We consider the following Problem: Strict Trs: { c(z, x, a()) -> f(b(b(f(z), z), x)) , b(y, b(z, a())) -> f(b(c(f(a()), y, z), z)) , f(c(c(z, a(), a()), x, a())) -> z} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: Arguments of following rules are not normal-forms: {f(c(c(z, a(), a()), x, a())) -> z} All above mentioned rules can be savely removed. We consider the following Problem: Strict Trs: { c(z, x, a()) -> f(b(b(f(z), z), x)) , b(y, b(z, a())) -> f(b(c(f(a()), y, z), z))} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: The weightgap principle applies, where following rules are oriented strictly: TRS Component: {c(z, x, a()) -> f(b(b(f(z), z), x))} Interpretation of nonconstant growth: ------------------------------------- The following argument positions are usable: Uargs(c) = {}, Uargs(f) = {1}, Uargs(b) = {1} We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation: Interpretation Functions: c(x1, x2, x3) = [1 0] x1 + [0 0] x2 + [0 0] x3 + [1] [0 0] [0 0] [0 0] [1] a() = [0] [0] f(x1) = [1 0] x1 + [0] [0 0] [1] b(x1, x2) = [1 0] x1 + [0 0] x2 + [0] [0 0] [1 0] [1] The strictly oriented rules are moved into the weak component. We consider the following Problem: Strict Trs: {b(y, b(z, a())) -> f(b(c(f(a()), y, z), z))} Weak Trs: {c(z, x, a()) -> f(b(b(f(z), z), x))} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: We consider the following Problem: Strict Trs: {b(y, b(z, a())) -> f(b(c(f(a()), y, z), z))} Weak Trs: {c(z, x, a()) -> f(b(b(f(z), z), x))} StartTerms: basic terms Strategy: innermost Certificate: YES(?,O(n^1)) Proof: The following argument positions are usable: Uargs(c) = {}, Uargs(f) = {1}, Uargs(b) = {1} We have the following constructor-based EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation: Interpretation Functions: c(x1, x2, x3) = [2 0] x1 + [1 0] x2 + [0 2] x3 + [0] [0 0] [0 1] [0 0] [0] a() = [0] [2] f(x1) = [1 0] x1 + [0] [0 0] [0] b(x1, x2) = [1 2] x1 + [1 0] x2 + [1] [0 0] [0 0] [0] Hurray, we answered YES(?,O(n^1))