Tool CaT
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^2)) |
|---|
| Input | SK90 4.12 |
|---|
stdout:
YES(?,O(n^2))
Problem:
+(0(),y) -> y
+(s(x),0()) -> s(x)
+(s(x),s(y)) -> s(+(s(x),+(y,0())))
Proof:
Complexity Transformation Processor:
strict:
+(0(),y) -> y
+(s(x),0()) -> s(x)
+(s(x),s(y)) -> s(+(s(x),+(y,0())))
weak:
Bounds Processor:
bound: 2
enrichment: match
automaton:
final states: {3}
transitions:
s1(7) -> 3*
s1(2) -> 11,7,5,6
s1(1) -> 11,7,5,6
+1(1,4) -> 5*
+1(6,5) -> 7*
+1(2,4) -> 5*
01() -> 4*
s2(2) -> 13,12
s2(1) -> 13,12
s2(13) -> 13,7
+0(1,2) -> 3*
+0(2,1) -> 3*
+0(1,1) -> 3*
+0(2,2) -> 3*
+2(1,10) -> 11*
+2(12,11) -> 13*
+2(2,10) -> 11*
s0(2) -> 3,1
s0(1) -> 3,1
02() -> 10*
00() -> 2*
1 -> 3*
2 -> 3*
4 -> 5*
10 -> 11*
problem:
strict:
+(s(x),0()) -> s(x)
+(0(),y) -> y
weak:
+(s(x),s(y)) -> s(+(s(x),+(y,0())))
Matrix Interpretation Processor:
dimension: 2
max_matrix:
[1 1]
[0 1]
interpretation:
[0]
[s](x0) = x0 + [1],
[1 1]
[+](x0, x1) = x0 + [0 1]x1,
[1]
[0] = [0]
orientation:
[1] [0]
+(s(x),0()) = x + [1] >= x + [1] = s(x)
[1 1] [1]
+(0(),y) = [0 1]y + [0] >= y = y
[1 1] [1] [1 1] [1]
+(s(x),s(y)) = x + [0 1]y + [2] >= x + [0 1]y + [2] = s(+(s(x),+(y,0())))
problem:
strict:
weak:
+(s(x),0()) -> s(x)
+(0(),y) -> y
+(s(x),s(y)) -> s(+(s(x),+(y,0())))
Qed
Tool IRC1
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^1)) |
|---|
| Input | SK90 4.12 |
|---|
stdout:
YES(?,O(n^1))
Tool IRC2
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^1)) |
|---|
| Input | SK90 4.12 |
|---|
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:
{ +(0(), y) -> y
, +(s(x), 0()) -> s(x)
, +(s(x), s(y)) -> s(+(s(x), +(y, 0())))}
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:
{ +(0(), y) -> y
, +(s(x), 0()) -> s(x)
, +(s(x), s(y)) -> s(+(s(x), +(y, 0())))}
Proof Output:
The following argument positions are usable:
Uargs(+) = {2}, Uargs(s) = {1}
We have the following constructor-restricted matrix interpretation:
Interpretation Functions:
+(x1, x2) = [1] x1 + [2] x2 + [1]
0() = [0]
s(x1) = [1] x1 + [4]Tool RC1
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^1)) |
|---|
| Input | SK90 4.12 |
|---|
stdout:
YES(?,O(n^1))
Tool RC2
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^1)) |
|---|
| Input | SK90 4.12 |
|---|
stdout:
YES(?,O(n^1))
'Fastest (timeout of 60.0 seconds)'
-----------------------------------
Answer: YES(?,O(n^1))
Input Problem: runtime-complexity with respect to
Rules:
{ +(0(), y) -> y
, +(s(x), 0()) -> s(x)
, +(s(x), s(y)) -> s(+(s(x), +(y, 0())))}
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:
{ +(0(), y) -> y
, +(s(x), 0()) -> s(x)
, +(s(x), s(y)) -> s(+(s(x), +(y, 0())))}
Proof Output:
The following argument positions are usable:
Uargs(+) = {1, 2}, Uargs(s) = {1}
We have the following constructor-restricted matrix interpretation:
Interpretation Functions:
+(x1, x2) = [1] x1 + [2] x2 + [1]
0() = [0]
s(x1) = [1] x1 + [4]Tool pair1rc
