Tool CaT
| Execution Time | Unknown |
|---|
| Answer | MAYBE |
|---|
| Input | SK90 2.13 |
|---|
stdout:
MAYBE
Problem:
double(0()) -> 0()
double(s(x)) -> s(s(double(x)))
+(x,0()) -> x
+(x,s(y)) -> s(+(x,y))
+(s(x),y) -> s(+(x,y))
double(x) -> +(x,x)
Proof:
OpenTool IRC1
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^1)) |
|---|
| Input | SK90 2.13 |
|---|
stdout:
YES(?,O(n^1))
Tool IRC2
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^1)) |
|---|
| Input | SK90 2.13 |
|---|
stdout:
YES(?,O(n^1))
'Fastest (timeout of 60.0 seconds)'
-----------------------------------
Answer: YES(?,O(n^1))
Input Problem: innermost runtime-complexity with respect to
Rules:
{ double(0()) -> 0()
, double(s(x)) -> s(s(double(x)))
, +(x, 0()) -> x
, +(x, s(y)) -> s(+(x, y))
, +(s(x), y) -> s(+(x, y))
, double(x) -> +(x, x)}
Proof Output:
'matrix-interpretation of dimension 1' proved the best result:
Details:
--------
'matrix-interpretation of dimension 1' succeeded with the following output:
'matrix-interpretation of dimension 1'
--------------------------------------
Answer: YES(?,O(n^1))
Input Problem: innermost runtime-complexity with respect to
Rules:
{ double(0()) -> 0()
, double(s(x)) -> s(s(double(x)))
, +(x, 0()) -> x
, +(x, s(y)) -> s(+(x, y))
, +(s(x), y) -> s(+(x, y))
, double(x) -> +(x, x)}
Proof Output:
The following argument positions are usable:
Uargs(double) = {}, Uargs(s) = {1}, Uargs(+) = {}
We have the following constructor-restricted matrix interpretation:
Interpretation Functions:
double(x1) = [6] x1 + [1]
0() = [1]
s(x1) = [1] x1 + [2]
+(x1, x2) = [2] x1 + [4] x2 + [0]Tool RC1
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^1)) |
|---|
| Input | SK90 2.13 |
|---|
stdout:
YES(?,O(n^1))
Tool RC2
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^1)) |
|---|
| Input | SK90 2.13 |
|---|
stdout:
YES(?,O(n^1))
'Fastest (timeout of 60.0 seconds)'
-----------------------------------
Answer: YES(?,O(n^1))
Input Problem: runtime-complexity with respect to
Rules:
{ double(0()) -> 0()
, double(s(x)) -> s(s(double(x)))
, +(x, 0()) -> x
, +(x, s(y)) -> s(+(x, y))
, +(s(x), y) -> s(+(x, y))
, double(x) -> +(x, x)}
Proof Output:
'matrix-interpretation of dimension 1' proved the best result:
Details:
--------
'matrix-interpretation of dimension 1' succeeded with the following output:
'matrix-interpretation of dimension 1'
--------------------------------------
Answer: YES(?,O(n^1))
Input Problem: runtime-complexity with respect to
Rules:
{ double(0()) -> 0()
, double(s(x)) -> s(s(double(x)))
, +(x, 0()) -> x
, +(x, s(y)) -> s(+(x, y))
, +(s(x), y) -> s(+(x, y))
, double(x) -> +(x, x)}
Proof Output:
The following argument positions are usable:
Uargs(double) = {}, Uargs(s) = {1}, Uargs(+) = {}
We have the following constructor-restricted matrix interpretation:
Interpretation Functions:
double(x1) = [6] x1 + [1]
0() = [1]
s(x1) = [1] x1 + [2]
+(x1, x2) = [2] x1 + [4] x2 + [0]Tool pair1rc
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^2)) |
|---|
| Input | SK90 2.13 |
|---|
stdout:
YES(?,O(n^2))
We consider the following Problem:
Strict Trs:
{ double(0()) -> 0()
, double(s(x)) -> s(s(double(x)))
, +(x, 0()) -> x
, +(x, s(y)) -> s(+(x, y))
, +(s(x), y) -> s(+(x, y))
, double(x) -> +(x, x)}
