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 use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { append#1(nil(), @l2) -> @l2 , appendAll2#1(nil()) -> nil() , appendAll3#1(nil()) -> nil() } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: Uargs(append) = {1, 2}, Uargs(::) = {2} 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 + [0] [nil] = [2] [appendAll](x1) = [1] x1 + [0] [appendAll#1](x1) = [1] x1 + [0] [appendAll2](x1) = [2] x1 + [0] [appendAll2#1](x1) = [2] x1 + [0] [appendAll3](x1) = [2] x1 + [1] [appendAll3#1](x1) = [2] x1 + [1] 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 + [0] >= [1] @l2 + [1] @x + [1] @xs + [0] = [::(@x, append(@xs, @l2))] [append#1(nil(), @l2)] = [1] @l2 + [2] > [1] @l2 + [0] = [@l2] [appendAll(@l)] = [1] @l + [0] >= [1] @l + [0] = [appendAll#1(@l)] [appendAll#1(::(@l1, @ls))] = [1] @l1 + [1] @ls + [0] >= [1] @l1 + [1] @ls + [0] = [append(@l1, appendAll(@ls))] [appendAll#1(nil())] = [2] >= [2] = [nil()] [appendAll2(@l)] = [2] @l + [0] >= [2] @l + [0] = [appendAll2#1(@l)] [appendAll2#1(::(@l1, @ls))] = [2] @l1 + [2] @ls + [0] >= [1] @l1 + [2] @ls + [0] = [append(appendAll(@l1), appendAll2(@ls))] [appendAll2#1(nil())] = [4] > [2] = [nil()] [appendAll3(@l)] = [2] @l + [1] >= [2] @l + [1] = [appendAll3#1(@l)] [appendAll3#1(::(@l1, @ls))] = [2] @l1 + [2] @ls + [1] >= [2] @l1 + [2] @ls + [1] = [append(appendAll2(@l1), appendAll3(@ls))] [appendAll3#1(nil())] = [5] > [2] = [nil()] We return to the main proof. 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)) , 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)) , appendAll3(@l) -> appendAll3#1(@l) , appendAll3#1(::(@l1, @ls)) -> append(appendAll2(@l1), appendAll3(@ls)) } Weak Trs: { append#1(nil(), @l2) -> @l2 , appendAll2#1(nil()) -> nil() , appendAll3#1(nil()) -> nil() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { appendAll#1(nil()) -> nil() } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: Uargs(append) = {1, 2}, Uargs(::) = {2} 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 + [0] [nil] = [2] [appendAll](x1) = [2] x1 + [0] [appendAll#1](x1) = [2] x1 + [0] [appendAll2](x1) = [2] x1 + [0] [appendAll2#1](x1) = [2] x1 + [0] [appendAll3](x1) = [2] x1 + [0] [appendAll3#1](x1) = [2] x1 + [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 + [0] >= [1] @l2 + [1] @x + [1] @xs + [0] = [::(@x, append(@xs, @l2))] [append#1(nil(), @l2)] = [1] @l2 + [2] > [1] @l2 + [0] = [@l2] [appendAll(@l)] = [2] @l + [0] >= [2] @l + [0] = [appendAll#1(@l)] [appendAll#1(::(@l1, @ls))] = [2] @l1 + [2] @ls + [0] >= [1] @l1 + [2] @ls + [0] = [append(@l1, appendAll(@ls))] [appendAll#1(nil())] = [4] > [2] = [nil()] [appendAll2(@l)] = [2] @l + [0] >= [2] @l + [0] = [appendAll2#1(@l)] [appendAll2#1(::(@l1, @ls))] = [2] @l1 + [2] @ls + [0] >= [2] @l1 + [2] @ls + [0] = [append(appendAll(@l1), appendAll2(@ls))] [appendAll2#1(nil())] = [4] > [2] = [nil()] [appendAll3(@l)] = [2] @l + [0] >= [2] @l + [0] = [appendAll3#1(@l)] [appendAll3#1(::(@l1, @ls))] = [2] @l1 + [2] @ls + [0] >= [2] @l1 + [2] @ls + [0] = [append(appendAll2(@l1), appendAll3(@ls))] [appendAll3#1(nil())] = [4] > [2] = [nil()] We return to the main proof. 