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: { a__f(X) -> f(X) , a__f(f(a())) -> a__f(g(f(a()))) , mark(f(X)) -> a__f(mark(X)) , mark(a()) -> a() , mark(g(X)) -> g(X) } 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: { mark(f(X)) -> a__f(mark(X)) , mark(a()) -> a() } 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). [a__f](x1) = [1] x1 + [2] [f](x1) = [1] x1 + [2] [a] = [1] [g](x1) = [1] x1 + [0] [mark](x1) = [2] x1 + [0] This order satisfies the following ordering constraints: [a__f(X)] = [1] X + [2] >= [1] X + [2] = [f(X)] [a__f(f(a()))] = [5] >= [5] = [a__f(g(f(a())))] [mark(f(X))] = [2] X + [4] > [2] X + [2] = [a__f(mark(X))] [mark(a())] = [2] > [1] = [a()] [mark(g(X))] = [2] X + [0] >= [1] X + [0] = [g(X)] 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: { a__f(X) -> f(X) , a__f(f(a())) -> a__f(g(f(a()))) , mark(g(X)) -> g(X) } Weak Trs: { mark(f(X)) -> a__f(mark(X)) , mark(a()) -> a() } 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: { mark(g(X)) -> g(X) } 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). [a__f](x1) = [1] x1 + [0] [f](x1) = [1] x1 + [0] [a] = [0] [g](x1) = [1] x1 + [0] [mark](x1) = [2] x1 + [2] This order satisfies the following ordering constraints: [a__f(X)] = [1] X + [0] >= [1] X + [0] = [f(X)] [a__f(f(a()))] = [0] >= [0] = [a__f(g(f(a())))] [mark(f(X))] = [2] X + [2] >= [2] X + [2] = [a__f(mark(X))] [mark(a())] = [2] > [0] = [a()] [mark(g(X))] = [2] X + [2] > [1] X + [0] = [g(X)] 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: { a__f(X) -> f(X) , a__f(f(a())) -> a__f(g(f(a()))) } Weak Trs: { mark(f(X)) -> a__f(mark(X)) , mark(a()) -> a() , mark(g(X)) -> g(X) } 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: { a__f(X) -> f(X) } 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). [a__f](x1) = [1] x1 + [2] [f](x1) = [1] x1 + [1] [a] = [1] [g](x1) = [1] x1 + [0] [mark](x1) = [2] x1 + [2] This order satisfies the following ordering constraints: [a__f(X)] = [1] X + [2] > [1] X + [1] = [f(X)] [a__f(f(a()))] = [4] >= [4] = [a__f(g(f(a())))] [mark(f(X))] = [2] X + [4] >= [2] X + [4] = [a__f(mark(X))] [mark(a())] = [4] > [1] = [a()] [mark(g(X))] = [2] X + [2] > [1] X + [0] = [g(X)] 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: { a__f(f(a())) -> a__f(g(f(a()))) } Weak Trs: { a__f(X) -> f(X) , mark(f(X)) -> a__f(mark(X)) , mark(a()) -> a() , mark(g(X)) -> g(X) } 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. Trs: { a__f(f(a())) -> a__f(g(f(a()))) } 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) and not(IDA(1)). [a__f](x1) = [1 1] x1 + [0] [0 0] [1] [f](x1) = [1 1] x1 + [0] [0 0] [1] [a] = [0] [0] [g](x1) = [1 0] x1 + [0] [0 0] [0] [mark](x1) = [1 1] x1 + [0] [0 0] [1] This order satisfies the following ordering constraints: [a__f(X)] = [1 1] X + [0] [0 0] [1] >= [1 1] X + [0] [0 0] [1] = [f(X)] [a__f(f(a()))] = [1] [1] > [0] [1] = [a__f(g(f(a())))] [mark(f(X))] = [1 1] X + [1] [0 0] [1] >= [1 1] X + [1] [0 0] [1] = [a__f(mark(X))] [mark(a())] = [0] [1] >= [0] [0] = [a()] [mark(g(X))] = [1 0] X + [0] [0 0] [1] >= [1 0] X + [0] [0 0] [0] = [g(X)] 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: { a__f(X) -> f(X) , a__f(f(a())) -> a__f(g(f(a()))) , mark(f(X)) -> a__f(mark(X)) , mark(a()) -> a() , mark(g(X)) -> g(X) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(n^1))