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