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)) , appendAll(@l) -> appendAll#1(@l) , appendAll#1(::(@l1, @ls)) -> append(@l1, appendAll(@ls)) , appendAll2(@l) -> appendAll2#1(@l) , appendAll2#1(::(@l1, @ls)) -> append(appendAll(@l1), appendAll2(@ls)) , appendAll3(@l) -> appendAll3#1(@l) , appendAll3#1(::(@l1, @ls)) -> append(appendAll2(@l1), appendAll3(@ls)) } Weak Trs: { append#1(nil(), @l2) -> @l2 , appendAll#1(nil()) -> nil() , appendAll2#1(nil()) -> nil() , appendAll3#1(nil()) -> nil() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { appendAll2(@l) -> appendAll2#1(@l) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: Uargs(append) = {1, 2}, Uargs(::) = {2} TcT has computed following constructor-based matrix interpretation satisfying not(EDA). [append](x1, x2) = [1] x1 + [1] x2 + [2] [append#1](x1, x2) = [1] x1 + [1] x2 + [2] [::](x1, x2) = [1] x1 + [1] x2 + [2] [nil] = [0] [appendAll](x1) = [1] x1 + [0] [appendAll#1](x1) = [1] x1 + [0] [appendAll2](x1) = [2] x1 + [2] [appendAll2#1](x1) = [2] x1 + [0] [appendAll3](x1) = [2] x1 + [0] [appendAll3#1](x1) = [2] x1 + [0] This order satisfies following ordering constraints [append(@l1, @l2)] = [1] @l1 + [1] @l2 + [2] >= [1] @l1 + [1] @l2 + [2] = [append#1(@l1, @l2)] [append#1(::(@x, @xs), @l2)] = [1] @l2 + [1] @x + [1] @xs + [4] >= [1] @l2 + [1] @x + [1] @xs + [4] = [::(@x, append(@xs, @l2))] [append#1(nil(), @l2)] = [1] @l2 + [2] > [1] @l2 + [0] = [@l2] [appendAll(@l)] = [1] @l + [0] >= [1] @l + [0] = [appendAll#1(@l)] [appendAll#1(::(@l1, @ls))] = [1] @l1 + [1] @ls + [2] >= [1] @l1 + [1] @ls + [2] = [append(@l1, appendAll(@ls))] [appendAll#1(nil())] = [0] >= [0] = [nil()] [appendAll2(@l)] = [2] @l + [2] > [2] @l + [0] = [appendAll2#1(@l)] [appendAll2#1(::(@l1, @ls))] = [2] @l1 + [2] @ls + [4] >= [1] @l1 + [2] @ls + [4] = [append(appendAll(@l1), appendAll2(@ls))] [appendAll2#1(nil())] = [0] >= [0] = [nil()] [appendAll3(@l)] = [2] @l + [0] >= [2] @l + [0] = [appendAll3#1(@l)] [appendAll3#1(::(@l1, @ls))] = [2] @l1 + [2] @ls + [4] >= [2] @l1 + [2] @ls + [4] = [append(appendAll2(@l1), appendAll3(@ls))] [appendAll3#1(nil())] = [0] >= [0] = [nil()] We return to the main proof. 