isNat(X)) -> c_5(isNat^#(X)) , 5: activate^#(n__s(X)) -> c_6(s^#(X)) , 6: activate^#(n__x(X1, X2)) -> c_7(x^#(X1, X2)) , 7: isNat^#(n__plus(V1, V2)) -> c_18(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) , 8: isNat^#(n__s(V1)) -> c_19(isNat^#(activate(V1)), activate^#(V1)) , 9: isNat^#(n__x(V1, V2)) -> c_20(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) , 10: U21^#(tt(), M, N) -> c_8(s^#(plus(activate(N), activate(M))), plus^#(activate(N), activate(M)), activate^#(N), activate^#(M)) , 11: U31^#(tt()) -> c_11(0^#()) , 12: U41^#(tt(), M, N) -> c_13(plus^#(x(activate(N), activate(M)), activate(N)), x^#(activate(N), activate(M)), activate^#(N), activate^#(M), activate^#(N)) , 13: and^#(tt(), X) -> c_15(activate^#(X)) , 14: activate^#(X) -> c_2() , 15: 0^#() -> c_12() , 16: plus^#(X1, X2) -> c_10() , 17: isNat^#(X) -> c_16() , 18: isNat^#(n__0()) -> c_17() , 19: s^#(X) -> c_9() , 20: x^#(X1, X2) -> c_14() } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { U11^#(tt(), N) -> c_1(activate^#(N)) , activate^#(n__isNat(X)) -> c_5(isNat^#(X)) , isNat^#(n__plus(V1, V2)) -> c_18(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) , isNat^#(n__s(V1)) -> c_19(isNat^#(activate(V1)), activate^#(V1)) , isNat^#(n__x(V1, V2)) -> c_20(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) , U21^#(tt(), M, N) -> c_8(s^#(plus(activate(N), activate(M))), plus^#(activate(N), activate(M)), activate^#(N), activate^#(M)) , U41^#(tt(), M, N) -> c_13(plus^#(x(activate(N), activate(M)), activate(N)), x^#(activate(N), activate(M)), activate^#(N), activate^#(M), activate^#(N)) , and^#(tt(), X) -> c_15(activate^#(X)) } Weak DPs: { activate^#(X) -> c_2() , activate^#(n__0()) -> c_3(0^#()) , activate^#(n__plus(X1, X2)) -> c_4(plus^#(X1, X2)) , activate^#(n__s(X)) -> c_6(s^#(X)) , activate^#(n__x(X1, X2)) -> c_7(x^#(X1, X2)) , 0^#() -> c_12() , plus^#(X1, X2) -> c_10() , isNat^#(X) -> c_16() , isNat^#(n__0()) -> c_17() , s^#(X) -> c_9() , x^#(X1, X2) -> c_14() , U31^#(tt()) -> c_11(0^#()) } Weak Trs: { U11(tt(), N) -> activate(N) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(X1, X2) , activate(n__isNat(X)) -> isNat(X) , activate(n__s(X)) -> s(X) , activate(n__x(X1, X2)) -> x(X1, X2) , U21(tt(), M, N) -> s(plus(activate(N), activate(M))) , s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) , 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) , 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: innermost runtime complexity Answer: YES(O(1),O(n^1)) The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { activate^#(X) -> c_2() , activate^#(n__0()) -> c_3(0^#()) , activate^#(n__plus(X1, X2)) -> c_4(plus^#(X1, X2)) , activate^#(n__s(X)) -> c_6(s^#(X)) , activate^#(n__x(X1, X2)) -> c_7(x^#(X1, X2)) , 0^#() -> c_12() , plus^#(X1, X2) -> c_10() , isNat^#(X) -> c_16() , isNat^#(n__0()) -> c_17() , s^#(X) -> c_9() , x^#(X1, X2) -> c_14() , U31^#(tt()) -> c_11(0^#()) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { U11^#(tt(), N) -> c_1(activate^#(N)) , activate^#(n__isNat(X)) -> c_5(isNat^#(X)) , isNat^#(n__plus(V1, V2)) -> c_18(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) , isNat^#(n__s(V1)) -> c_19(isNat^#(activate(V1)), activate^#(V1)) , isNat^#(n__x(V1, V2)) -> c_20(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) , U21^#(tt(), M, N) -> c_8(s^#(plus(activate(N), activate(M))), plus^#(activate(N), activate(M)), activate^#(N), activate^#(M)) , U41^#(tt(), M, N) -> c_13(plus^#(x(activate(N), activate(M)), activate(N)), x^#(activate(N), activate(M)), activate^#(N), activate^#(M), activate^#(N)) , and^#(tt(), X) -> c_15(activate^#(X)) } Weak Trs: { U11(tt(), N) -> activate(N) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(X1, X2) , activate(n__isNat(X)) -> isNat(X) , activate(n__s(X)) -> s(X) , activate(n__x(X1, X2)) -> x(X1, X2) , U21(tt(), M, N) -> s(plus(activate(N), activate(M))) , s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) , 