YES(O(1),O(n^2)) We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^2)). Strict Trs: { append(@l1, @l2) -> append#1(@l1, @l2) , append#1(::(@x, @xs), @l2) -> ::(@x, append(@xs, @l2)) , append#1(nil(), @l2) -> @l2 , appendAll(@l) -> appendAll#1(@l) , appendAll#1(::(@l1, @ls)) -> append(@l1, appendAll(@ls)) , appendAll#1(nil()) -> nil() , appendAll2(@l) -> appendAll2#1(@l) , appendAll2#1(::(@l1, @ls)) -> append(appendAll(@l1), appendAll2(@ls)) , appendAll2#1(nil()) -> nil() , appendAll3(@l) -> appendAll3#1(@l) , appendAll3#1(::(@l1, @ls)) -> append(appendAll2(@l1), appendAll3(@ls)) , appendAll3#1(nil()) -> nil() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) We add following weak dependency pairs: Strict DPs: { append^#(@l1, @l2) -> c_1(append#1^#(@l1, @l2)) , append#1^#(::(@x, @xs), @l2) -> c_2(append^#(@xs, @l2)) , append#1^#(nil(), @l2) -> c_3() , appendAll^#(@l) -> c_4(appendAll#1^#(@l)) , appendAll#1^#(::(@l1, @ls)) -> c_5(append^#(@l1, appendAll(@ls))) , appendAll#1^#(nil()) -> c_6() , appendAll2^#(@l) -> c_7(appendAll2#1^#(@l)) , appendAll2#1^#(::(@l1, @ls)) -> c_8(append^#(appendAll(@l1), appendAll2(@ls))) , appendAll2#1^#(nil()) -> c_9() , appendAll3^#(@l) -> c_10(appendAll3#1^#(@l)) , appendAll3#1^#(::(@l1, @ls)) -> c_11(append^#(appendAll2(@l1), appendAll3(@ls))) , appendAll3#1^#(nil()) -> c_12() } and mark the set of starting terms. We apply the transformation 'usablerules' on the sub-problem: Strict DPs: { append^#(@l1, @l2) -> c_1(append#1^#(@l1, @l2)) , append#1^#(::(@x, @xs), @l2) -> c_2(append^#(@xs, @l2)) , append#1^#(nil(), @l2) -> c_3() , appendAll^#(@l) -> c_4(appendAll#1^#(@l)) , appendAll#1^#(::(@l1, @ls)) -> c_5(append^#(@l1, appendAll(@ls))) , appendAll#1^#(nil()) -> c_6() , appendAll2^#(@l) -> c_7(appendAll2#1^#(@l)) , appendAll2#1^#(::(@l1, @ls)) -> c_8(append^#(appendAll(@l1), appendAll2(@ls))) , appendAll2#1^#(nil()) -> c_9() , appendAll3^#(@l) -> c_10(appendAll3#1^#(@l)) , appendAll3#1^#(::(@l1, @ls)) -> c_11(append^#(appendAll2(@l1), appendAll3(@ls))) , appendAll3#1^#(nil()) -> c_12() } Strict Trs: { append(@l1, @l2) -> append#1(@l1, @l2) , append#1(::(@x, @xs), @l2) -> ::(@x, append(@xs, @l2)) , append#1(nil(), @l2) -> @l2 , appendAll(@l) -> appendAll#1(@l) , appendAll#1(::(@l1, @ls)) -> append(@l1, appendAll(@ls)) , appendAll#1(nil()) -> nil() , appendAll2(@l) -> appendAll2#1(@l) , appendAll2#1(::(@l1, @ls)) -> append(appendAll(@l1), appendAll2(@ls)) , appendAll2#1(nil()) -> nil() , appendAll3(@l) -> appendAll3#1(@l) , appendAll3#1(::(@l1, @ls)) -> append(appendAll2(@l1), appendAll3(@ls)) , appendAll3#1(nil()) -> nil() } StartTerms: