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: { from(X) -> cons(X, n__from(s(X))) , from(X) -> n__from(X) , sel(s(X), cons(Y, Z)) -> sel(X, activate(Z)) , sel(0(), cons(X, Y)) -> X , activate(X) -> X , activate(n__from(X)) -> from(X) } Obligation: runtime complexity Answer: YES(O(1),O(n^1)) We add the following weak dependency pairs: Strict DPs: { from^#(X) -> c_1(X, X) , from^#(X) -> c_2(X) , sel^#(s(X), cons(Y, Z)) -> c_3(sel^#(X, activate(Z))) , sel^#(0(), cons(X, Y)) -> c_4(X) , activate^#(X) -> c_5(X) , activate^#(n__from(X)) -> c_6(from^#(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: { from^#(X) -> c_1(X, X) , from^#(X) -> c_2(X) , sel^#(s(X), cons(Y, Z)) -> c_3(sel^#(X, activate(Z))) , sel^#(0(), cons(X, Y)) -> c_4(X) , activate^#(X) -> c_5(X) , activate^#(n__from(X)) -> c_6(from^#(X)) } Strict Trs: { from(X) -> cons(X, n__from(s(X))) , from(X) -> n__from(X) , sel(s(X), cons(Y, Z)) -> sel(X, activate(Z)) , sel(0(), cons(X, Y)) -> X , activate(X) -> X , activate(n__from(X)) -> from(X) } Obligation: runtime complexity Answer: YES(O(1),O(n^1)) We replace rewrite rules by usable rules: Strict Usable Rules: { from(X) -> cons(X, n__from(s(X))) , from(X) -> n__from(X) , activate(X) -> X , activate(n__from(X)) -> from(X) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { from^#(X) -> c_1(X, X) , from^#(X) -> c_2(X) , sel^#(s(X), cons(Y, Z)) -> c_3(sel^#(X, activate(Z))) , sel^#(0(), cons(X, Y)) -> c_4(X) , activate^#(X) -> c_5(X) , activate^#(n__from(X)) -> c_6(from^#(X)) } Strict Trs: { from(X) -> cons(X, n__from(s(X))) , from(X) -> n__from(X) , activate(X) -> X , activate(n__from(X)) -> from(X) } Obligation: runtime complexity Answer: YES(O(1),O(n^1)) The weightgap principle applies (using the following constant growth matrix-interpretation) The following argument positions are usable: Uargs(sel^#) = {2}, Uargs(c_3) = {1}, Uargs(c_6) = {1} TcT has computed the following constructor-restricted matrix interpretation. [from](x1) = [1] [0] [cons](x1, x2) = [1 0] x2 + [0] [0 0] [0] [n__from](x1) = [0] [0] [s](x1) = [1 2] x1 + [0] [0 0] [0] [0] = [0] [0] [activate](x1) = [1 0] x1 + [2] [0 2] [0] [from^#](x1) = [0 0] x1 + [1] [1 2] [1] [c_1](x1, x2) = [0 0] x1 + [0 0] x2 + [1] [1 2] [2 2] [1] [c_2](x1) = [0 0] x1 + [2] [1 1] [1] [sel^#](x1, x2) = [1 0] x2 + [0] [0 0] [0] [c_3](x1) = [1 0] x1 + [2] [0 1] [2] [c_4](x1) = [1] [0] [activate^#](x1) = [1 1] x1 + [2] [1 1] [2] [c_5](x1) = [1 1] x1 + [1] [1 1] [1] [c_6](x1) = [1 0] x1 + [2] [0 1] [2] The following symbols are considered usable {from, activate, from^#, sel^#, activate^#} The order satisfies the following ordering constraints: [from(X)] = [1] [0] > [0] [0] = [cons(X, n__from(s(X)))] [from(X)] = [1] [0] > [0] [0] = [n__from(X)] [activate(X)] = [1 0] X + [2] [0 2] [0] > [1 0] X + [0] [0 1] [0] = [X] [activate(n__from(X))] = [2] [0] > [1] [0] = [from(X)] [from^#(X)] = [0 0] X + [1] [1 2] [1] ? [0 0] X + [1] [3 4] [1] = [c_1(X, X)] [from^#(X)] = [0 0] X + [1] [1 