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) , 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: innermost runtime complexity Answer: YES(O(1),O(n^1)) Due to missing edges in the dependency-graph, the right-hand sides of following rules could be simplified: { U21^#(tt(), M, N) -> c_8(s^#(plus(activate(N), activate(M))), plus^#(activate(N), activate(M)), activate^#(N), activate^#(M)) , U41^#(tt(), M, N) -> c_13(plus^#(x(activate(N), activate(M)), activate(N)), x^#(activate(N), activate(M)), activate^#(N), activate^#(M), activate^#(N)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { U11^#(tt(), N) -> c_1(activate^#(N)) , activate^#(n__isNat(X)) -> c_2(isNat^#(X)) , isNat^#(n__plus(V1, V2)) -> c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) , isNat^#(n__s(V1)) -> c_4(isNat^#(activate(V1)), activate^#(V1)) , isNat^#(n__x(V1, V2)) -> c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) , U21^#(tt(), M, N) -> c_6(activate^#(N), activate^#(M)) , U41^#(tt(), M, N) -> c_7(activate^#(N), activate^#(M), activate^#(N)) , and^#(tt(), X) -> c_8(activate^#(X)) } Weak Trs: { U11(tt(), N) -> activate(N) , activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(X1, X2) , activate(n__isNat(X)) -> isNat(X) , activate(n__s(X)) -> s(X) , activate(n__x(X1, X2)) -> x(X1, X2) , U21(tt(), M, N) -> s(plus(activate(N), activate(M))) , s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) , 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) , 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: innermost runtime complexity Answer: YES(O(1),O(n^1)) We replace rewrite rules by usable rules: Weak Usable Rules: { activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(X1, X2) , activate(n__isNat(X)) -> isNat(X) , activate(n__s(X)) -> s(X) , activate(n__x(X1, X2)) -> x(X1, X2) , s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) , 0() -> n__0() , x(X1, X2) -> n__x(X1, X2) , 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))) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { U11^#(tt(), N) -> c_1(activate^#(N)) , activate^#(n__isNat(X)) -> c_2(isNat^#(X)) , isNat^#(n__plus(V1, V2)) -> c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) , isNat^#(n__s(V1)) -> c_4(isNat^#(activate(V1)), activate^#(V1)) , isNat^#(n__x(V1, V2)) -> c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) , U21^#(tt(), M, N) -> c_6(activate^#(N), activate^#(M)) , U41^#(tt(), M, N) -> c_7(activate^#(N), activate^#(M), activate^#(N)) , and^#(tt(), X) -> c_8(activate^#(X)) } Weak Trs: { activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(X1, X2) , activate(n__isNat(X)) -> isNat(X) , activate(n__s(X)) -> s(X) , activate(n__x(X1, X2)) -> x(X1, X2) , s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) , 0() -> n__0() , x(X1, X2) -> n__x(X1, X2) , 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: innermost runtime complexity Answer: YES(O(1),O(n^1)) Consider the dependency graph 1: U11^#(tt(), N) -> c_1(activate^#(N)) -->_1 activate^#(n__isNat(X)) -> c_2(isNat^#(X)) :2 2: activate^#(n__isNat(X)) -> c_2(isNat^#(X)) -->_1 isNat^#(n__x(V1, V2)) -> c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) :5 -->_1 isNat^#(n__s(V1)) -> c_4(isNat^#(activate(V1)), activate^#(V1)) :4 -->_1 isNat^#(n__plus(V1, V2)) -> c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) :3 3: isNat^#(n__plus(V1, V2)) -> c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) -->_1 and^#(tt(), X) -> c_8(activate^#(X)) :8 -->_2 isNat^#(n__x(V1, V2)) -> c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) :5 -->_2 isNat^#(n__s(V1)) -> c_4(isNat^#(activate(V1)), activate^#(V1)) :4 -->_2 isNat^#(n__plus(V1, V2)) -> c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) :3 -->_4 activate^#(n__isNat(X)) -> c_2(isNat^#(X)) :2 -->_3 activate^#(n__isNat(X)) -> c_2(isNat^#(X)) :2 4: isNat^#(n__s(V1)) -> c_4(isNat^#(activate(V1)), activate^#(V1)) -->_1 