We consider the following Problem: Strict Trs: { a(a(f(b(), a(x)))) -> f(a(a(a(x))), b()) , a(a(x)) -> f(b(), a(f(a(x), b()))) , f(a(x), b()) -> f(b(), a(x))} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: We consider the following Problem: Strict Trs: { a(a(f(b(), a(x)))) -> f(a(a(a(x))), b()) , a(a(x)) -> f(b(), a(f(a(x), b()))) , f(a(x), b()) -> f(b(), a(x))} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: The weightgap principle applies, where following rules are oriented strictly: TRS Component: {a(a(f(b(), a(x)))) -> f(a(a(a(x))), b())} Interpretation of nonconstant growth: ------------------------------------- We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation: Interpretation Functions: a(x1) = [1 0] x1 + [2] [0 0] [0] f(x1, x2) = [1 1] x1 + [1 0] x2 + [1] [0 0] [0 0] [0] b() = [0] [1] The strictly oriented rules are moved into the weak component. We consider the following Problem: Strict Trs: { a(a(x)) -> f(b(), a(f(a(x), b()))) , f(a(x), b()) -> f(b(), a(x))} Weak Trs: {a(a(f(b(), a(x)))) -> f(a(a(a(x))), b())} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: The weightgap principle applies, where following rules are oriented strictly: TRS Component: {a(a(x)) -> f(b(), a(f(a(x), b())))} Interpretation of nonconstant growth: ------------------------------------- We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation: Interpretation Functions: a(x1) = [1 2] x1 + [0] [0 0] [2] f(x1, x2) = [1 0] x1 + [1 0] x2 + [0] [0 0] [0 1] [0] b() = [0] [0] The strictly oriented rules are moved into the weak component. We consider the following Problem: Strict Trs: {f(a(x), b()) -> f(b(), a(x))} Weak Trs: { a(a(x)) -> f(b(), a(f(a(x), b()))) , a(a(f(b(), a(x)))) -> f(a(a(a(x))), b())} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: We consider the following Problem: Strict Trs: {f(a(x), b()) -> f(b(), a(x))} Weak Trs: { a(a(x)) -> f(b(), a(f(a(x), b()))) , a(a(f(b(), a(x)))) -> f(a(a(a(x))), b())} 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: { a_0(2) -> 1 , f_0(2, 2) -> 1 , b_0() -> 2} Hurray, we answered YES(?,O(n^1))