basic terms Strategy: innermost We apply the transformation 'weightgap of dimension 2, cbits 3' on the sub-problem: Strict DPs: { append^#(@l1, @l2) -> c_1(append#1^#(@l1, @l2)) , append#1^#(::(@x, @xs), @l2) -> c_2(append^#(@xs, @l2)) , append#1^#(nil(), @l2) -> c_3() , appendAll^#(@l) -> c_4(appendAll#1^#(@l)) , appendAll#1^#(::(@l1, @ls)) -> c_5(append^#(@l1, appendAll(@ls))) , appendAll#1^#(nil()) -> c_6() , appendAll2^#(@l) -> c_7(appendAll2#1^#(@l)) , appendAll2#1^#(::(@l1, @ls)) -> c_8(append^#(appendAll(@l1), appendAll2(@ls))) , appendAll2#1^#(nil()) -> c_9() , appendAll3^#(@l) -> c_10(appendAll3#1^#(@l)) , appendAll3#1^#(::(@l1, @ls)) -> c_11(append^#(appendAll2(@l1), appendAll3(@ls))) , appendAll3#1^#(nil()) -> c_12() } Strict Trs: { append(@l1, @l2) -> append#1(@l1, @l2) , append#1(::(@x, @xs), @l2) -> ::(@x, append(@xs, @l2)) , append#1(nil(), @l2) -> @l2 , appendAll(@l) -> appendAll#1(@l) , appendAll#1(::(@l1, @ls)) -> append(@l1, appendAll(@ls)) , appendAll#1(nil()) -> nil() , appendAll2(@l) -> appendAll2#1(@l) , appendAll2#1(::(@l1, @ls)) -> append(appendAll(@l1), appendAll2(@ls)) , appendAll2#1(nil()) -> nil() , appendAll3(@l) -> appendAll3#1(@l) , appendAll3#1(::(@l1, @ls)) -> append(appendAll2(@l1), appendAll3(@ls)) , appendAll3#1(nil()) -> nil() } StartTerms: basic terms Strategy: innermost The weightgap principle applies (using the following constant growth matrix-interpretation) The following argument positions are usable: Uargs(append) = {1, 2}, Uargs(::) = {2}, Uargs(append^#) = {1, 2}, Uargs(c_1) = {1}, Uargs(c_2) = {1}, Uargs(c_4) = {1}, Uargs(c_5) = {1}, Uargs(c_7) = {1}, Uargs(c_8) = {1}, Uargs(c_10) = {1}, Uargs(c_11) = {1} TcT has computed following constructor-restricted matrix interpretation. [append](x1, x2) = [1 1] x1 + [1 0] x2 + [1] [0 1] [0 1] [2] [append#1](x1, x2) = [1 1] x1 + [1 0] x2 + [0] [0 1] [0 1] [2] [::](x1, x2) = [1 2] x1 + [1 0] x2 + [2] [0 1] [0 1] [2] [nil] = [2] [0] [appendAll](x1) = [2 0] x1 + [1] [0 2] [0] [appendAll#1](x1) = [2 0] x1 + [0] [0 2] [0] [appendAll2](x1) = [2 0] x1 + [1] [0 2] [0] [appendAll2#1](x1) = [2 0] x1 + [0] [0 2] [0] [appendAll3](x1) = [2 0] x1 + [1] [2 0] [0] [appendAll3#1](x1) = [2 0] x1 + [0] [2 0] [0] [append^#](x1, x2) = [1 0] x1 + [1 0] x2 + [0] [0 0] [0 0] [0] [c_1](x1) = [1 0] x1 + [1] [0 1] [1] [append#1^#](x1, x2) = [1 0] x1 + [1 0] x2 + [2] [1 2] [2 2] [1] [c_2](x1) = [1 0] x1 + [2] [0 1] [2] [c_3] = [1] [1] [appendAll^#](x1) = [2 2] x1 + [2] [1 1] [1] [c_4](x1) = [1 0] x1 + [1] [0 1] [1] [appendAll#1^#](x1) = [2 0] x1 + [2] [1 1] [2] [c_5](x1) = [1 0] x1 + [1] [0 1] [2] [c_6] = [1] [1] [appendAll2^#](x1) = [2 