We consider the following Problem: Strict Trs: { f(0(), 1(), x) -> f(s(x), x, x) , f(x, y, s(z)) -> s(f(0(), 1(), z))} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^2)) Proof: We consider the following Problem: Strict Trs: { f(0(), 1(), x) -> f(s(x), x, x) , f(x, y, s(z)) -> s(f(0(), 1(), z))} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^2)) Proof: The weightgap principle applies, where following rules are oriented strictly: TRS Component: {f(0(), 1(), x) -> f(s(x), x, x)} Interpretation of nonconstant growth: ------------------------------------- The following argument positions are usable: Uargs(f) = {}, Uargs(s) = {1} We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation: Interpretation Functions: f(x1, x2, x3) = [0 2] x1 + [0 0] x2 + [0 1] x3 + [1] [0 0] [0 0] [0 0] [1] 0() = [0] [2] 1() = [0] [0] s(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(x, y, s(z)) -> s(f(0(), 1(), z))} Weak Trs: {f(0(), 1(), x) -> f(s(x), x, x)} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^2)) Proof: We consider the following Problem: Strict Trs: {f(x, y, s(z)) -> s(f(0(), 1(), z))} Weak Trs: {f(0(), 1(), x) -> f(s(x), x, x)} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^2)) Proof: The following argument positions are usable: Uargs(f) = {}, Uargs(s) = {1} We have the following restricted polynomial interpretation: Interpretation Functions: [f](x1, x2, x3) = x3^2 [0]() = 2 [1]() = 2 [s](x1) = 2 + x1 Hurray, we answered YES(?,O(n^2))