Tool CaT
Execution Time | Unknown |
---|
Answer | MAYBE |
---|
Input | AG01 3.52 |
---|
stdout:
MAYBE
Problem:
f(0(),1(),x) -> f(s(x),x,x)
f(x,y,s(z)) -> s(f(0(),1(),z))
Proof:
OpenTool IRC1
Execution Time | Unknown |
---|
Answer | MAYBE |
---|
Input | AG01 3.52 |
---|
stdout:
MAYBE
Tool IRC2
Execution Time | Unknown |
---|
Answer | YES(?,O(n^1)) |
---|
Input | AG01 3.52 |
---|
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:
{ f(0(), 1(), x) -> f(s(x), x, x)
, f(x, y, s(z)) -> s(f(0(), 1(), z))}
Proof Output:
'wdg' proved the best result:
Details:
--------
'wdg' succeeded with the following output:
'wdg'
-----
Answer: YES(?,O(n^1))
Input Problem: innermost runtime-complexity with respect to
Rules:
{ f(0(), 1(), x) -> f(s(x), x, x)
, f(x, y, s(z)) -> s(f(0(), 1(), z))}
Proof Output:
Transformation Details:
-----------------------
We have computed the following set of weak (innermost) dependency pairs:
{ 1: f^#(0(), 1(), x) -> c_0(f^#(s(x), x, x))
, 2: f^#(x, y, s(z)) -> c_1(f^#(0(), 1(), z))}
Following Dependency Graph (modulo SCCs) was computed. (Answers to
subproofs are indicated to the right.)
->{1,2} [ YES(?,O(n^1)) ]
Sub-problems:
-------------
* Path {1,2}: YES(?,O(n^1))
-------------------------
The usable rules of this path are empty.
The weightgap principle applies, using the following adequate RMI:
The following argument positions are usable:
Uargs(f) = {}, Uargs(s) = {}, Uargs(f^#) = {}, Uargs(c_0) = {1},
Uargs(c_1) = {1}
We have the following constructor-restricted matrix interpretation:
Interpretation Functions:
f(x1, x2, x3) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [0]
[0 0] [0 0] [0 0] [0]
0() = [0]
[0]
1() = [0]
[0]
s(x1) = [0 0] x1 + [0]
[0 0] [0]
f^#(x1, x2, x3) = [3 3] x1 + [0 0] x2 + [3 3] x3 + [0]
[3 3] [3 3] [3 3] [0]
c_0(x1) = [1 0] x1 + [0]
[0 1] [0]
c_1(x1) = [1 0] x1 + [0]
[0 1] [0]
We apply the sub-processor on the resulting sub-problem:
'matrix-interpretation of dimension 2'
--------------------------------------
Answer: YES(?,O(n^1))
Input Problem: innermost DP runtime-complexity with respect to
Strict Rules:
{ f^#(0(), 1(), x) -> c_0(f^#(s(x), x, x))
, f^#(x, y, s(z)) -> c_1(f^#(0(), 1(), z))}
Weak Rules: {}
Proof Output:
The following argument positions are usable:
Uargs(s) = {}, Uargs(f^#) = {}, Uargs(c_0) = {1}, Uargs(c_1) = {1}
We have the following constructor-restricted matrix interpretation:
Interpretation Functions:
0() = [1]
[0]
1() = [4]
[0]
s(x1) = [0 0] x1 + [0]
[0 1] [2]
f^#(x1, x2, x3) = [1 0] x1 + [0 0] x2 + [0 2] x3 + [0]
[0 0] [2 0] [0 0] [0]
c_0(x1) = [1 0] x1 + [0]
[0 0] [7]
c_1(x1) = [1 0] x1 + [0]
[0 0] [0]Tool RC1
Execution Time | Unknown |
---|
Answer | MAYBE |
---|
Input | AG01 3.52 |
---|
stdout:
MAYBE
Tool RC2
Execution Time | Unknown |
---|
Answer | YES(?,O(n^1)) |
---|
Input | AG01 3.52 |
---|
stdout:
YES(?,O(n^1))
'Fastest (timeout of 60.0 seconds)'
-----------------------------------
Answer: YES(?,O(n^1))
Input Problem: runtime-complexity with respect to
Rules:
{ f(0(), 1(), x) -> f(s(x), x, x)
, f(x, y, s(z)) -> s(f(0(), 1(), z))}
Proof Output:
'wdg' proved the best result:
Details:
--------
'wdg' succeeded with the following output:
'wdg'
-----
Answer: YES(?,O(n^1))
Input Problem: runtime-complexity with respect to
Rules:
{ f(0(), 1(), x) -> f(s(x), x, x)
, f(x, y, s(z)) -> s(f(0(), 1(), z))}
Proof Output:
Transformation Details:
-----------------------
We have computed the following set of weak (innermost) dependency pairs:
{ 1: f^#(0(), 1(), x) -> c_0(f^#(s(x), x, x))
, 2: f^#(x, y, s(z)) -> c_1(f^#(0(), 1(), z))}
Following Dependency Graph (modulo SCCs) was computed. (Answers to
subproofs are indicated to the right.)
