We consider the following Problem: Strict Trs: { f(f(X)) -> f(g(f(g(f(X))))) , f(g(f(X))) -> f(g(X))} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: We consider the following Problem: Strict Trs: { f(f(X)) -> f(g(f(g(f(X))))) , f(g(f(X))) -> f(g(X))} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: The weightgap principle applies, where following rules are oriented strictly: TRS Component: {f(g(f(X))) -> f(g(X))} Interpretation of nonconstant growth: ------------------------------------- We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation: Interpretation Functions: f(x1) = [1 0] x1 + [2] [0 0] [1] g(x1) = [1 0] x1 + [0] [0 0] [0] The strictly oriented rules are moved into the weak component. We consider the following Problem: Strict Trs: {f(f(X)) -> f(g(f(g(f(X)))))} Weak Trs: {f(g(f(X))) -> f(g(X))} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: The weightgap principle applies, where following rules are oriented strictly: TRS Component: {f(f(X)) -> f(g(f(g(f(X)))))} Interpretation of nonconstant growth: ------------------------------------- We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation: Interpretation Functions: f(x1) = [1 2] x1 + [0] [0 0] [1] g(x1) = [1 0] x1 + [0] [0 0] [0] The strictly oriented rules are moved into the weak component. We consider the following Problem: Weak Trs: { f(f(X)) -> f(g(f(g(f(X))))) , f(g(f(X))) -> f(g(X))} StartTerms: basic terms Strategy: innermost Certificate: YES(O(1),O(1)) Proof: We consider the following Problem: Weak Trs: { f(f(X)) -> f(g(f(g(f(X))))) , f(g(f(X))) -> f(g(X))} StartTerms: basic terms Strategy: innermost Certificate: YES(O(1),O(1)) Proof: Empty rules are trivially bounded Hurray, we answered YES(?,O(n^1))