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^1)) |
|---|
| Input | SK90 4.12 |
|---|
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ +(0(), y) -> y
, +(s(x), 0()) -> s(x)
, +(s(x), s(y)) -> s(+(s(x), +(y, 0())))}
StartTerms: basic terms
Strategy: none
Certificate: YES(?,O(n^1))
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:
{ +(0(), y) -> y
, +(s(x), 0()) -> s(x)
, +(s(x), s(y)) -> s(+(s(x), +(y, 0())))}
StartTerms: basic terms
Strategy: none
1) 'Fastest' proved the goal fastest:
'Sequentially' proved the goal fastest:
'Fastest' succeeded:
'matrix-interpretation of dimension 2 (timeout of 100.0 seconds)' proved the goal fastest:
We have the following constructor-restricted (at most 1 in the main diagonals) matrix interpretation:
Interpretation Functions:
+(x1, x2) = [1 1] x1 + [2 0] x2 + [0]
[0 0] [0 1] [0]
0() = [0]
[1]
s(x1) = [1 1] x1 + [2]
[0 0] [1]
Hurray, we answered YES(?,O(n^1))Tool pair2rc
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^1)) |
|---|
| Input | SK90 4.12 |
|---|
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ +(0(), y) -> y
, +(s(x), 0()) -> s(x)
, +(s(x), s(y)) -> s(+(s(x), +(y, 0())))}
StartTerms: basic terms
Strategy: none
Certificate: YES(?,O(n^1))
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:
{ +(0(), y) -> y
, +(s(x), 0()) -> s(x)
, +(s(x), s(y)) -> s(+(s(x), +(y, 0())))}
StartTerms: basic terms
Strategy: none
1) 'Fastest' proved the goal fastest:
'Sequentially' proved the goal fastest:
'Fastest' succeeded:
'matrix-interpretation of dimension 2 (timeout of 100.0 seconds)' proved the goal fastest:
We have the following constructor-restricted (at most 1 in the main diagonals) matrix interpretation:
Interpretation Functions:
+(x1, x2) = [1 1] x1 + [2 0] x2 + [0]
[0 0] [0 1] [0]
0() = [0]
[1]
s(x1) = [1 1] x1 + [2]
[0 0] [1]
Hurray, we answered YES(?,O(n^1))Tool pair3irc
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^1)) |
|---|
| Input | SK90 4.12 |
|---|
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ +(0(), y) -> y
, +(s(x), 0()) -> s(x)
, +(s(x), s(y)) -> s(+(s(x), +(y, 0())))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
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:
{ +(0(), y) -> y
, +(s(x), 0()) -> s(x)
, +(s(x), s(y)) -> s(+(s(x), +(y, 0())))}
StartTerms: basic terms
Strategy: innermost
1) 'Fastest' proved the goal 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(+) = {2}, Uargs(s) = {1}
We have the following constructor-restricted (at most 1 in the main diagonals) matrix interpretation:
Interpretation Functions:
+(x1, x2) = [1 1] x1 + [2 0] x2 + [0]
[0 0] [0 1] [0]
0() = [0]
[1]
s(x1) = [1 1] x1 + [2]
[0 0] [1]
Hurray, we answered YES(?,O(n^1))Tool pair3rc
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^1)) |
|---|
| Input | SK90 4.12 |
|---|
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ +(0(), y) -> y
, +(s(x), 0()) -> s(x)
, +(s(x), s(y)) -> s(+(s(x), +(y, 0())))}