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)) , appendAll(@l) -> appendAll#1(@l) , appendAll#1(::(@l1, @ls)) -> append(@l1, appendAll(@ls)) , appendAll2#1(::(@l1, @ls)) -> append(appendAll(@l1), appendAll2(@ls)) , appendAll3(@l) -> appendAll3#1(@l) , appendAll3#1(::(@l1, @ls)) -> append(appendAll2(@l1), appendAll3(@ls)) } Weak Trs: { append#1(nil(), @l2) -> @l2 , appendAll#1(nil()) -> nil() , appendAll2(@l) -> appendAll2#1(@l) , appendAll2#1(nil()) -> nil() , appendAll3#1(nil()) -> nil() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { appendAll#1(::(@l1, @ls)) -> append(@l1, appendAll(@ls)) , appendAll2#1(::(@l1, @ls)) -> append(appendAll(@l1), appendAll2(@ls)) , appendAll3(@l) -> appendAll3#1(@l) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: Uargs(append) = {1, 2}, Uargs(::) = {2} TcT has computed following constructor-based matrix interpretation satisfying not(EDA). [append](x1, x2) = [1] x1 + [1] x2 + [2] [append#1](x1, x2) = [1] x1 + [1] x2 + [2] [::](x1, x2) = [1] x1 + [1] x2 + [2] [nil] = [0] [appendAll](x1) = [2] x1 + [0] [appendAll#1](x1) = [2] x1 + [0] [appendAll2](x1) = [2] x1 + [0] [appendAll2#1](x1) = [2] x1 + [0] [appendAll3](x1) = [2] x1 + [2] [appendAll3#1](x1) = [2] x1 + [0] This order satisfies following ordering constraints [append(@l1, @l2)] = [1] @l1 + [1] @l2 + [2] >= [1] @l1 + [1] @l2 + [2] = [append#1(@l1, @l2)] [append#1(::(@x, @xs), @l2)] = [1] @l2 + [1] @x + [1] @xs + [4] >= [1] @l2 + [1] @x + [1] @xs + [4] = [::(@x, append(@xs, @l2))] [append#1(nil(), @l2)] = [1] @l2 + [2] > [1] @l2 + [0] = [@l2] [appendAll(@l)] = [2] @l + [0] >= [2] @l + [0] = [appendAll#1(@l)] [appendAll#1(::(@l1, @ls))] = [2] @l1 + [2] @ls + [4] > [1] @l1 + [2] @ls + [2] = [append(@l1, appendAll(@ls))] [appendAll#1(nil())] = [0] >= [0] = [nil()] [appendAll2(@l)] = [2] @l + [0] >= [2] @l + [0] = [appendAll2#1(@l)] [appendAll2#1(::(@l1, @ls))] = [2] @l1 + [2] @ls + [4] > [2] @l1 + [2] @ls + [2] = [append(appendAll(@l1), appendAll2(@ls))] [appendAll2#1(nil())] = [0] >= [0] = [nil()] [appendAll3(@l)] = [2] @l + [2] > [2] @l + [0] = [appendAll3#1(@l)] [appendAll3#1(::(@l1, @ls))] = [2] @l1 + [2] @ls + [4] >= [2] @l1 + [2] @ls + [4] = [append(appendAll2(@l1), appendAll3(@ls))] [appendAll3#1(nil())] = [0] >= [0] = [nil()] We return to the main proof. 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)) , appendAll(@l) -> appendAll#1(@l) , appendAll3#1(::(@l1, @ls)) -> append(appendAll2(@l1), appendAll3(@ls)) } Weak Trs: { append#1(nil(), @l2) -> @l2 , 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(nil()) -> nil() } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { appendAll(@l) -> appendAll#1(@l) , appendAll3#1(::(@l1, @ls)) -> append(appendAll2(@l1), appendAll3(@ls)) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: Uargs(append) = {1, 2}, Uargs(::) = {2} 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 + [2] [nil] = [0] [appendAll](x1) = [1] x1 + [1] [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) = [2] x1 + [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 + [2] >= [1] @l2 + [1] @x + [1] @xs + [2] = [::(@x, append(@xs, @l2))] [append#1(nil(), @l2)] = [1] @l2 + [0] >= [1] @l2 + [0] = [@l2] [appendAll(@l)] = [1] @l + [1] > [1] @l + [0] = [appendAll#1(@l)] [appendAll#1(::(@l1, @ls))] = [1] @l1 + [1] @ls + [2] > [1] @l1 + [1] @ls + [1] = [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 + [2] > [1] @l1 + [1] @ls + [1] = [append(appendAll(@l1), appendAll2(@ls))] [appendAll2#1(nil())] = [0] >= [0] = [nil()] [appendAll3(@l)] = [2] @l + [0] >= [2] @l + [0] = [appendAll3#1(@l)] [appendAll3#1(::(@l1, @ls))] = [2] @l1 + [2] @ls + [4] > [1] @l1 + [2] @ls + [0] = [append(appendAll2(@l1), appendAll3(@ls))] [appendAll3#1(nil())] = [0] >= [0] = [nil()] We return to the main proof. 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)) } Weak Trs: { 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 use the processor 'matrix interpretation of dimension 2' to orient following rules strictly. Trs: { append#1(::(@x, @xs), @l2) -> ::(@x, append(@xs, @l2)) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^2)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: Uargs(append) = {1, 2}, Uargs(::) = {2} TcT has computed following constructor-based matrix interpretation satisfying not(EDA). [append](x1, x2) = [1 1] x1 + [1 0] x2 + [2] [0 1] [0 1] [2] [append#1](x1, x2) = [1 1] x1 + [1 0] x2 + [2] [0 1] [0 1] [2] [::](x1, x2) = [1 1] x1 + [1 0] x2 + [1] [0 1] [0 1] [2] [nil] = [0] [0] [appendAll](x1) = [1 2] x1 + [1] [0 2] [2] [appendAll#1](x1) = [1 2] x1 + [0] [0 2] [0] [appendAll2](x1) = [2 2] x1 + [0] [0 2] [0] [appendAll2#1](x1) = [2 2] x1 + [0] [0 2] [0] [appendAll3](x1) = [2 2] x1 + [0] [2 0] [0] [appendAll3#1](x1) = [2 2] x1 + [0] [2 0] [0] This order satisfies following ordering constraints [append(@l1, @l2)] = [1 1] @l1 + [1 0] @l2 + [2] [0 1] [0 1] [2] >= [1 1] @l1 + [1 0] @l2 + [2] [0 1] [0 1] [2] = [append#1(@l1, @l2)] [append#1(::(@x, @xs), @l2)] = [1 0] @l2 + [1 2] @x + [1 1] @xs + [5] [0 1] [0 1] [0 1] [4] > [1 0] @l2 + [1 1] @x + [1 1] @xs + [3] [0 1] [0 1] [0 1] [4] = [::(@x, append(@xs, @l2))] [append#1(nil(), @l2)] = [1 0] @l2 + [2] [0 1] [2] > [1 0] @l2 + [0] [0 1] [0] = [@l2] [appendAll(@l)] = [1 2] @l + [1] [0 2] [2] > [1 2] @l + [0] [0 2] [0] = [appendAll#1(@l)] [appendAll#1(::(@l1, @ls))] = [1 3] @l1 + [1 2] @ls + [5] [0 2] [0 2] [4] > [1 1] @l1 + [1 2] @ls + [3] [0 1] [0 2] [4] = [append(@l1, appendAll(@ls))] [appendAll#1(nil())] = [0] [0] >= [0] [0] = [nil()] [appendAll2(@l)] = [2 2] @l + [0] [0 2] [0] >= [2 2] @l + [0] [0 2] [0] = [appendAll2#1(@l)] [appendAll2#1(::(@l1, @ls))] = [2 4] @l1 + [2 2] @ls + [6] [0 2] [0 2] [4] > [1 4] @l1 + [2 2] @ls + [5] [0 2] [0 2] [4] = [append(appendAll(@l1), appendAll2(@ls))] [appendAll2#1(nil())] = [0] [0] >= [0] [0] = [nil()] [appendAll3(@l)] = [2 2] @l + [0] [2 0] [0] >= [2 2] @l + [0] [2 0] [0] = [appendAll3#1(@l)] [appendAll3#1(::(@l1, @ls))] = [2 4] @l1 + [2 2] @ls + [6] [2 2] [2 0] [2] > [2 4] @l1 + [2 2] @ls + [2] [0 2] [2 0] [2] = [append(appendAll2(@l1), appendAll3(@ls))] [appendAll3#1(nil())] = [0] [0] >= [0] [0] = [nil()] We return to the main proof. 