1] x1 + [1] [1 1] [2] [c_7](x1) = [1 0] x1 + [1] [0 1] [1] [appendAll2#1^#](x1) = [2 0] x1 + [2] [1 1] [2] [c_8](x1) = [1 0] x1 + [2] [0 1] [2] [c_9] = [1] [0] [appendAll3^#](x1) = [2 2] x1 + [1] [1 1] [1] [c_10](x1) = [1 0] x1 + [1] [0 1] [1] [appendAll3#1^#](x1) = [2 0] x1 + [2] [1 2] [1] [c_11](x1) = [1 0] x1 + [2] [0 1] [2] [c_12] = [1] [1] This order satisfies following ordering constraints Further, it can be verified that all rules not oriented are covered by the weightgap condition. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { append^#(@l1, @l2) -> c_1(append#1^#(@l1, @l2)) , appendAll^#(@l) -> c_4(appendAll#1^#(@l)) , appendAll2^#(@l) -> c_7(appendAll2#1^#(@l)) , appendAll3^#(@l) -> c_10(appendAll3#1^#(@l)) } Weak DPs: { append#1^#(::(@x, @xs), @l2) -> c_2(append^#(@xs, @l2)) , append#1^#(nil(), @l2) -> c_3() , appendAll#1^#(::(@l1, @ls)) -> c_5(append^#(@l1, appendAll(@ls))) , appendAll#1^#(nil()) -> c_6() , appendAll2#1^#(::(@l1, @ls)) -> c_8(append^#(appendAll(@l1), appendAll2(@ls))) , appendAll2#1^#(nil()) -> c_9() , appendAll3#1^#(::(@l1, @ls)) -> c_11(append^#(appendAll2(@l1), appendAll3(@ls))) , appendAll3#1^#(nil()) -> c_12() } Weak Trs: { append(@l1, @l2) -> append#1(@l1, @l2) , append#1(::(@x, @xs), @l2) -> ::(@x, append(@xs, @l2)) , append#1(nil(), @l2) -> @l2 , appendAll(@l) -> appendAll#1(@l) , appendAll#1(::(@l1, @ls)) -> append(@l1, appendAll(@ls)) , appendAll#1(nil()) -> nil() , appendAll2(@l) -> appendAll2#1(@l) , appendAll2#1(::(@l1, @ls)) -> append(appendAll(@l1), appendAll2(@ls)) , appendAll2#1(nil()) -> nil() , appendAll3(@l) -> appendAll3#1(@l) , appendAll3#1(::(@l1, @ls)) -> append(appendAll2(@l1), appendAll3(@ls)) , appendAll3#1(nil()) -> nil() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We estimate the number of application of {2,3,4} by applications of Pre({2,3,4}) = {}. Here rules are labeled as follows: DPs: { 1: append^#(@l1, @l2) -> c_1(append#1^#(@l1, @l2)) , 2: appendAll^#(@l) -> c_4(appendAll#1^#(@l)) , 3: appendAll2^#(@l) -> c_7(appendAll2#1^#(@l)) , 4: appendAll3^#(@l) -> c_10(appendAll3#1^#(@l)) , 5: append#1^#(::(@x, @xs), @l2) -> c_2(append^#(@xs, @l2)) , 6: append#1^#(nil(), @l2) -> c_3() , 7: appendAll#1^#(::(@l1, @ls)) -> c_5(append^#(@l1, appendAll(@ls))) , 8: appendAll#1^#(nil()) -> c_6() , 9: appendAll2#1^#(::(@l1, @ls)) -> c_8(append^#(appendAll(@l1), appendAll2(@ls))) , 10: appendAll2#1^#(nil()) -> c_9() , 11: appendAll3#1^#(::(@l1, @ls)) -> c_11(append^#(appendAll2(@l1), appendAll3(@ls))) , 12: appendAll3#1^#(nil()) -> c_12() } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { