Tool CaT
Execution Time | Unknown |
---|
Answer | YES(?,O(n^1)) |
---|
Input | SK90 2.14 |
---|
stdout:
YES(?,O(n^1))
Problem:
double(0()) -> 0()
double(s(x)) -> s(s(double(x)))
half(0()) -> 0()
half(s(0())) -> 0()
half(s(s(x))) -> s(half(x))
-(x,0()) -> x
-(s(x),s(y)) -> -(x,y)
if(0(),y,z) -> y
if(s(x),y,z) -> z
half(double(x)) -> x
Proof:
Bounds Processor:
bound: 1
enrichment: match
automaton:
final states: {6,5,4,3}
transitions:
-1(1,2) -> 5*
-1(2,1) -> 5*
-1(1,1) -> 5*
-1(2,2) -> 5*
s1(12) -> 12,4
s1(7) -> 8*
s1(8) -> 7,3
half1(2) -> 12*
half1(1) -> 12*
01() -> 12,7,4,3
double1(2) -> 7*
double1(1) -> 7*
double0(2) -> 3*
double0(1) -> 3*
00() -> 1*
s0(2) -> 2*
s0(1) -> 2*
half0(2) -> 4*
half0(1) -> 4*
-0(1,2) -> 5*
-0(2,1) -> 5*
-0(1,1) -> 5*
-0(2,2) -> 5*
if0(1,1,1) -> 6*
if0(2,2,1) -> 6*
if0(1,1,2) -> 6*
if0(2,2,2) -> 6*
if0(1,2,1) -> 6*
if0(2,1,1) -> 6*
if0(1,2,2) -> 6*
if0(2,1,2) -> 6*
1 -> 6,5
2 -> 6,5
problem:
QedTool IRC1
Execution Time | Unknown |
---|
Answer | YES(?,O(n^1)) |
---|
Input | SK90 2.14 |
---|
stdout:
YES(?,O(n^1))
Tool IRC2
Execution Time | Unknown |
---|
Answer | YES(?,O(n^1)) |
---|
Input | SK90 2.14 |
---|
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)))
, half(0()) -> 0()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x))
, -(x, 0()) -> x
, -(s(x), s(y)) -> -(x, y)
, if(0(), y, z) -> y
, if(s(x), y, z) -> z
, half(double(x)) -> x}
Proof Output:
'Bounds with minimal-enrichment and initial automaton 'match'' proved the best result:
Details:
--------
'Bounds with minimal-enrichment and initial automaton 'match'' succeeded with the following output:
'Bounds with minimal-enrichment and initial automaton 'match''
--------------------------------------------------------------
Answer: YES(?,O(n^1))
Input Problem: innermost runtime-complexity with respect to
Rules:
{ double(0()) -> 0()
, double(s(x)) -> s(s(double(x)))
, half(0()) -> 0()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x))
, -(x, 0()) -> x
, -(s(x), s(y)) -> -(x, y)
, if(0(), y, z) -> y
, if(s(x), y, z) -> z
, half(double(x)) -> x}
Proof Output:
The problem is match-bounded by 1.
The enriched problem is compatible with the following automaton:
{ double_0(2) -> 1
, double_1(2) -> 4
, 0_0() -> 1
, 0_0() -> 2
, 0_1() -> 1
, 0_1() -> 3
, 0_1() -> 4
, s_0(2) -> 1
, s_0(2) -> 2
, s_1(3) -> 1
, s_1(3) -> 3
, s_1(3) -> 4
, s_1(4) -> 3
, half_0(2) -> 1
, half_1(2) -> 3
, -_0(2, 2) -> 1
, -_1(2, 2) -> 1
, if_0(2, 2, 2) -> 1}Tool RC1
Execution Time | Unknown |
---|
Answer | YES(?,O(n^1)) |
---|
Input | SK90 2.14 |
---|
stdout:
YES(?,O(n^1))
Tool RC2
Execution Time | Unknown |
---|
Answer | YES(?,O(n^1)) |
---|
Input | SK90 2.14 |
---|
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)))
, half(0()) -> 0()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x))
, -(x, 0()) -> x
, -(s(x), s(y)) -> -(x, y)
, if(0(), y, z) -> y
, if(s(x), y, z) -> z
, half(double(x)) -> x}
Proof Output:
'Bounds with minimal-enrichment and initial automaton 'match'' proved the best result:
Details:
--------
'Bounds with minimal-enrichment and initial automaton 'match'' succeeded with the following output:
'Bounds with minimal-enrichment and initial automaton 'match''
--------------------------------------------------------------
Answer: YES(?,O(n^1))
Input Problem: runtime-complexity with respect to
Rules:
{ double(0()) -> 0()
, double(s(x)) -> s(s(double(x)))
, half(0()) -> 0()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x))
, -(x, 0()) -> x
, -(s(x), s(y)) -> -(x, y)
, if(0(), y, z) -> y
, if(s(x), y, z) -> z
, half(double(x)) -> x}
Proof Output:
The problem is match-bounded by 1.