isNat^#(n__x(V1, V2)) -> c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) :5 -->_1 isNat^#(n__s(V1)) -> c_4(isNat^#(activate(V1)), activate^#(V1)) :4 -->_1 isNat^#(n__plus(V1, V2)) -> c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) :3 -->_2 activate^#(n__isNat(X)) -> c_2(isNat^#(X)) :2 5: isNat^#(n__x(V1, V2)) -> c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) -->_1 and^#(tt(), X) -> c_8(activate^#(X)) :8 -->_2 isNat^#(n__x(V1, V2)) -> c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) :5 -->_2 isNat^#(n__s(V1)) -> c_4(isNat^#(activate(V1)), activate^#(V1)) :4 -->_2 isNat^#(n__plus(V1, V2)) -> c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) :3 -->_4 activate^#(n__isNat(X)) -> c_2(isNat^#(X)) :2 -->_3 activate^#(n__isNat(X)) -> c_2(isNat^#(X)) :2 6: U21^#(tt(), M, N) -> c_6(activate^#(N), activate^#(M)) -->_2 activate^#(n__isNat(X)) -> c_2(isNat^#(X)) :2 -->_1 activate^#(n__isNat(X)) -> c_2(isNat^#(X)) :2 7: U41^#(tt(), M, N) -> c_7(activate^#(N), activate^#(M), activate^#(N)) -->_3 activate^#(n__isNat(X)) -> c_2(isNat^#(X)) :2 -->_2 activate^#(n__isNat(X)) -> c_2(isNat^#(X)) :2 -->_1 activate^#(n__isNat(X)) -> c_2(isNat^#(X)) :2 8: and^#(tt(), X) -> c_8(activate^#(X)) -->_1 activate^#(n__isNat(X)) -> c_2(isNat^#(X)) :2 Following roots of the dependency graph are removed, as the considered set of starting terms is closed under reduction with respect to these rules (modulo compound contexts). { U11^#(tt(), N) -> c_1(activate^#(N)) , U21^#(tt(), M, N) -> c_6(activate^#(N), activate^#(M)) , U41^#(tt(), M, N) -> c_7(activate^#(N), activate^#(M), activate^#(N)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { activate^#(n__isNat(X)) -> c_2(isNat^#(X)) , isNat^#(n__plus(V1, V2)) -> c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) , isNat^#(n__s(V1)) -> c_4(isNat^#(activate(V1)), activate^#(V1)) , isNat^#(n__x(V1, V2)) -> c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) , and^#(tt(), X) -> c_8(activate^#(X)) } Weak Trs: { activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(X1, X2) , activate(n__isNat(X)) -> isNat(X) , activate(n__s(X)) -> s(X) , activate(n__x(X1, X2)) -> x(X1, X2) , s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) , 0() -> n__0() , x(X1, X2) -> n__x(X1, X2) , 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: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 2' to orient following rules strictly. DPs: { 3: isNat^#(n__s(V1)) -> c_4(isNat^#(activate(V1)), activate^#(V1)) , 5: and^#(tt(), X) -> c_8(activate^#(X)) } Trs: { activate(n__plus(X1, X2)) -> plus(X1, X2) , activate(n__x(X1, X2)) -> x(X1, X2) , 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))) } Sub-proof: ---------- The following argument positions are usable: Uargs(c_2) = {1}, Uargs(c_3) = {1, 2, 3, 4}, Uargs(c_4) = {1, 2}, Uargs(c_5) = {1, 2, 3, 4}, Uargs(c_8) = {1} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA) and not(IDA(1)). [U11](x1, x2) = [7 7] x1 + [7 7] x2 + [0] [0 0] [0 0] [0] [tt] = [0] [2] [activate](x1) = [2 0] x1 + [0] [0 1] [0] [U21](x1, x2, x3) = [7 7] x1 + [7 7] x2 + [7 7] x3 + [0] [0 0] [0 0] [0 0] [0] [s](x1) = [0 0] x1 + [0] [1 1] [4] [plus](x1, x2) = [0 0] x1 + [0 0] x2 + [1] [1 1] [1 1] [4] [U31](x1) = [7 7] x1 + [0] [0 0] [0] [0] = [0] [0] [U41](x1, x2, x3) = [7 7] x1 + [7 7] x2 + [7 7] x3 + [0] [0 0] [0 0] [0 0] [0] [x](x1, x2) = [0 0] x1 + [0 0] x2 + [5] [1 1] [1 1] [4] [and](x1, x2) = [2 0] x2 + [0] [0 1] [0] [isNat](x1) = [0 2] x1 + [0] [0 0] [2] [n__0] = [0] [0] [n__plus](x1, x2) = [0 0] x1 + [0 0] x2 + [1] [1 1] [1 1] [4] [n__isNat](x1) = [0 1] x1 + [0] [0 0] [2] [n__s](x1) = [0 0] x1 + [0] [1 1] [4] [n__x](x1, x2) = [0 0] x1 + [0 0] x2 + [5] [1 1] [1 1] [4] [U11^#](x1, x2) = [7 7] x1 + [7 7] x2 + [0] [0 0] [0 0] [0] [c_1](x1) = [7 7] x1 + [0] [0 0] [0] [activate^#](x1) = [1 0] x1 + [0] [0 4] [0] [c_2] = [0] [0] [c_3](x1) = [7 7] x1 + [0] [0 0] [0] [0^#] = [0] [0] [c_4](x1) = [7 7] x1 + [0] [0 