append^#(@l1, @l2) -> c_1(append#1^#(@l1, @l2)) } Weak DPs: { append#1^#(::(@x, @xs), @l2) -> c_2(append^#(@xs, @l2)) , append#1^#(nil(), @l2) -> c_3() , appendAll^#(@l) -> c_4(appendAll#1^#(@l)) , appendAll#1^#(::(@l1, @ls)) -> c_5(append^#(@l1, appendAll(@ls))) , appendAll#1^#(nil()) -> c_6() , appendAll2^#(@l) -> c_7(appendAll2#1^#(@l)) , appendAll2#1^#(::(@l1, @ls)) -> c_8(append^#(appendAll(@l1), appendAll2(@ls))) , appendAll2#1^#(nil()) -> c_9() , appendAll3^#(@l) -> c_10(appendAll3#1^#(@l)) , appendAll3#1^#(::(@l1, @ls)) -> c_11(append^#(appendAll2(@l1), appendAll3(@ls))) , appendAll3#1^#(nil()) -> c_12() } Weak Trs: { append(@l1, @l2) -> append#1(@l1, @l2) , append#1(::(@x, @xs), @l2) -> ::(@x, append(@xs, @l2)) , append#1(nil(), @l2) -> @l2 , appendAll(@l) -> appendAll#1(@l) , appendAll#1(::(@l1, @ls)) -> append(@l1, appendAll(@ls)) , appendAll#1(nil()) -> nil() , appendAll2(@l) -> appendAll2#1(@l) , appendAll2#1(::(@l1, @ls)) -> append(appendAll(@l1), appendAll2(@ls)) , appendAll2#1(nil()) -> nil() , appendAll3(@l) -> appendAll3#1(@l) , appendAll3#1(::(@l1, @ls)) -> append(appendAll2(@l1), appendAll3(@ls)) , appendAll3#1(nil()) -> nil() } 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. { append#1^#(nil(), @l2) -> c_3() , appendAll#1^#(nil()) -> c_6() , appendAll2#1^#(nil()) -> c_9() , appendAll3#1^#(nil()) -> c_12() } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { append^#(@l1, @l2) -> c_1(append#1^#(@l1, @l2)) } Weak DPs: { append#1^#(::(@x, @xs), @l2) -> c_2(append^#(@xs, @l2)) , appendAll^#(@l) -> c_4(appendAll#1^#(@l)) , appendAll#1^#(::(@l1, @ls)) -> c_5(append^#(@l1, appendAll(@ls))) , appendAll2^#(@l) -> c_7(appendAll2#1^#(@l)) , appendAll2#1^#(::(@l1, @ls)) -> c_8(append^#(appendAll(@l1), appendAll2(@ls))) , appendAll3^#(@l) -> c_10(appendAll3#1^#(@l)) , appendAll3#1^#(::(@l1, @ls)) -> c_11(append^#(appendAll2(@l1), appendAll3(@ls))) } Weak Trs: { append(@l1, @l2) -> append#1(@l1, @l2) , append#1(::(@x, @xs), @l2) -> ::(@x, append(@xs, @l2)) , append#1(nil(), @l2) -> @l2 , appendAll(@l) -> appendAll#1(@l) , appendAll#1(::(@l1, @ls)) -> append(@l1, appendAll(@ls)) , appendAll#1(nil()) -> nil() , appendAll2(@l) -> appendAll2#1(@l) , appendAll2#1(::(@l1, @ls)) -> append(appendAll(@l1), appendAll2(@ls)) , appendAll2#1(nil()) -> nil() , appendAll3(@l) -> appendAll3#1(@l) , appendAll3#1(::(@l1, @ls)) -> append(appendAll2(@l1), appendAll3(@ls)) , appendAll3#1(nil()) -> nil() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) Consider the dependency graph 1: append^#(@l1, @l2) -> c_1(append#1^#(@l1, @l2)) -->_1 append#1^#(::(@x, @xs), @l2) -> c_2(append^#(@xs, @l2)) :2 