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: { app(nil(), xs) -> nil() , app(cons(x, xs), ys) -> cons(x, app(xs, ys)) } 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. Trs: { app(cons(x, xs), ys) -> cons(x, app(xs, ys)) } 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: ---------- TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [app](x1, x2) = [2] x1 + [3] x2 + [0] [nil] = [0] [cons](x1, x2) = [1] x1 + [1] x2 + [2] This order satisfies the following ordering constraints: [app(nil(), xs)] = [3] xs + [0] >= [0] = [nil()] [app(cons(x, xs), ys)] = [2] xs + [2] x + [3] ys + [4] > [2] xs + [1] x + [3] ys + [2] = [cons(x, app(xs, ys))] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { app(nil(), xs) -> nil() } Weak Trs: { app(cons(x, xs), ys) -> cons(x, app(xs, ys)) } 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. Trs: { app(nil(), xs) -> 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: ---------- TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [app](x1, x2) = [2] x1 + [3] x2 + [1] [nil] = [0] [cons](x1, x2) = [1] x1 + [1] x2 + [0] This order satisfies the following ordering constraints: [app(nil(), xs)] = [3] xs + [1] > [0] = [nil()] [app(cons(x, xs), ys)] = [2] xs + [2] x + [3] ys + [1] >= [2] xs + [1] x + [3] ys + [1] = [cons(x, app(xs, ys))] 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: { app(nil(), xs) -> nil() , app(cons(x, xs), ys) -> cons(x, app(xs, ys)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(n^1))