0] [0] [plus^#](x1, x2) = [7 7] x1 + [7 7] x2 + [0] [0 0] [0 0] [0] [c_5](x1) = [7 7] x1 + [0] [0 0] [0] [isNat^#](x1) = [0 1] x1 + [0] [1 0] [7] [c_6](x1) = [7 7] x1 + [0] [0 0] [0] [s^#](x1) = [7 7] x1 + [0] [0 0] [0] [c_7](x1) = [7 7] x1 + [0] [0 0] [0] [x^#](x1, x2) = [7 7] x1 + [7 7] x2 + [0] [0 0] [0 0] [0] [U21^#](x1, x2, x3) = [7 7] x1 + [7 7] x2 + [7 7] x3 + [0] [0 0] [0 0] [0 0] [0] [c_8](x1, x2, x3, x4) = [7 7] x1 + [7 7] x2 + [7 7] x3 + [7 7] x4 + [0] [0 0] [0 0] [0 0] [0 0] [0] [c_9] = [0] [0] [c_10] = [0] [0] [U31^#](x1) = [7 7] x1 + [0] [0 0] [0] [c_11](x1) = [7 7] x1 + [0] [0 0] [0] [c_12] = [0] [0] [U41^#](x1, x2, x3) = [7 7] x1 + [7 7] x2 + [7 7] x3 + [0] [0 0] [0 0] [0 0] [0] [c_13](x1, x2, x3, x4, x5) = [7 7] x1 + [7 7] x2 + [7 7] x3 + [7 7] x4 + [7 7] x5 + [0] [0 0] [0 0] [0 0] [0 0] [0 0] [0] [c_14] = [0] [0] [and^#](x1, x2) = [0 2] x1 + [1 0] x2 + [0] [0 0] [4 4] [0] [c_15](x1) = [7 7] x1 + [0] [0 0] [0] [c_16] = [0] [0] [c_17] = [0] [0] [c_18](x1, x2, x3, x4) = [7 7] x1 + [7 7] x2 + [7 7] x3 + [7 7] x4 + [0] [0 0] [0 0] [0 0] [0 0] [0] [c_19](x1, x2) = [7 7] x1 + [7 7] x2 + [0] [0 0] [0 0] [0] [c_20](x1, x2, x3, x4) = [7 7] x1 + [7 7] x2 + [7 7] x3 + [7 7] x4 + [0] [0 0] [0 0] [0 0] [0 0] [0] [c] = [0] [0] [c_1](x1) = [7 7] x1 + [0] [0 0] [0] [c_2](x1) = [1 0] x1 + [0] [0 0] [7] [c_3](x1, x2, x3, x4) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [1 0] x4 + [0] [0 0] [0 0] [0 0] [0 0] [7] [c_4](x1, x2) = [1 0] x1 + [1 0] x2 + [1] [0 0] [0 0] [3] [c_5](x1, x2, x3, x4) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [1 0] x4 + [0] [0 0] [0 0] [0 0] [0 0] [3] [c_6](x1, x2) = [7 7] x1 + [7 7] x2 + [0] [0 0] [0 0] [0] [c_7](x1, x2, x3) = [7 7] x1 + [7 7] x2 + [7 7] x3 + [0] [0 0] [0 0] [0 0] [0] [c_8](x1) = [1 0] x1 + [3] [0 0] [0] The following symbols are considered usable {activate, s, plus, 0, x, and, isNat, activate^#, isNat^#, and^#} The order satisfies the following ordering constraints: [activate(X)] = [2 0] X + [0] [0 1] [0] >= [1 0] X + [0] [0 1] [0] = [X] [activate(n__0())] = [0] [0] >= [0] [0] = [0()] [activate(n__plus(X1, X2))] = [0 0] X1 + [0 0] X2 + [2] [1 1] [1 1] [4] > [0 0] X1 + [0 0] X2 + [1] [1 1] [1 1] [4] = [plus(X1, X2)] [activate(n__isNat(X))] = [0 2] X + [0] [0 0] [2] >= [0 2] X + [0] [0 0] [2] = [isNat(X)] [activate(n__s(X))] = [0 0] X + [0] [1 1] [4] >= [0 0] X + [0] [1 1] [4] = [s(X)] [activate(n__x(X1, X2))] = [0 0] X1 + [0 0] X2 + [10] [1 1] [1 1] [4] > [0 0] X1 + [0 0] X2 + [5] [1 1] [1 1] [4] = [x(X1, X2)] [s(X)] = [0 0] X + [0] [1 1] [4] >= [0 0] X + [0] [1 1] [4] = [n__s(X)] [plus(X1, X2)] = [0 0] X1 + [0 0] X2 + [1] [1 1] [1 1] [4] >= [0 0] X1 + [0 0] X2 + [1] [1 1] [1 1] [4] = [n__plus(X1, X2)] [0()] = [0] [0] >= [0] [0] = [n__0()] [x(X1, X2)] = [0 0] X1 + [0 0] X2 + [5] [1 1] [1 1] [4] >= [0 0] X1 + [0 0] X2 + [5] [1 1] [1 1] [4] = [n__x(X1, X2)] [and(tt(), X)] = [2 0] X + [0] [0 1] [0] >= [2 0] X + [0] [0 1] [0] = [activate(X)] [isNat(X)] = [0 2] X + [0] [0 0] [2] >= [0 1] X + [0] [0 0] [2] = [n__isNat(X)] [isNat(n__0())] = [0] [2] >= [0] [2] = [tt()] [isNat(n__plus(V1, V2))] = [2 2] V1 + [2 2] V2 + [8] [0 0] [0 0] [2] > [0 2] V2 + [0] [0 0] [2] = [and(isNat(activate(V1)), n__isNat(activate(V2)))] [isNat(n__s(V1))] = [2 2] V1 + [8] [0 0] [2] > [0 2] V1 + [0] [0 0] [2] = [isNat(activate(V1))] [isNat(n__x(V1, V2))] = [2 2] V1 + [2 2] V2 + [8] [0 0] [0 0] [2] > [0 2] V2 + [0] [0 0] [2] = [and(isNat(activate(V1)), n__isNat(activate(V2)))] [activate^#(n__isNat(X))] = [0 1] X + [0] [0 0] [8] >= [0 1] X + [0] [0 0] [7] = [c_2(isNat^#(X))] [isNat^#(n__plus(V1, V2))] = [1 1] V1 + [1 1] V2 + [4] [0 0] [0 0] [8] >= [1 1] V1 + [1 1] V2 + [4] [0 0] [0 0] [7] = [c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2))] [isNat^#(n__s(V1))] = [1 1] V1 + [4] [0 0] [7] > [1 1] V1 + [1] [0 0] [3] = [c_4(isNat^#(activate(V1)), activate^#(V1))] [isNat^#(n__x(V1, V2))] = [1 1] V1 + [1 1] V2 + [4] [0 0] [0 0] [12] >= [1 1] V1 + [1 1] V2 + [4] [0 0] [0 0] [3] = [c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2))] [and^#(tt(), X)] = [1 0] X + [4] [4 4] [0] > [1 0] X + [3] [0 0] [0] = [c_8(activate^#(X))] The strictly oriented rules are moved into the weak component. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { activate^#(n__isNat(X)) -> c_2(isNat^#(X)) , isNat^#(n__plus(V1, V2)) -> c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) , isNat^#(n__x(V1, V2)) -> c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) } Weak DPs: { isNat^#(n__s(V1)) -> c_4(isNat^#(activate(V1)), activate^#(V1)) , and^#(tt(), X) -> c_8(activate^#(X)) } Weak Trs: { activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(X1, X2) , activate(n__isNat(X)) -> isNat(X) , activate(n__s(X)) -> s(X) , activate(n__x(X1, X2)) -> x(X1, X2) , s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) , 0() -> n__0() , x(X1, X2) -> n__x(X1, X2) , 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: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 2' to orient following rules strictly. DPs: { 1: activate^#(n__isNat(X)) -> c_2(isNat^#(X)) , 4: isNat^#(n__s(V1)) -> c_4(isNat^#(activate(V1)), activate^#(V1)) } Trs: { activate(n__isNat(X)) -> isNat(X) , activate(n__s(X)) -> s(X) , s(X) -> n__s(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))) } Sub-proof: ---------- The following argument positions are usable: Uargs(c_2) = {1}, Uargs(c_3) = {1, 2, 3, 4}, Uargs(c_4) = {1, 2}, Uargs(c_5) = {1, 2, 3, 4}, Uargs(c_8) = {1} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA) and not(IDA(1)). [U11](x1, x2) = [7 7] x1 + [7 7] x2 + [0] [0 0] [0 0] [0] [tt] = [0] [0] [activate](x1) = [2 0] x1 + [0] [0 1] [0] [U21](x1, x2, x3) = [7 7] x1 + [7 7] x2 + [7 7] x3 + [0] [0 0] [0 0] [0 0] [0] [s](x1) = [0 0] x1 + [5] [1 1] [4] [plus](x1, x2) = [0 0] x1 + [0 0] x2 + [0] [1 1] [1 1] [4] [U31](x1) = [7 7] x1 + [0] [0 0] [0] [0] = [0] [0] [U41](x1, x2, x3) = [7 7] x1 + [7 7] x2 + [7 7] x3 + [0] [0 0] [0 0] [0 0] [0] [x](x1, x2) = [0 0] x1 + [0 0] x2 + [0] [1 1] [1 1] [4] [and](x1, x2) = [2 0] x2 + [0] [0 2] [0] [isNat](x1) = [0 2] x1 + [4] [0 0] [0] [n__0] = [0] [0] [n__plus](x1, x2) = [0 0] x1 + [0 0] x2 + [0] [1 1] [1 1] [4] [n__isNat](x1) = [0 1] x1 + [4] [0 0] [0] [n__s](x1) = [0 0] x1 + [4] [1 1] [4] [n__x](x1, x2) = [0 0] x1 + [0 0] x2 + [0] [1 1] [1 1] [4] [U11^#](x1, x2) = [7 7] x1 + [7 7] x2 + [0] [0 0] [0 0] [0] [c_1](x1) = [7 7] x1 + [0] [0 0] [0] [activate^#](x1) = [1 0] x1 + [0] [0 0] [4] [c_2] = [0] [0] [c_3](x1) = [7 7] x1 + [0] [0 0] [0] [0^#] = [0] [0] [c_4](x1) = [7 7] x1 + [0] [0 0] [0] [plus^#](x1, x2) = [7 7] x1 + [7 7] x2 + [0] [0 0] [0 0] [0] [c_5](x1) = [7 7] x1 + [0] [0 0] [0] [isNat^#](x1) = [0 1] x1 + [0] [0 0] [0] [c_6](x1) = [7 7] x1 + [0] [0 0] [0] [s^#](x1) = [7 7] x1 + [0] [0 0] [0] [c_7](x1) = [7 7] x1 + [0] [0 0] [0] [x^#](x1, x2) = [7 7] x1 + [7 7] x2 + [0] [0 0] [0 0] [0] [U21^#](x1, x2, x3) = [7 7] x1 + [7 7] x2 + [7 7] x3 + [0] [0 0] [0 0] [0 0] [0] [c_8](x1, x2, x3, x4) = [7 7] x1 + [7 7] x2 + [7 7] x3 + [7 7] x4 + [0] [0 0] [0 0] [0 0] [0 0] [0] [c_9] = [0] [0] [c_10] = [0] [0] [U31^#](x1) = [7 7] x1 + [0] [0 0] [0] [c_11](x1) = [7 7] x1 + [0] [0 0] [0] [c_12] = [0] [0] [U41^#](x1, x2, x3) = [7 7] x1 + [7 7] x2 + [7 7] x3 + [0] [0 0] [0 0] [0 0] [0] [c_13](x1, x2, x3, x4, x5) = [7 7] x1 + [7 7] x2 + [7 7] x3 + [7 7] x4 + [7 7] x5 + [0] [0 0] [0 0] [0 0] [0 0] [0 0] [0] [c_14] = [0] [0] [and^#](x1, x2) = [1 0] x2 + [0] [4 0] [4] [c_15](x1) = [7 7] x1 + [0] [0 0] [0] [c_16] = [0] [0] [c_17] = [0] [0] [c_18](x1, x2, x3, x4) = [7 7] x1 + [7 7] x2 + [7 7] x3 + [7 7] x4 + [0] [0 0] [0 0] [0 0] [0 0] [0] [c_19](x1, x2) = [7 7] x1 + [7 7] x2 + [0] [0 0] [0 0] [0] [c_20](x1, x2, x3, x4) = [7 7] x1 + [7 7] x2 + [7 7] x3 + [7 7] x4 + [0] [0 0] [0 0] [0 0] [0 0] [0] [c] = [0] [0] [c_1](x1) = [7 7] x1 + [0] [0 0] [0] [c_2](x1) = [1 0] x1 + [1] [0 