StartTerms: basic terms
Strategy: none
Certificate: YES(?,O(n^2))
Application of 'pair1 (timeout of 60.0 seconds)':
-------------------------------------------------
The processor is not applicable, reason is:
Input problem is not restricted to innermost rewriting
We abort the transformation and continue with the subprocessor on the problem
Strict Trs:
{ double(0()) -> 0()
, double(s(x)) -> s(s(double(x)))
, +(x, 0()) -> x
, +(x, s(y)) -> s(+(x, y))
, +(s(x), y) -> s(+(x, y))
, double(x) -> +(x, x)}
StartTerms: basic terms
Strategy: none
1) 'Fastest' proved the goal fastest:
'Sequentially' proved the goal fastest:
'Fastest' succeeded:
'matrix-interpretation of dimension 3 (timeout of 100.0 seconds)' proved the goal fastest:
The following argument positions are usable:
Uargs(double) = {}, Uargs(s) = {1}, Uargs(+) = {}
We have the following constructor-restricted (at most 2 in the main diagonals) matrix interpretation:
Interpretation Functions:
double(x1) = [2 0 2] x1 + [1]
[0 1 2] [0]
[0 0 2] [2]
0() = [0]
[0]
[2]
s(x1) = [1 0 0] x1 + [1]
[0 0 1] [0]
[0 0 1] [2]
+(x1, x2) = [2 0 0] x1 + [0 0 2] x2 + [0]
[0 1 1] [0 0 1] [0]
[0 0 1] [0 0 1] [2]
Hurray, we answered YES(?,O(n^2))Tool pair2rc
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^2)) |
|---|
| Input | SK90 2.13 |
|---|
stdout:
YES(?,O(n^2))
We consider the following Problem:
Strict Trs:
{ double(0()) -> 0()
, double(s(x)) -> s(s(double(x)))
, +(x, 0()) -> x
, +(x, s(y)) -> s(+(x, y))
, +(s(x), y) -> s(+(x, y))
, double(x) -> +(x, x)}
StartTerms: basic terms
Strategy: none
Certificate: YES(?,O(n^2))
Application of 'pair2 (timeout of 60.0 seconds)':
-------------------------------------------------
The processor is not applicable, reason is:
Input problem is not restricted to innermost rewriting
We abort the transformation and continue with the subprocessor on the problem
Strict Trs:
{ double(0()) -> 0()
, double(s(x)) -> s(s(double(x)))
, +(x, 0()) -> x
, +(x, s(y)) -> s(+(x, y))
, +(s(x), y) -> s(+(x, y))
, double(x) -> +(x, x)}
StartTerms: basic terms
Strategy: none
1) 'Fastest' proved the goal fastest:
'Sequentially' proved the goal fastest:
'Fastest' succeeded:
'matrix-interpretation of dimension 3 (timeout of 100.0 seconds)' proved the goal fastest:
The following argument positions are usable:
Uargs(double) = {}, Uargs(s) = {1}, Uargs(+) = {}
We have the following constructor-restricted (at most 2 in the main diagonals) matrix interpretation:
Interpretation Functions:
double(x1) = [2 0 2] x1 + [1]
[0 1 2] [0]
[0 0 2] [2]
0() = [0]
[0]
[2]
s(x1) = [1 0 0] x1 + [1]
[0 0 1] [0]
[0 0 1] [2]
+(x1, x2) = [2 0 0] x1 + [0 0 2] x2 + [0]
[0 1 1] [0 0 1] [0]
[0 0 1] [0 0 1] [2]
Hurray, we answered YES(?,O(n^2))Tool pair3irc
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^2)) |
|---|
| Input | SK90 2.13 |
|---|
stdout:
YES(?,O(n^2))
We consider the following Problem:
Strict Trs:
{ double(0()) -> 0()
, double(s(x)) -> s(s(double(x)))
, +(x, 0()) -> x
, +(x, s(y)) -> s(+(x, y))
, +(s(x), y) -> s(+(x, y))
, double(x) -> +(x, x)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^2))
Application of 'pair3 (timeout of 60.0 seconds)':
-------------------------------------------------
The input problem contains no overlaps that give rise to inapplicable rules.
