We consider the following Problem: Strict Trs: { +(*(x, y), *(a(), y)) -> *(+(x, a()), y) , *(*(x, y), z) -> *(x, *(y, z))} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: We consider the following Problem: Strict Trs: { +(*(x, y), *(a(), y)) -> *(+(x, a()), y) , *(*(x, y), z) -> *(x, *(y, z))} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: The weightgap principle applies, where following rules are oriented strictly: TRS Component: {+(*(x, y), *(a(), y)) -> *(+(x, a()), y)} Interpretation of nonconstant growth: ------------------------------------- The following argument positions are usable: Uargs(+) = {}, Uargs(*) = {2} We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation: Interpretation Functions: +(x1, x2) = [0 0] x1 + [1 0] x2 + [1] [0 0] [1 0] [1] *(x1, x2) = [0 0] x1 + [1 0] x2 + [0] [0 0] [0 0] [1] a() = [0] [0] The strictly oriented rules are moved into the weak component. We consider the following Problem: Strict Trs: {*(*(x, y), z) -> *(x, *(y, z))} Weak Trs: {+(*(x, y), *(a(), y)) -> *(+(x, a()), y)} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: The weightgap principle applies, where following rules are oriented strictly: TRS Component: {*(*(x, y), z) -> *(x, *(y, z))} Interpretation of nonconstant growth: ------------------------------------- The following argument positions are usable: Uargs(+) = {}, Uargs(*) = {2} We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation: Interpretation Functions: +(x1, x2) = [0 0] x1 + [1 0] x2 + [1] [0 0] [0 1] [0] *(x1, x2) = [0 3] x1 + [1 0] x2 + [0] [0 1] [0 1] [3] a() = [0] [0] The strictly oriented rules are moved into the weak component. We consider the following Problem: Weak Trs: { *(*(x, y), z) -> *(x, *(y, z)) , +(*(x, y), *(a(), y)) -> *(+(x, a()), y)} StartTerms: basic terms Strategy: innermost Certificate: YES(O(1),O(1)) Proof: We consider the following Problem: Weak Trs: { *(*(x, y), z) -> *(x, *(y, z)) , +(*(x, y), *(a(), y)) -> *(+(x, a()), y)} StartTerms: basic terms Strategy: innermost Certificate: YES(O(1),O(1)) Proof: Empty rules are trivially bounded Hurray, we answered YES(?,O(n^1))