0] [3] [c_3](x1, x2, x3, x4) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [1 0] x4 + [0] [0 0] [0 0] [0 0] [0 0] [0] [c_4](x1, x2) = [1 0] x1 + [1 0] x2 + [3] [0 0] [0 0] [0] [c_5](x1, x2, x3, x4) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [1 0] x4 + [0] [0 0] [0 0] [0 0] [0 0] [0] [c_6](x1, x2) = [7 7] x1 + [7 7] x2 + [0] [0 0] [0 0] [0] [c_7](x1, x2, x3) = [7 7] x1 + [7 7] x2 + [7 7] x3 + [0] [0 0] [0 0] [0 0] [0] [c_8](x1) = [1 0] x1 + [0] [0 0] [3] The following symbols are considered usable {activate, s, plus, 0, x, and, isNat, activate^#, isNat^#, and^#} The order satisfies the following ordering constraints: [activate(X)] = [2 0] X + [0] [0 1] [0] >= [1 0] X + [0] [0 1] [0] = [X] [activate(n__0())] = [0] [0] >= [0] [0] = [0()] [activate(n__plus(X1, X2))] = [0 0] X1 + [0 0] X2 + [0] [1 1] [1 1] [4] >= [0 0] X1 + [0 0] X2 + [0] [1 1] [1 1] [4] = [plus(X1, X2)] [activate(n__isNat(X))] = [0 2] X + [8] [0 0] [0] > [0 2] X + [4] [0 0] [0] = [isNat(X)] [activate(n__s(X))] = [0 0] X + [8] [1 1] [4] > [0 0] X + [5] [1 1] [4] = [s(X)] [activate(n__x(X1, X2))] = [0 0] X1 + [0 0] X2 + [0] [1 1] [1 1] [4] >= [0 0] X1 + [0 0] X2 + [0] [1 1] [1 1] [4] = [x(X1, X2)] [s(X)] = [0 0] X + [5] [1 1] [4] > [0 0] X + [4] [1 1] [4] = [n__s(X)] [plus(X1, X2)] = [0 0] X1 + [0 0] X2 + [0] [1 1] [1 1] [4] >= [0 0] X1 + [0 0] X2 + [0] [1 1] [1 1] [4] = [n__plus(X1, X2)] [0()] = [0] [0] >= [0] [0] = [n__0()] [x(X1, X2)] = [0 0] X1 + [0 0] X2 + [0] [1 1] [1 1] [4] >= [0 0] X1 + [0 0] X2 + [0] [1 1] [1 1] [4] = [n__x(X1, X2)] [and(tt(), X)] = [2 0] X + [0] [0 2] [0] >= [2 0] X + [0] [0 1] [0] = [activate(X)] [isNat(X)] = [0 2] X + [4] [0 0] [0] >= [0 1] X + [4] [0 0] [0] = [n__isNat(X)] [isNat(n__0())] = [4] [0] > [0] [0] = [tt()] [isNat(n__plus(V1, V2))] = [2 2] V1 + [2 2] V2 + [12] [0 0] [0 0] [0] > [0 2] V2 + [8] [0 0] [0] = [and(isNat(activate(V1)), n__isNat(activate(V2)))] [isNat(n__s(V1))] = [2 2] V1 + [12] [0 0] [0] > [0 2] V1 + [4] [0 0] [0] = [isNat(activate(V1))] [isNat(n__x(V1, V2))] = [2 2] V1 + [2 2] V2 + [12] [0 0] [0 0] [0] > [0 2] V2 + [8] [0 0] [0] = [and(isNat(activate(V1)), n__isNat(activate(V2)))] [activate^#(n__isNat(X))] = [0 1] X + [4] [0 0] [4] > [0 1] X + [1] [0 0] [3] = [c_2(isNat^#(X))] [isNat^#(n__plus(V1, V2))] = [1 1] V1 + [1 1] V2 + [4] [0 0] [0 0] [0] >= [1 1] V1 + [1 1] V2 + [4] [0 0] [0 0] [0] = [c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2))] [isNat^#(n__s(V1))] = [1 1] V1 + [4] [0 0] [0] > [1 1] V1 + [3] [0 0] [0] = [c_4(isNat^#(activate(V1)), activate^#(V1))] [isNat^#(n__x(V1, V2))] = [1 1] V1 + [1 1] V2 + [4] [0 0] [0 0] [0] >= [1 1] V1 + [1 1] V2 + [4] [0 0] [0 0] [0] = [c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2))] [and^#(tt(), X)] = [1 0] X + [0] [4 0] [4] >= [1 0] X + [0] [0 0] [3] = [c_8(activate^#(X))] The strictly oriented rules are moved into the weak component. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { isNat^#(n__plus(V1, V2)) -> c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) , isNat^#(n__x(V1, V2)) -> c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) } Weak DPs: { activate^#(n__isNat(X)) -> c_2(isNat^#(X)) , isNat^#(n__s(V1)) -> c_4(isNat^#(activate(V1)), activate^#(V1)) , and^#(tt(), X) -> c_8(activate^#(X)) } Weak Trs: { activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(X1, X2) , activate(n__isNat(X)) -> isNat(X) , activate(n__s(X)) -> s(X) , activate(n__x(X1, X2)) -> x(X1, X2) , s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) , 0() -> n__0() , x(X1, X2) -> n__x(X1, X2) , 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: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 2' to orient following rules strictly. DPs: { 1: isNat^#(n__plus(V1, V2)) -> c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) , 2: isNat^#(n__x(V1, V2)) -> c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) } Sub-proof: ---------- The following argument positions are usable: Uargs(c_2) = {1}, Uargs(c_3) = {1, 2, 3, 4}, Uargs(c_4) = {1, 2}, Uargs(c_5) = {1, 2, 3, 4}, Uargs(c_8) = {1} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA) and not(IDA(1)). [U11](x1, x2) = [7 7] x1 + [7 7] x2 + [0] [0 0] [0 0] [0] [tt] = [0] [0] [activate](x1) = [1 0] x1 + [0] [0 2] [0] [U21](x1, x2, x3) = [7 7] x1 + [7 7] x2 + [7 7] x3 + [0] [0 0] [0 0] [0 0] [0] [s](x1) = [1 1] x1 + [0] [0 0] [3] [plus](x1, x2) = [1 1] x1 + [1 1] x2 + [2] [0 0] [0 0] [0] [U31](x1) = [7 7] x1 + [0] [0 0] [0] [0] = [0] [0] [U41](x1, x2, x3) = [7 7] x1 + [7 7] x2 + [7 7] x3 + [0] [0 0] [0 0] [0 0] [0] [x](x1, x2) = [1 1] x1 + [1 1] x2 + [2] [0 0] [0 0] [4] [and](x1, x2) = [1 0] x2 + [0] [0 2] [0] [isNat](x1) = [0 0] x1 + [0] [2 0] [0] [n__0] = [0] [0] [n__plus](x1, x2) = [1 1] x1 + [1 1] x2 + [2] [0 0] [0 0] [0] [n__isNat](x1) = [0 0] x1 + [0] [1 0] [0] [n__s](x1) = [1 1] x1 + [0] [0 0] [2] [n__x](x1, x2) = [1 1] x1 + [1 1] x2 + [2] [0 0] [0 0] [2] [U11^#](x1, x2) = [7 7] x1 + [7 7] x2 + [0] [0 0] [0 0] [0] [c_1](x1) = [7 7] x1 + [0] [0 0] [0] [activate^#](x1) = [0 2] x1 + [0] [0 0] [4] [c_2] = [0] [0] [c_3](x1) = [7 7] x1 + [0] [0 0] [0] [0^#] = [0] [0] [c_4](x1) = [7 7] x1 + [0] [0 0] [0] [plus^#](x1, x2) = [7 7] x1 + [7 7] x2 + [0] [0 0] [0 0] [0] [c_5](x1) = [7 7] x1 + [0] [0 0] [0] [isNat^#](x1) = [2 0] x1 + [0] [0 0] [0] [c_6](x1) = [7 7] x1 + [0] [0 0] [0] [s^#](x1) = [7 7] x1 + [0] [0 0] [0] [c_7](x1) = [7 7] x1 + [0] [0 0] [0] [x^#](x1, x2) = [7 7] x1 + [7 7] x2 + [0] [0 0] [0 0] [0] [U21^#](x1, x2, x3) = [7 7] x1 + [7 7] x2 + [7 7] x3 + [0] [0 0] [0 0] [0 0] [0] [c_8](x1, x2, x3, x4) = [7 7] x1 + [7 7] x2 + [7 7] x3 + [7 7] x4 + [0] [0 0] [0 0] [0 0] [0 0] [0] [c_9] = [0] [0] [c_10] = [0] [0] [U31^#](x1) = [7 7] x1 + [0] [0 0] [0] [c_11](x1) = [7 7] x1 + [0] [0 0] [0] [c_12] = [0] [0] [U41^#](x1, x2, x3) = [7 7] x1 + [7 7] x2 + [7 7] x3 + [0] [0 0] [0 0] [0 0] [0] [c_13](x1, x2, x3, x4, x5) = [7 7] x1 + [7 7] x2 + [7 7] x3 + [7 7] x4 + [7 7] x5 + [0] [0 0] [0 0] [0 0] [0 0] [0 0] [0] [c_14] = [0] [0] [and^#](x1, x2) = [4 2] x2 + [0] [0 4] [4] [c_15](x1) = [7 7] x1 + [0] [0 0] [0] [c_16] = [0] [0] [c_17] = [0] [0] [c_18](x1, x2, x3, x4) = [7 7] x1 + [7 7] x2 + [7 7] x3 + [7 7] x4 + [0] [0 0] [0 0] [0 0] [0 0] [0] [c_19](x1, x2) = [7 7] x1 + [7 7] x2 + [0] [0 0] [0 0] [0] [c_20](x1, x2, x3, x4) = [7 7] x1 + [7 7] x2 + [7 7] x3 + [7 7] x4 + [0] [0 0] [0 0] [0 0] [0 0] [0] [c] = [0] [0] [c_1](x1) = [7 7] x1 + [0] [0 0] [0] [c_2](x1) = [1 0] x1 + [0] [0 0] [3] [c_3](x1, x2, x3, x4) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [1 0] x4 + [1] [0 0] [0 0] [0 0] [0 0] [0] [c_4](x1, x2) = [1 0] x1 + [1 0] x2 + [0] [0 0] [0 0] [0] [c_5](x1, x2, x3, x4) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [1 0] x4 + [3] [0 0] [0 0] [0 0] [0 0] [0] [c_6](x1, x2) = [7 7] x1 + [7 7] x2 + [0] [0 0] [0 0] [0] [c_7](x1, x2, x3) = [7 7] x1 + [7 7] x2 + [7 7] x3 + [0] [0 0] [0 0] [0 0] [0] [c_8](x1) = [1 0] x1 + [0] [0 0] [3] The following symbols are considered usable {activate, s, plus, 0, x, and, isNat, activate^#, isNat^#, and^#} The order satisfies the following ordering constraints: [activate(X)] = [1 0] X + [0] [0 2] [0] >= [1 0] X + [0] [0 1] [0] = [X] [activate(n__0())] = [0] [0] >= [0] [0] = [0()] [activate(n__plus(X1, X2))] = [1 1] X1 + [1 1] X2 + [2] [0 0] [0 0] [0] >= [1 1] X1 + [1 1] X2 + [2] [0 0] [0 0] [0] = [plus(X1, X2)] [activate(n__isNat(X))] = [0 0] X + [0] [2 0] [0] >= [0 0] X + [0] [2 0] [0] = [isNat(X)] [activate(n__s(X))] = [1 1] X + [0] [0 0] [4] >= [1 1] X + [0] [0 0] [3] = [s(X)] [activate(n__x(X1, X2))] = [1 1] X1 + [1 1] X2 + [2] [0 0] [0 0] [4] >= [1 1] X1 + [1 1] X2 + [2] [0 0] [0 0] [4] = [x(X1, X2)] [s(X)] = [1 1] X + [0] [0 0] [3] >= [1 1] X + [0] [0 0] [2] = [n__s(X)] [plus(X1, X2)] = [1 1] X1 + [1 1] X2 + [2] [0 0] [0 0] [0] >= [1 1] X1 + [1 1] X2 + [2] [0 0] [0 0] [0] = [n__plus(X1, X2)] [0()] = [0] [0] >= [0] [0] = [n__0()] [x(X1, X2)] = [1 1] X1 + [1 1] X2 + [2] [0 0] [0 0] [4] >= [1 1] X1 + [1 1] X2 + [2] [0 0] [0 0] [2] = [n__x(X1, X2)] [and(tt(), X)] = [1 0] X + [0] [0 2] [0] >= [1 0] X + [0] [0 2] [0] = [activate(X)] [isNat(X)] = [0 0] X + [0] [2 0] [0] >= [0 0] X + [0] [1 0] [0] = [n__isNat(X)] [isNat(n__0())] = [0] [0] >= [0] [0] = [tt()] [isNat(n__plus(V1, V2))] = [0 0] V1 + [0 0] V2 + [0] [2 2] [2 2] [4] >= [0 0] V2 + [0] [2 0] [0] = [and(isNat(activate(V1)), n__isNat(activate(V2)))] [isNat(n__s(V1))] = [0 0] V1 + [0] [2 2] [0] >= [0 0] V1 + [0] [2 0] [0] = [isNat(activate(V1))] [isNat(n__x(V1, V2))] = [0 0] V1 + [0 0] V2 + [0] [2 2] [2 2] [4] >= [0 0] V2 + [0] [2 0] [0] = [and(isNat(activate(V1)), n__isNat(activate(V2)))] [activate^#(n__isNat(X))] = [2 0] X + [0] [0 0] [4] >= [2 0] X + [0] [0 0] [3] = [c_2(isNat^#(X))] [isNat^#(n__plus(V1, V2))] = [2 2] V1 + [2 2] V2 + [4] [0 0] [0 0] [0] > [2 2] V1 + [2 2] V2 + [1] [0 0] [0 0] [0] = [c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2))] [isNat^#(n__s(V1))] = [2 2] V1 + [0] [0 0] [0] >= [2 2] V1 + [0] [0 0] [0] = [c_4(isNat^#(activate(V1)), activate^#(V1))] [isNat^#(n__x(V1, V2))] = [2 2] V1 + [2 2] V2 + [4] [0 0] [0 0] [0] > [2 2] V1 + [2 2] V2 + [3] [0 0] [0 0] [0] = [c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2))] [and^#(tt(), X)] = [4 2] X + [0] [0 4] [4] >= [0 2] X + [0] [0 0] [3] = [c_8(activate^#(X))] We return to the main proof. Consider the set of all dependency pairs : { 1: isNat^#(n__plus(V1, V2)) -> c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) , 2: isNat^#(n__x(V1, V2)) -> c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) , 3: activate^#(n__isNat(X)) -> c_2(isNat^#(X)) , 4: isNat^#(n__s(V1)) -> c_4(isNat^#(activate(V1)), activate^#(V1)) , 5: and^#(tt(), X) -> c_8(activate^#(X)) } Processor 'matrix interpretation of dimension 2' induces the complexity certificate YES(?,O(n^1)) on application of dependency pairs {1,2}. These cover all (indirect) predecessors of dependency pairs {1,2,5}, their number of application is equally bounded. The dependency pairs are shifted into the weak component. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak DPs: { activate^#(n__isNat(X)) -> c_2(isNat^#(X)) , isNat^#(n__plus(V1, V2)) -> c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) , isNat^#(n__s(V1)) -> c_4(isNat^#(activate(V1)), activate^#(V1)) , isNat^#(n__x(V1, V2)) -> c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) , and^#(tt(), X) -> c_8(activate^#(X)) } Weak Trs: { activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(X1, X2) , activate(n__isNat(X)) -> isNat(X) , activate(n__s(X)) -> s(X) , activate(n__x(X1, X2)) -> x(X1, X2) , s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) , 0() -> n__0() , x(X1, X2) -> n__x(X1, X2) , 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: innermost runtime complexity Answer: YES(O(1),O(1)) The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { activate^#(n__isNat(X)) -> c_2(isNat^#(X)) , isNat^#(n__plus(V1, V2)) -> c_3(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) , isNat^#(n__s(V1)) -> c_4(isNat^#(activate(V1)), activate^#(V1)) , isNat^#(n__x(V1, V2)) -> c_5(and^#(isNat(activate(V1)), n__isNat(activate(V2))), isNat^#(activate(V1)), activate^#(V1), activate^#(V2)) , and^#(tt(), X) -> c_8(activate^#(X)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak Trs: { activate(X) -> X , activate(n__0()) -> 0() , activate(n__plus(X1, X2)) -> plus(X1, X2) , activate(n__isNat(X)) -> isNat(X) , activate(n__s(X)) -> s(X) , activate(n__x(X1, X2)) -> x(X1, X2) , s(X) -> n__s(X) , plus(X1, X2) -> n__plus(X1, X2) , 0() -> n__0() , x(X1, X2) -> n__x(X1, X2) , 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: innermost runtime complexity Answer: YES(O(1),O(1)) No rule is usable, rules are removed from the input problem. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Rules: Empty Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(n^1))