2: append#1^#(::(@x, @xs), @l2) -> c_2(append^#(@xs, @l2)) -->_1 append^#(@l1, @l2) -> c_1(append#1^#(@l1, @l2)) :1 3: appendAll^#(@l) -> c_4(appendAll#1^#(@l)) -->_1 appendAll#1^#(::(@l1, @ls)) -> c_5(append^#(@l1, appendAll(@ls))) :4 4: appendAll#1^#(::(@l1, @ls)) -> c_5(append^#(@l1, appendAll(@ls))) -->_1 append^#(@l1, @l2) -> c_1(append#1^#(@l1, @l2)) :1 5: appendAll2^#(@l) -> c_7(appendAll2#1^#(@l)) -->_1 appendAll2#1^#(::(@l1, @ls)) -> c_8(append^#(appendAll(@l1), appendAll2(@ls))) :6 6: appendAll2#1^#(::(@l1, @ls)) -> c_8(append^#(appendAll(@l1), appendAll2(@ls))) -->_1 append^#(@l1, @l2) -> c_1(append#1^#(@l1, @l2)) :1 7: appendAll3^#(@l) -> c_10(appendAll3#1^#(@l)) -->_1 appendAll3#1^#(::(@l1, @ls)) -> c_11(append^#(appendAll2(@l1), appendAll3(@ls))) :8 8: appendAll3#1^#(::(@l1, @ls)) -> c_11(append^#(appendAll2(@l1), appendAll3(@ls))) -->_1 append^#(@l1, @l2) -> c_1(append#1^#(@l1, @l2)) :1 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). { appendAll^#(@l) -> c_4(appendAll#1^#(@l)) , appendAll2^#(@l) -> c_7(appendAll2#1^#(@l)) , appendAll3^#(@l) -> c_10(appendAll3#1^#(@l)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { append^#(@l1, @l2) -> c_1(append#1^#(@l1, @l2)) } Weak DPs: { append#1^#(::(@x, @xs), @l2) -> c_2(append^#(@xs, @l2)) , appendAll#1^#(::(@l1, @ls)) -> c_5(append^#(@l1, appendAll(@ls))) , appendAll2#1^#(::(@l1, @ls)) -> c_8(append^#(appendAll(@l1), appendAll2(@ls))) , appendAll3#1^#(::(@l1, @ls)) -> c_11(append^#(appendAll2(@l1), appendAll3(@ls))) } Weak Trs: { append(@l1, @l2) -> append#1(@l1, @l2) , append#1(::(@x, @xs), @l2) -> ::(@x, append(@xs, @l2)) , append#1(nil(), @l2) -> @l2 , appendAll(@l) -> appendAll#1(@l) , appendAll#1(::(@l1, @ls)) -> append(@l1, appendAll(@ls)) , appendAll#1(nil()) -> nil() , appendAll2(@l) -> appendAll2#1(@l) , appendAll2#1(::(@l1, @ls)) -> append(appendAll(@l1), appendAll2(@ls)) , appendAll2#1(nil()) -> nil() , appendAll3(@l) -> appendAll3#1(@l) , appendAll3#1(::(@l1, @ls)) -> append(appendAll2(@l1), appendAll3(@ls)) , appendAll3#1(nil()) -> nil() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. DPs: { 2: append#1^#(::(@x, @xs), @l2) -> c_2(append^#(@xs, @l2)) , 3: appendAll#1^#(::(@l1, @ls)) -> c_5(append^#(@l1, appendAll(@ls))) , 4: appendAll2#1^#(::(@l1, @ls)) -> c_8(append^#(appendAll(@l1), appendAll2(@ls))) , 5: appendAll3#1^#(::(@l1, @ls)) -> c_11(append^#(appendAll2(@l1), appendAll3(@ls))) } Trs: { appendAll#1(::(@l1, @ls)) -> append(@l1, appendAll(@ls)) , appendAll2#1(::(@l1, @ls)) -> append(appendAll(@l1), appendAll2(@ls)) } Sub-proof: ---------- The following argument