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) } Weak Trs: { 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 use the processor 'matrix interpretation of dimension 2' to orient following rules strictly. Trs: { append(@l1, @l2) -> append#1(@l1, @l2) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^2)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: Uargs(append) = {1, 2}, Uargs(::) = {2} TcT has computed following constructor-based matrix interpretation satisfying not(EDA). [append](x1, x2) = [1 1] x1 + [1 0] x2 + [1] [0 2] [0 1] [0] [append#1](x1, x2) = [1 1] x1 + [1 0] x2 + [0] [0 2] [0 1] [0] [::](x1, x2) = [1 2] x1 + [1 0] x2 + [0] [0 1] [0 1] [1] [nil] = [0] [0] [appendAll](x1) = [1 1] x1 + [0] [0 2] [0] [appendAll#1](x1) = [1 1] x1 + [0] [0 2] [0] [appendAll2](x1) = [1 1] x1 + [1] [1 2] [0] [appendAll2#1](x1) = [1 1] x1 + [1] [1 2] [0] [appendAll3](x1) = [2 2] x1 + [2] [2 0] [0] [appendAll3#1](x1) = [2 2] x1 + [2] [2 0] [0] This order satisfies following ordering constraints [append(@l1, @l2)] = [1 1] @l1 + [1 0] @l2 + [1] [0 2] [0 1] [0] > [1 1] @l1 + [1 0] @l2 + [0] [0 2] [0 1] [0] = [append#1(@l1, @l2)] [append#1(::(@x, @xs), @l2)] = [1 0] @l2 + [1 3] @x + [1 1] @xs + [1] [0 1] [0 2] [0 2] [2] >= [1 0] @l2 + [1 2] @x + [1 1] @xs + [1] [0 1] [0 1] [0 2] [1] = [::(@x, append(@xs, @l2))] [append#1(nil(), @l2)] = [1 0] @l2 + [0] [0 1] [0] >= [1 0] @l2 + [0] [0 1] [0] = [@l2] [appendAll(@l)] = [1 1] @l + [0] [0 2] [0] >= [1 1] @l + [0] [0 2] [0] = [appendAll#1(@l)] [appendAll#1(::(@l1, @ls))] = [1 3] @l1 + [1 1] @ls + [1] [0 2] [0 2] [2] >= [1 1] @l1 + [1 1] @ls + [1] [0 2] [0 2] [0] = [append(@l1, appendAll(@ls))] [appendAll#1(nil())] = [0] [0] >= [0] [0] = [nil()] [appendAll2(@l)] = [1 1] @l + [1] [1 2] [0] >= [1 1] @l + [1] [1 2] [0] = [appendAll2#1(@l)] [appendAll2#1(::(@l1, @ls))] = [1 3] @l1 + [1 1] @ls + [2] [1 4] [1 2] [2] >= [1 3] @l1 + [1 1] @ls + [2] [0 4] [1 2] [0] = [append(appendAll(@l1), appendAll2(@ls))] [appendAll2#1(nil())] = [1] [0] > [0] [0] = [nil()] [appendAll3(@l)] = [2 2] @l + [2] [2 0] [0] >= [2 2] @l + [2] [2 0] [0] = [appendAll3#1(@l)] [appendAll3#1(::(@l1, @ls))] = [2 6] @l1 + [2 2] @ls + [4] [2 4] [2 0] [0] >= [2 3] @l1 + [2 2] @ls + [4] [2 4] [2 0] [0] = [append(appendAll2(@l1), appendAll3(@ls))] [appendAll3#1(nil())] = [2] [0] > [0] [0] = [nil()] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). 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(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(n^2))