StartTerms: basic terms
Strategy: none
Certificate: YES(?,O(n^1))
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:
{ +(0(), y) -> y
, +(s(x), 0()) -> s(x)
, +(s(x), s(y)) -> s(+(s(x), +(y, 0())))}
StartTerms: basic terms
Strategy: none
1) 'dp' proved the goal fastest:
We have computed the following dependency pairs
Strict Dependency Pairs:
{ +^#(0(), y) -> c_1(y)
, +^#(s(x), 0()) -> c_2(x)
, +^#(s(x), s(y)) -> c_3(+^#(s(x), +(y, 0())))}
We consider the following Problem:
Strict DPs:
{ +^#(0(), y) -> c_1(y)
, +^#(s(x), 0()) -> c_2(x)
, +^#(s(x), s(y)) -> c_3(+^#(s(x), +(y, 0())))}
Strict Trs:
{ +(0(), y) -> y
, +(s(x), 0()) -> s(x)
, +(s(x), s(y)) -> s(+(s(x), +(y, 0())))}
StartTerms: basic terms
Strategy: none
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(+) = {1, 2}, Uargs(s) = {1}, Uargs(+^#) = {2},
Uargs(c_1) = {1}, Uargs(c_2) = {}, Uargs(c_3) = {1}
We have the following constructor-restricted (at most 1 in the main diagonals) matrix interpretation:
Interpretation Functions:
+(x1, x2) = [1 1] x1 + [2 0] x2 + [0]
[0 0] [0 1] [0]
0() = [0]
[1]
s(x1) = [1 1] x1 + [2]
[0 0] [1]
+^#(x1, x2) = [0 0] x1 + [1 0] x2 + [2]
[0 0] [0 0] [0]
c_1(x1) = [1 0] x1 + [1]
[0 0] [0]
c_2(x1) = [0 0] x1 + [1]
[0 0] [0]
c_3(x1) = [1 0] x1 + [1]
[0 0] [0]
Hurray, we answered YES(?,O(n^1))Tool rc
| Execution Time | Unknown |
|---|
| Answer | YES(?,O(n^1)) |
|---|
| Input | SK90 4.12 |
|---|
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ +(0(), y) -> y
, +(s(x), 0()) -> s(x)
, +(s(x), s(y)) -> s(+(s(x), +(y, 0())))}
StartTerms: basic terms
Strategy: none
Certificate: YES(?,O(n^1))
Application of 'rc (timeout of 60.0 seconds)':
----------------------------------------------
'Fastest' proved the goal fastest:
'Sequentially' proved the goal fastest:
'Fastest' succeeded:
'matrix-interpretation of dimension 2 (timeout of 100.0 seconds)' proved the goal fastest:
We have the following constructor-restricted (at most 1 in the main diagonals) matrix interpretation:
Interpretation Functions:
+(x1, x2) = [1 1] x1 + [2 0] x2 + [0]
[0 0] [0 1] [0]
0() = [0]
[1]
s(x1) = [1 1] x1 + [2]
[0 0] [1]
Hurray, we answered YES(?,O(n^1))Tool tup3irc
| Execution Time | 37.058884ms |
|---|
| Answer | YES(?,O(n^1)) |
|---|
| Input | SK90 4.12 |
|---|
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ +(0(), y) -> y
, +(s(x), 0()) -> s(x)
, +(s(x), s(y)) -> s(+(s(x), +(y, 0())))}
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:
{ +(0(), y) -> y
, +(s(x), 0()) -> s(x)
, +(s(x), s(y)) -> s(+(s(x), +(y, 0())))}
StartTerms: basic terms
Strategy: innermost
1) 'Fastest' proved the goal 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(+) = {2}, Uargs(s) = {1}
We have the following constructor-restricted (at most 1 in the main diagonals) matrix interpretation:
Interpretation Functions:
+(x1, x2) = [1 1] x1 + [2 0] x2 + [0]
[0 0] [0 1] [0]
0() = [0]
[1]
s(x1) = [1 1] x1 + [2]
[0 0] [1]
Hurray, we answered YES(?,O(n^1))