->{1,2} [ YES(?,O(n^1)) ]
Sub-problems:
-------------
* Path {1,2}: YES(?,O(n^1))
-------------------------
The usable rules of this path are empty.
The weightgap principle applies, using the following adequate RMI:
The following argument positions are usable:
Uargs(f) = {}, Uargs(s) = {}, Uargs(f^#) = {}, Uargs(c_0) = {1},
Uargs(c_1) = {1}
We have the following constructor-restricted matrix interpretation:
Interpretation Functions:
f(x1, x2, x3) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [0]
[0 0] [0 0] [0 0] [0]
0() = [0]
[0]
1() = [0]
[0]
s(x1) = [0 0] x1 + [0]
[0 0] [0]
f^#(x1, x2, x3) = [3 3] x1 + [0 0] x2 + [3 3] x3 + [0]
[3 3] [3 3] [3 3] [0]
c_0(x1) = [1 0] x1 + [0]
[0 1] [0]
c_1(x1) = [1 0] x1 + [0]
[0 1] [0]
We apply the sub-processor on the resulting sub-problem:
'matrix-interpretation of dimension 2'
--------------------------------------
Answer: YES(?,O(n^1))
Input Problem: DP runtime-complexity with respect to
Strict Rules:
{ f^#(0(), 1(), x) -> c_0(f^#(s(x), x, x))
, f^#(x, y, s(z)) -> c_1(f^#(0(), 1(), z))}
Weak Rules: {}
Proof Output:
The following argument positions are usable:
Uargs(s) = {}, Uargs(f^#) = {}, Uargs(c_0) = {1}, Uargs(c_1) = {1}
We have the following constructor-restricted matrix interpretation:
Interpretation Functions:
0() = [1]
[0]
1() = [4]
[0]
s(x1) = [0 0] x1 + [0]
[0 1] [2]
f^#(x1, x2, x3) = [1 0] x1 + [0 0] x2 + [0 2] x3 + [0]
[0 0] [2 0] [0 0] [0]
c_0(x1) = [1 0] x1 + [0]
[0 0] [7]
c_1(x1) = [1 0] x1 + [0]
[0 0] [0]Tool pair1rc
Execution Time | Unknown |
---|
Answer | YES(?,O(n^1)) |
---|
Input | AG01 3.52 |
---|
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ f(0(), 1(), x) -> f(s(x), x, x)
, f(x, y, s(z)) -> s(f(0(), 1(), z))}
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:
{ f(0(), 1(), x) -> f(s(x), x, x)
, f(x, y, s(z)) -> s(f(0(), 1(), z))}
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:
The following argument positions are usable:
Uargs(f) = {}, Uargs(s) = {1}
We have the following constructor-restricted (at most 1 in the main diagonals) matrix interpretation:
Interpretation Functions:
f(x1, x2, x3) = [0 1] x1 + [0 0] x2 + [2 0] x3 + [0]
[0 0] [0 0] [0 0] [0]
0() = [0]
[1]
1() = [0]
[0]
s(x1) = [1 0] x1 + [2]
[0 0] [0]
Hurray, we answered YES(?,O(n^1))Tool pair2rc
Execution Time | Unknown |
---|
Answer | YES(?,O(n^1)) |
---|
Input | AG01 3.52 |
---|
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ f(0(), 1(), x) -> f(s(x), x, x)
, f(x, y, s(z)) -> s(f(0(), 1(), z))}
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:
{ f(0(), 1(), x) -> f(s(x), x, x)
, f(x, y, s(z)) -> s(f(0(), 1(), z))}
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:
The following argument positions are usable:
Uargs(f) = {}, Uargs(s) = {1}
We have the following constructor-restricted (at most 1 in the main diagonals) matrix interpretation:
Interpretation Functions:
f(x1, x2, x3) = [0 1] x1 + [0 0] x2 + [2 0] x3 + [0]
[0 0] [0 0] [0 0] [0]
0() = [0]
