We consider the following Problem: Strict Trs: { f(a(), f(b(), f(a(), x))) -> f(a(), f(b(), f(b(), f(a(), x)))) , f(b(), f(b(), f(b(), x))) -> f(b(), f(b(), x))} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: We consider the following Problem: Strict Trs: { f(a(), f(b(), f(a(), x))) -> f(a(), f(b(), f(b(), f(a(), x)))) , f(b(), f(b(), f(b(), x))) -> f(b(), f(b(), 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(b(), f(b(), f(b(), x))) -> f(b(), f(b(), x))} Interpretation of nonconstant growth: ------------------------------------- The following argument positions are usable: Uargs(f) = {} We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation: Interpretation Functions: f(x1, x2) = [0 2] x1 + [1 0] x2 + [0] [0 0] [0 0] [1] a() = [0] [0] b() = [0] [2] The strictly oriented rules are moved into the weak component. We consider the following Problem: Strict Trs: {f(a(), f(b(), f(a(), x))) -> f(a(), f(b(), f(b(), f(a(), x))))} Weak Trs: {f(b(), f(b(), f(b(), x))) -> f(b(), f(b(), x))} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: We consider the following Problem: Strict Trs: {f(a(), f(b(), f(a(), x))) -> f(a(), f(b(), f(b(), f(a(), x))))} Weak Trs: {f(b(), f(b(), f(b(), x))) -> f(b(), f(b(), x))} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: The problem is match-bounded by 0. The enriched problem is compatible with the following automaton: { f_0(2, 2) -> 1 , a_0() -> 2 , b_0() -> 2} Hurray, we answered YES(?,O(n^1))