positions are usable: Uargs(c_1) = {1}, Uargs(c_2) = {1}, Uargs(c_5) = {1}, Uargs(c_8) = {1}, Uargs(c_11) = {1} TcT has computed following constructor-based matrix interpretation satisfying not(EDA). [append](x1, x2) = [1] x1 + [1] x2 + [0] [append#1](x1, x2) = [1] x1 + [1] x2 + [0] [::](x1, x2) = [1] x1 + [1] x2 + [1] [nil] = [0] [appendAll](x1) = [1] x1 + [0] [appendAll#1](x1) = [1] x1 + [0] [appendAll2](x1) = [1] x1 + [0] [appendAll2#1](x1) = [1] x1 + [0] [appendAll3](x1) = [2] x1 + [0] [appendAll3#1](x1) = [0] [append^#](x1, x2) = [1] x1 + [0] [c_1](x1) = [1] x1 + [0] [append#1^#](x1, x2) = [1] x1 + [0] [c_2](x1) = [1] x1 + [0] [c_3] = [0] [appendAll^#](x1) = [1] x1 + [0] [c_4](x1) = [1] x1 + [0] [appendAll#1^#](x1) = [2] x1 + [1] [c_5](x1) = [2] x1 + [0] [c_6] = [0] [appendAll2^#](x1) = [1] x1 + [0] [c_7](x1) = [1] x1 + [0] [appendAll2#1^#](x1) = [2] x1 + [2] [c_8](x1) = [2] x1 + [1] [c_9] = [0] [appendAll3^#](x1) = [1] x1 + [0] [c_10](x1) = [1] x1 + [0] [appendAll3#1^#](x1) = [2] x1 + [1] [c_11](x1) = [2] x1 + [0] [c_12] = [0] This order satisfies following ordering constraints [append(@l1, @l2)] = [1] @l1 + [1] @l2 + [0] >= [1] @l1 + [1] @l2 + [0] = [append#1(@l1, @l2)] [append#1(::(@x, @xs), @l2)] = [1] @l2 + [1] @x + [1] @xs + [1] >= [1] @l2 + [1] @x + [1] @xs + [1] = [::(@x, append(@xs, @l2))] [append#1(nil(), @l2)] = [1] @l2 + [0] >= [1] @l2 + [0] = [@l2] [appendAll(@l)] = [1] @l + [0] >= [1] @l + [0] = [appendAll#1(@l)] [appendAll#1(::(@l1, @ls))] = [1] @l1 + [1] @ls + [1] > [1] @l1 + [1] @ls + [0] = [append(@l1, appendAll(@ls))] [appendAll#1(nil())] = [0] >= [0] = [nil()] [appendAll2(@l)] = [1] @l + [0] >= [1] @l + [0] = [appendAll2#1(@l)] [appendAll2#1(::(@l1, @ls))] = [1] @l1 + [1] @ls + [1] > [1] @l1 + [1] @ls + [0] = [append(appendAll(@l1), appendAll2(@ls))] [appendAll2#1(nil())] = [0] >= [0] = [nil()] [append^#(@l1, @l2)] = [1] @l1 + [0] >= [1] @l1 + [0] = [c_1(append#1^#(@l1, @l2))] [append#1^#(::(@x, @xs), @l2)] = [1] @x + [1] @xs + [1] > [1] @xs + [0] = [c_2(append^#(@xs, @l2))] [appendAll#1^#(::(@l1, @ls))] = [2] @l1 + [2] @ls + [3] > [2] @l1 + [0] = [c_5(append^#(@l1, appendAll(@ls)))] [appendAll2#1^#(::(@l1, @ls))] = [2] @l1 + [2] @ls + [4] > [2] @l1 + [1] = [c_8(append^#(appendAll(@l1), appendAll2(@ls)))] [appendAll3#1^#(::(@l1, @ls))] = [2] @l1 + [2] @ls + [3] > [2] @l1 + [0] = [c_11(append^#(appendAll2(@l1), appendAll3(@ls)))] Consider the set of all dependency pairs DPs: { 1: append^#(@l1, @l2) -> c_1(append#1^#(@l1, @l2)) , 2: append#1^#(::(@x, @xs), @l2) -> c_2(append^#(@xs, @l2)) , 3: appendAll#1^#(::(@l1, @ls)) -> c_5(append^#(@l1, appendAll(@ls))) , 