2] [1] ? [0 0] X + [2] [1 1] [1] = [c_2(X)] [sel^#(s(X), cons(Y, Z))] = [1 0] Z + [0] [0 0] [0] ? [1 0] Z + [4] [0 0] [2] = [c_3(sel^#(X, activate(Z)))] [sel^#(0(), cons(X, Y))] = [1 0] Y + [0] [0 0] [0] ? [1] [0] = [c_4(X)] [activate^#(X)] = [1 1] X + [2] [1 1] [2] > [1 1] X + [1] [1 1] [1] = [c_5(X)] [activate^#(n__from(X))] = [2] [2] ? [0 0] X + [3] [1 2] [3] = [c_6(from^#(X))] 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: { from^#(X) -> c_1(X, X) , from^#(X) -> c_2(X) , sel^#(s(X), cons(Y, Z)) -> c_3(sel^#(X, activate(Z))) , sel^#(0(), cons(X, Y)) -> c_4(X) , activate^#(n__from(X)) -> c_6(from^#(X)) } Weak DPs: { activate^#(X) -> c_5(X) } Weak Trs: { from(X) -> cons(X, n__from(s(X))) , from(X) -> n__from(X) , activate(X) -> X , activate(n__from(X)) -> from(X) } Obligation: runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. DPs: { 5: activate^#(n__from(X)) -> c_6(from^#(X)) , 6: activate^#(X) -> c_5(X) } Sub-proof: ---------- The following argument positions are usable: Uargs(c_3) = {1}, Uargs(c_6) = {1} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [from](x1) = [1] x1 + [0] [cons](x1, x2) = [1] x1 + [0] [n__from](x1) = [1] x1 + [0] [s](x1) = [1] x1 + [0] [sel](x1, x2) = [7] x1 + [7] x2 + [0] [0] = [7] [activate](x1) = [1] x1 + [0] [from^#](x1) = [1] x1 + [0] [c_1](x1, x2) = [1] x2 + [0] [c_2](x1) = [1] x1 + [0] [sel^#](x1, x2) = [0] [c_3](x1) = [4] x1 + [0] [c_4](x1) = [0] [activate^#](x1) = [5] x1 + [7] [c_5](x1) = [5] x1 + [6] [c_6](x1) = [4] x1 + [1] The following symbols are considered usable {from, activate, from^#, sel^#, activate^#} The order satisfies the following ordering constraints: [from(X)] = [1] X + [0] >= [1] X + [0] = [cons(X, n__from(s(X)))] [from(X)] = [1] X + [0] >= [1] X + [0] = [n__from(X)] [activate(X)] = [1] X + [0] >= [1] X + [0] = [X] [activate(n__from(X))] = [1] X + [0] >= [1] X + [0] = [from(X)] [from^#(X)] = [1] X + [0] >= [1] X + [0] = [c_1(X, X)] [from^#(X)] = [1] X + [0] >= [1] X + [0] = [c_2(X)] [sel^#(s(X), cons(Y, Z))] = [0] >= [0] = [c_3(sel^#(X, activate(Z)))] [sel^#(0(), cons(X, Y))] = [0] >= [0] = [c_4(X)] [activate^#(X)] = [5] X + [7] > [5] X + [6] = [c_5(X)] [activate^#(n__from(X))] = [5] X + [7] > [4] X + [1] = [c_6(from^#(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: { from^#(X) -> c_1(X, X) , from^#(X) -> c_2(X) , sel^#(s(X), cons(Y, Z)) -> c_3(sel^#(X, activate(Z))) , sel^#(0(), cons(X, Y)) -> c_4(X) } Weak DPs: { activate^#(X) -> c_5(X) , activate^#(n__from(X)) -> c_6(from^#(X)) } Weak Trs: { from(X) -> cons(X, n__from(s(X))) , from(X) -> n__from(X) , activate(X) -> X , activate(n__from(X)) -> from(X) } Obligation: runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. DPs: { 1: from^#(X) -> c_1(X, X) , 2: from^#(X) -> c_2(X) , 5: activate^#(X) -> c_5(X) , 6: activate^#(n__from(X)) -> c_6(from^#(X)) } Sub-proof: ---------- The following argument positions are usable: Uargs(c_3) = {1}, Uargs(c_6) = {1} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [from](x1) = [1] x1 + [0] [cons](x1, x2) = [1] x1 + [0] [n__from](x1) = [1] x1 + [0] [s](x1) = [1] x1 + [0] [sel](x1, x2) = [7] x1 + [7] x2 + [0] [0] = [7] [activate](x1) = [1] x1 + [0] [from^#](x1) = [7] x1 + [1] [c_1](x1, x2) = [3] x1 + [3] x2 + [0] [c_2](x1) = [7] x1 + [0] [sel^#](x1, x2) = [0] [c_3](x1) = [4] x1 + [0] [c_4](x1) = [0] [activate^#](x1) = [7] x1 + [7] [c_5](x1) = [7] x1 + [6] [c_6](x1) = [1] x1 + [0] The following symbols are considered usable {from, activate, from^#, sel^#, activate^#} The order satisfies the following ordering constraints: [from(X)] = [1] X + [0] >= [1] X + [0] = [cons(X, n__from(s(X)))] [from(X)] = [1] X + [0] >= [1] X + [0] = [n__from(X)] [activate(X)] = [1] X + [0] >= [1] X + [0] = [X] [activate(n__from(X))] = [1] X + [0] >= [1] X + [0] = [from(X)] [from^#(X)] = [7] X + [1] > [6] X + [0] = [c_1(X, X)] [from^#(X)] = [7] X + [1] > [7] X + [0] = [c_2(X)] [sel^#(s(X), cons(Y, Z))] = [0] >= [0] = [c_3(sel^#(X, activate(Z)))] [sel^#(0(), cons(X, Y))] = [0] >= [0] = [c_4(X)] [activate^#(X)] = [7] X + [7] > [7] X + [6] = [c_5(X)] [activate^#(n__from(X))] = [7] X + [7] > [7] X + [1] = [c_6(from^#(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: { sel^#(s(X), cons(Y, Z)) -> c_3(sel^#(X, activate(Z))) , sel^#(0(), cons(X, Y)) -> c_4(X) } Weak DPs: { from^#(X) -> c_1(X, X) , from^#(X) -> c_2(X) , activate^#(X) -> c_5(X) , activate^#(n__from(X)) -> c_6(from^#(X)) } Weak Trs: { from(X) -> cons(X, n__from(s(X))) , from(X) -> n__from(X) , activate(X) -> X , activate(n__from(X)) -> from(X) } Obligation: runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. DPs: { 1: sel^#(s(X), cons(Y, Z)) -> c_3(sel^#(X, activate(Z))) , 5: activate^#(X) -> c_5(X) , 6: activate^#(n__from(X)) -> c_6(from^#(X)) } Trs: { from(X) -> cons(X, n__from(s(X))) , from(X) -> n__from(X) , activate(X) -> X , activate(n__from(X)) -> from(X) } Sub-proof: ---------- The following argument positions are usable: Uargs(c_3) = {1}, Uargs(c_6) = {1} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [from](x1) = [1] x1 + [4] [cons](x1, x2) = [1] x1 + [0] [n__from](x1) = [1] x1 + [1] [s](x1) = [1] x1 + [2] [sel](x1, x2) = [7] x1 + [7] x2 + [0] [0] = [0] [activate](x1) = [1] x1 + [7] [from^#](x1) = [0] [c_1](x1, x2) = [0] [c_2](x1) = [0] [sel^#](x1, x2) = [4] x1 + [0] [c_3](x1) = [1] x1 + [1] [c_4](x1) = [0] [activate^#](x1) = [7] [c_5](x1) = [6] [c_6](x1) = [4] x1 + [1] The following symbols are considered usable {from, activate, from^#, sel^#, activate^#} The order satisfies the following ordering constraints: [from(X)] = [1] X + [4] > [1] X + [0] = [cons(X, n__from(s(X)))] [from(X)] = [1] X + [4] > [1] X + [1] = [n__from(X)] [activate(X)] = [1] X + [7] > [1] X + [0] = [X] [activate(n__from(X))] = [1] X + [8] > [1] X + [4] = [from(X)] [from^#(X)] = [0] >= [0] = [c_1(X, X)] [from^#(X)] = [0] >= [0] = [c_2(X)] [sel^#(s(X), cons(Y, Z))] = [4] X + [8] > [4] X + [1] = [c_3(sel^#(X, activate(Z)))] [sel^#(0(), cons(X, Y))] = [0] >= [0] = [c_4(X)] [activate^#(X)] = [7] > [6] = [c_5(X)] [activate^#(n__from(X))] = [7] > [1] = [c_6(from^#(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(1)). Strict DPs: { sel^#(0(), cons(X, Y)) -> c_4(X) } Weak DPs: { from^#(X) -> c_1(X, X) , from^#(X) -> c_2(X) , sel^#(s(X), cons(Y, Z)) -> c_3(sel^#(X, activate(Z))) , activate^#(X) -> c_5(X) , activate^#(n__from(X)) -> c_6(from^#(X)) } Weak Trs: { from(X) -> cons(X, n__from(s(X))) , from(X) -> n__from(X) , activate(X) -> X , activate(n__from(X)) -> from(X) } Obligation: runtime complexity Answer: YES(O(1),O(1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. DPs: { 1: sel^#(0(), cons(X, Y)) -> c_4(X) , 5: activate^#(X) -> c_5(X) , 6: activate^#(n__from(X)) -> c_6(from^#(X)) } Sub-proof: ---------- The following argument positions are usable: Uargs(c_3) = {1}, Uargs(c_6) = {1} TcT has computed the following constructor-restricted matrix interpretation. Note that the diagonal of the component-wise maxima of interpretation-entries (of constructors) contains no more than 0 non-zero entries. [from](x1) = [0] [cons](x1, x2) = [0] [n__from](x1) = [0] [s](x1) = [7] [sel](x1, x2) = [0] [0] = [7] [activate](x1) = [4] x1 + [0] [from^#](x1) = [0] [c_1](x1, x2) = [0] [c_2](x1) = [0] [sel^#](x1, x2) = [1] [c_3](x1) = [1] x1 + [0] [c_4](x1) = [0] [activate^#](x1) = [7] x1 + [7] [c_5](x1) = [7] x1 + [5] [c_6](x1) = [4] x1 + [1] The following symbols are considered usable {from, activate, from^#, sel^#, activate^#} The order satisfies the following ordering constraints: [from(X)] = [0] >= [0] = [cons(X, n__from(s(X)))] [from(X)] = [0] >= [0] = [n__from(X)] [activate(X)] = [4] X + [0] >= [1] X + [0] = [X] [activate(n__from(X))] = [0] >= [0] = [from(X)] [from^#(X)] = [0] >= [0] = [c_1(X, X)] [from^#(X)] = [0] >= [0] = [c_2(X)] [sel^#(s(X), cons(Y, Z))] = [1] >= [1] = [c_3(sel^#(X, activate(Z)))] [sel^#(0(), cons(X, Y))] = [1] > [0] = [c_4(X)] [activate^#(X)] = [7] X + [7] > [7] X + [5] = [c_5(X)] [activate^#(n__from(X))] = [7] > [1] = [c_6(from^#(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(1)). Weak DPs: { from^#(X) -> c_1(X, X) , from^#(X) -> c_2(X) , sel^#(s(X), cons(Y, Z)) -> c_3(sel^#(X, activate(Z))) , sel^#(0(), cons(X, Y)) -> c_4(X) , activate^#(X) -> c_5(X) , activate^#(n__from(X)) -> c_6(from^#(X)) } Weak Trs: { from(X) -> cons(X, n__from(s(X))) , from(X) -> n__from(X) , activate(X) -> X , activate(n__from(X)) -> from(X) } Obligation: 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. { from^#(X) -> c_1(X, X) , from^#(X) -> c_2(X) , sel^#(s(X), cons(Y, Z)) -> c_3(sel^#(X, activate(Z))) , sel^#(0(), cons(X, Y)) -> c_4(X) , activate^#(X) -> c_5(X) , activate^#(n__from(X)) -> c_6(from^#(X)) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak Trs: { from(X) -> cons(X, n__from(s(X))) , from(X) -> n__from(X) , activate(X) -> X , activate(n__from(X)) -> from(X) } Obligation: 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: runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(n^1))