We abort the transformation and continue with the subprocessor on the problem
Strict Trs:
{ double(0()) -> 0()
, double(s(x)) -> s(s(double(x)))
, +(x, 0()) -> x
, +(x, s(y)) -> s(+(x, y))
, +(s(x), y) -> s(+(x, y))
, double(x) -> +(x, x)}
StartTerms: basic terms
Strategy: innermost
1) 'Fastest' proved the goal fastest:
'Sequentially' proved the goal fastest:
'Fastest' succeeded:
'matrix-interpretation of dimension 3 (timeout of 100.0 seconds)' proved the goal fastest:
The following argument positions are usable:
Uargs(double) = {}, Uargs(s) = {1}, Uargs(+) = {}
We have the following constructor-restricted (at most 2 in the main diagonals) matrix interpretation:
Interpretation Functions:
double(x1) = [2 0 2] x1 + [1]
[0 1 2] [0]
[0 0 2] [2]
0() = [0]
[0]
[2]
s(x1) = [1 0 0] x1 + [1]
[0 0 1] [0]
[0 0 1] [2]
+(x1, x2) = [2 0 0] x1 + [0 0 2] x2 + [0]
[0 1 1] [0 0 1] [0]
[0 0 1] [0 0 1] [2]
Hurray, we answered YES(?,O(n^2))Tool pair3rc
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^2)) |
|---|
| Input | SK90 2.13 |
|---|
stdout:
YES(?,O(n^2))
We consider the following Problem:
Strict Trs:
{ double(0()) -> 0()
, double(s(x)) -> s(s(double(x)))
, +(x, 0()) -> x
, +(x, s(y)) -> s(+(x, y))
, +(s(x), y) -> s(+(x, y))
, double(x) -> +(x, x)}
StartTerms: basic terms
Strategy: none
Certificate: YES(?,O(n^2))
Application of 'pair3 (timeout of 60.0 seconds)':
-------------------------------------------------
The processor is not applicable, reason is:
Input problem is not restricted to innermost rewriting
We abort the transformation and continue with the subprocessor on the problem
Strict Trs:
{ double(0()) -> 0()
, double(s(x)) -> s(s(double(x)))
, +(x, 0()) -> x
, +(x, s(y)) -> s(+(x, y))
, +(s(x), y) -> s(+(x, y))
, double(x) -> +(x, x)}
StartTerms: basic terms
Strategy: none
1) 'Fastest' proved the goal fastest:
'Sequentially' proved the goal fastest:
'Fastest' succeeded:
'matrix-interpretation of dimension 3 (timeout of 100.0 seconds)' proved the goal fastest:
The following argument positions are usable:
Uargs(double) = {}, Uargs(s) = {1}, Uargs(+) = {}
We have the following constructor-restricted (at most 2 in the main diagonals) matrix interpretation:
Interpretation Functions:
double(x1) = [2 0 2] x1 + [1]
[0 1 2] [0]
[0 0 2] [2]
0() = [0]
[0]
[2]
s(x1) = [1 0 0] x1 + [1]
[0 0 1] [0]
[0 0 1] [2]
+(x1, x2) = [2 0 0] x1 + [0 0 2] x2 + [0]
[0 1 1] [0 0 1] [0]
[0 0 1] [0 0 1] [2]
Hurray, we answered YES(?,O(n^2))Tool rc
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^2)) |
|---|
| Input | SK90 2.13 |
|---|
stdout:
YES(?,O(n^2))
We consider the following Problem:
Strict Trs:
{ double(0()) -> 0()
, double(s(x)) -> s(s(double(x)))
, +(x, 0()) -> x
, +(x, s(y)) -> s(+(x, y))
, +(s(x), y) -> s(+(x, y))
, double(x) -> +(x, x)}
StartTerms: basic terms
Strategy: none
Certificate: YES(?,O(n^2))
Application of 'rc (timeout of 60.0 seconds)':
----------------------------------------------
'Fastest' proved the goal fastest:
'Sequentially' proved the goal fastest:
'Fastest' succeeded:
'matrix-interpretation of dimension 3 (timeout of 100.0 seconds)' proved the goal fastest:
The following argument positions are usable:
Uargs(double) = {}, Uargs(s) = {1}, Uargs(+) = {}
We have the following constructor-restricted (at most 2 in the main diagonals) matrix interpretation:
Interpretation Functions:
double(x1) = [2 0 2] x1 + [1]
[0 1 2] [0]
[0 0 2] [2]
0() = [0]
[0]
[2]
s(x1) = [1 0 0] x1 + [1]
[0 0 1] [0]
[0 0 1] [2]
+(x1, x2) = [2 0 0] x1 + [0 0 2] x2 + [0]
[0 1 1] [0 0 1] [0]
[0 0 1] [0 0 1] [2]
Hurray, we answered YES(?,O(n^2))Tool tup3irc
| Execution Time | 2.134795ms |
|---|
| Answer | YES(?,O(n^1)) |
|---|
| Input | SK90 2.13 |
|---|
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ double(0()) -> 0()
, double(s(x)) -> s(s(double(x)))
, +(x, 0()) -> x
, +(x, s(y)) -> s(+(x, y))
, +(s(x), y) -> s(+(x, y))
, double(x) -> +(x, x)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Application of 'tup3 (timeout of 60.0 seconds)':
------------------------------------------------
The input problem contains no overlaps that give rise to inapplicable rules.