[1]
1() = [0]
[0]
s(x1) = [1 0] x1 + [2]
[0 0] [0]
Hurray, we answered YES(?,O(n^1))Tool pair3irc
Execution Time | Unknown |
---|
Answer | YES(?,O(n^1)) |
---|
Input | AG01 3.52 |
---|
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ f(0(), 1(), x) -> f(s(x), x, x)
, f(x, y, s(z)) -> s(f(0(), 1(), z))}
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:
{ f(0(), 1(), x) -> f(s(x), x, x)
, f(x, y, s(z)) -> s(f(0(), 1(), z))}
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(f) = {}, Uargs(s) = {1}
We have the following constructor-restricted (at most 1 in the main diagonals) matrix interpretation:
Interpretation Functions:
f(x1, x2, x3) = [0 1] x1 + [0 0] x2 + [2 0] x3 + [0]
[0 0] [0 0] [0 0] [0]
0() = [0]
[1]
1() = [0]
[0]
s(x1) = [1 0] x1 + [2]
[0 0] [0]
Hurray, we answered YES(?,O(n^1))Tool pair3rc
Execution Time | Unknown |
---|
Answer | YES(?,O(n^1)) |
---|
Input | AG01 3.52 |
---|
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ f(0(), 1(), x) -> f(s(x), x, x)
, f(x, y, s(z)) -> s(f(0(), 1(), z))}
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:
{ f(0(), 1(), x) -> f(s(x), x, x)
, f(x, y, s(z)) -> s(f(0(), 1(), z))}
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:
The following argument positions are usable:
Uargs(f) = {}, Uargs(s) = {1}
We have the following constructor-restricted (at most 1 in the main diagonals) matrix interpretation:
Interpretation Functions:
f(x1, x2, x3) = [0 1] x1 + [0 0] x2 + [2 0] x3 + [0]
[0 0] [0 0] [0 0] [0]
0() = [0]
[1]
1() = [0]
[0]
s(x1) = [1 0] x1 + [2]
[0 0] [0]
Hurray, we answered YES(?,O(n^1))Tool rc
Execution Time | Unknown |
---|
Answer | YES(?,O(n^1)) |
---|
Input | AG01 3.52 |
---|
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ f(0(), 1(), x) -> f(s(x), x, x)
, f(x, y, s(z)) -> s(f(0(), 1(), z))}
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:
The following argument positions are usable:
Uargs(f) = {}, Uargs(s) = {1}
We have the following constructor-restricted (at most 1 in the main diagonals) matrix interpretation:
Interpretation Functions:
f(x1, x2, x3) = [0 1] x1 + [0 0] x2 + [2 0] x3 + [0]
[0 0] [0 0] [0 0] [0]
0() = [0]
[1]
1() = [0]
[0]
s(x1) = [1 0] x1 + [2]
[0 0] [0]
Hurray, we answered YES(?,O(n^1))Tool tup3irc
Execution Time | 0.35439515ms |
---|
Answer | YES(?,O(n^1)) |
---|
Input | AG01 3.52 |
---|
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ f(0(), 1(), x) -> f(s(x), x, x)
, f(x, y, s(z)) -> s(f(0(), 1(), z))}
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:
{ f(0(), 1(), x) -> f(s(x), x, x)
, f(x, y, s(z)) -> s(f(0(), 1(), z))}
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(f) = {}, Uargs(s) = {1}
We have the following constructor-restricted (at most 1 in the main diagonals) matrix interpretation:
Interpretation Functions:
f(x1, x2, x3) = [0 1] x1 + [0 0] x2 + [2 0] x3 + [0]
[0 0] [0 0] [0 0] [0]
0() = [0]
[1]
1() = [0]
[0]
s(x1) = [1 0] x1 + [2]
[0 0] [0]
Hurray, we answered YES(?,O(n^1))