4: appendAll2#1^#(::(@l1, @ls)) -> c_8(append^#(appendAll(@l1), appendAll2(@ls))) , 5: appendAll3#1^#(::(@l1, @ls)) -> c_11(append^#(appendAll2(@l1), appendAll3(@ls))) } Processor 'matrix interpretation of dimension 1' induces the complexity certificate YES(?,O(n^1)) on application of dependency pairs {2,3,4,5}. These cover all (indirect) predecessors of dependency pairs {1,2,3,4,5}, their number of application is equally bounded. The dependency pairs are shifted into the corresponding weak component(s). We apply the transformation 'removetails' on the sub-problem: Weak DPs: { append^#(@l1, @l2) -> c_1(append#1^#(@l1, @l2)) , append#1^#(::(@x, @xs), @l2) -> c_2(append^#(@xs, @l2)) , appendAll#1^#(::(@l1, @ls)) -> c_5(append^#(@l1, appendAll(@ls))) , appendAll2#1^#(::(@l1, @ls)) -> c_8(append^#(appendAll(@l1), appendAll2(@ls))) , appendAll3#1^#(::(@l1, @ls)) -> c_11(append^#(appendAll2(@l1), appendAll3(@ls))) } Weak Trs: { append(@l1, @l2) -> append#1(@l1, @l2) , append#1(::(@x, @xs), @l2) -> ::(@x, append(@xs, @l2)) , append#1(nil(), @l2) -> @l2 , appendAll(@l) -> appendAll#1(@l) , appendAll#1(::(@l1, @ls)) -> append(@l1, appendAll(@ls)) , appendAll#1(nil()) -> nil() , appendAll2(@l) -> appendAll2#1(@l) , appendAll2#1(::(@l1, @ls)) -> append(appendAll(@l1), appendAll2(@ls)) , appendAll2#1(nil()) -> nil() , appendAll3(@l) -> appendAll3#1(@l) , appendAll3#1(::(@l1, @ls)) -> append(appendAll2(@l1), appendAll3(@ls)) , appendAll3#1(nil()) -> nil() } StartTerms: basic terms Strategy: innermost The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { append^#(@l1, @l2) -> c_1(append#1^#(@l1, @l2)) , append#1^#(::(@x, @xs), @l2) -> c_2(append^#(@xs, @l2)) , appendAll#1^#(::(@l1, @ls)) -> c_5(append^#(@l1, appendAll(@ls))) , appendAll2#1^#(::(@l1, @ls)) -> c_8(append^#(appendAll(@l1), appendAll2(@ls))) , appendAll3#1^#(::(@l1, @ls)) -> c_11(append^#(appendAll2(@l1), appendAll3(@ls))) } We apply the transformation 'usablerules' on the sub-problem: Weak Trs: { append(@l1, @l2) -> append#1(@l1, @l2) , append#1(::(@x, @xs), @l2) -> ::(@x, append(@xs, @l2)) , append#1(nil(), @l2) -> @l2 , appendAll(@l) -> appendAll#1(@l) , appendAll#1(::(@l1, @ls)) -> append(@l1, appendAll(@ls)) , appendAll#1(nil()) -> nil() , appendAll2(@l) -> appendAll2#1(@l) , appendAll2#1(::(@l1, @ls)) -> append(appendAll(@l1), appendAll2(@ls)) , appendAll2#1(nil()) -> nil() , appendAll3(@l) -> appendAll3#1(@l) , appendAll3#1(::(@l1, @ls)) -> append(appendAll2(@l1), appendAll3(@ls)) , appendAll3#1(nil()) -> nil() } StartTerms: basic terms Strategy: innermost 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^2))