We abort the transformation and continue with the subprocessor on the problem
Strict Trs:
{ double(0()) -> 0()
, double(s(x)) -> s(s(double(x)))
, +(x, 0()) -> x
, +(x, s(y)) -> s(+(x, y))
, +(s(x), y) -> s(+(x, y))
, double(x) -> +(x, x)}
StartTerms: basic terms
Strategy: innermost
1) 'dp' proved the goal fastest:
We have computed the following dependency pairs
Strict Dependency Pairs:
{ double^#(0()) -> c_1()
, double^#(s(x)) -> c_2(double^#(x))
, +^#(x, 0()) -> c_3()
, +^#(x, s(y)) -> c_4(+^#(x, y))
, +^#(s(x), y) -> c_5(+^#(x, y))
, double^#(x) -> c_6(+^#(x, x))}
We consider the following Problem:
Strict DPs:
{ double^#(0()) -> c_1()
, double^#(s(x)) -> c_2(double^#(x))
, +^#(x, 0()) -> c_3()
, +^#(x, s(y)) -> c_4(+^#(x, y))
, +^#(s(x), y) -> c_5(+^#(x, y))
, double^#(x) -> c_6(+^#(x, x))}
Weak Trs:
{ double(0()) -> 0()
, double(s(x)) -> s(s(double(x)))
, +(x, 0()) -> x
, +(x, s(y)) -> s(+(x, y))
, +(s(x), y) -> s(+(x, y))
, double(x) -> +(x, x)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Application of 'Fastest':
-------------------------
'Sequentially' proved the goal fastest:
'Fastest' succeeded:
'matrix-interpretation of dimension 2 (timeout of 100.0 seconds)' proved the goal fastest:
The following argument positions are usable:
Uargs(double) = {}, Uargs(s) = {}, Uargs(+) = {},
Uargs(double^#) = {}, Uargs(c_2) = {1}, Uargs(+^#) = {},
Uargs(c_4) = {1}, Uargs(c_5) = {1}, Uargs(c_6) = {1}
We have the following constructor-restricted (at most 1 in the main diagonals) matrix interpretation:
Interpretation Functions:
double(x1) = [2 0] x1 + [2]
[0 2] [0]
0() = [1]
[0]
s(x1) = [1 0] x1 + [2]
[0 0] [0]
+(x1, x2) = [1 0] x1 + [1 0] x2 + [2]
[0 2] [0 0] [0]
double^#(x1) = [2 0] x1 + [1]
[0 0] [0]
c_1() = [0]
[0]
c_2(x1) = [1 0] x1 + [0]
[0 0] [0]
+^#(x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [0]
c_3() = [0]
[0]
c_4(x1) = [1 0] x1 + [1]
[0 0] [0]
c_5(x1) = [1 0] x1 + [1]
[0 0] [0]
c_6(x1) = [1 0] x1 + [0]
[0 0] [0]
Hurray, we answered YES(?,O(n^1))