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))