The enriched problem is compatible with the following automaton:
{ double_0(2) -> 1
, double_1(2) -> 4
, 0_0() -> 1
, 0_0() -> 2
, 0_1() -> 1
, 0_1() -> 3
, 0_1() -> 4
, s_0(2) -> 1
, s_0(2) -> 2
, s_1(3) -> 1
, s_1(3) -> 3
, s_1(3) -> 4
, s_1(4) -> 3
, half_0(2) -> 1
, half_1(2) -> 3
, -_0(2, 2) -> 1
, -_1(2, 2) -> 1
, if_0(2, 2, 2) -> 1}Tool pair1rc
Execution Time | Unknown |
---|
Answer | YES(?,O(n^1)) |
---|
Input | SK90 2.14 |
---|
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ double(0()) -> 0()
, double(s(x)) -> s(s(double(x)))
, half(0()) -> 0()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x))
, -(x, 0()) -> x
, -(s(x), s(y)) -> -(x, y)
, if(0(), y, z) -> y
, if(s(x), y, z) -> z
, half(double(x)) -> x}
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:
{ double(0()) -> 0()
, double(s(x)) -> s(s(double(x)))
, half(0()) -> 0()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x))
, -(x, 0()) -> x
, -(s(x), s(y)) -> -(x, y)
, if(0(), y, z) -> y
, if(s(x), y, z) -> z
, half(double(x)) -> x}
StartTerms: basic terms
Strategy: none
1) 'Fastest' proved the goal fastest:
'Fastest' proved the goal fastest:
'Bounds with minimal-enrichment and initial automaton 'match'' proved the goal fastest:
The problem is match-bounded by 1.
The enriched problem is compatible with the following automaton:
{ double_0(2) -> 1
, double_1(2) -> 4
, 0_0() -> 1
, 0_0() -> 2
, 0_1() -> 1
, 0_1() -> 3
, 0_1() -> 4
, s_0(2) -> 1
, s_0(2) -> 2
, s_1(3) -> 1
, s_1(3) -> 3
, s_1(3) -> 4
, s_1(4) -> 3
, half_0(2) -> 1
, half_1(2) -> 3
, -_0(2, 2) -> 1
, -_1(2, 2) -> 1
, if_0(2, 2, 2) -> 1}
Hurray, we answered YES(?,O(n^1))Tool pair2rc
Execution Time | Unknown |
---|
Answer | YES(?,O(n^1)) |
---|
Input | SK90 2.14 |
---|
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ double(0()) -> 0()
, double(s(x)) -> s(s(double(x)))
, half(0()) -> 0()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x))
, -(x, 0()) -> x
, -(s(x), s(y)) -> -(x, y)
, if(0(), y, z) -> y
, if(s(x), y, z) -> z
, half(double(x)) -> x}
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:
{ double(0()) -> 0()
, double(s(x)) -> s(s(double(x)))
, half(0()) -> 0()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x))
, -(x, 0()) -> x
, -(s(x), s(y)) -> -(x, y)
, if(0(), y, z) -> y
, if(s(x), y, z) -> z
, half(double(x)) -> x}
StartTerms: basic terms
Strategy: none
1) 'Fastest' proved the goal fastest:
'Fastest' proved the goal fastest:
'Bounds with minimal-enrichment and initial automaton 'match'' proved the goal fastest:
The problem is match-bounded by 1.
The enriched problem is compatible with the following automaton:
{ double_0(2) -> 1
, double_1(2) -> 4
, 0_0() -> 1
, 0_0() -> 2
, 0_1() -> 1
, 0_1() -> 3
, 0_1() -> 4
, s_0(2) -> 1
, s_0(2) -> 2
, s_1(3) -> 1
, s_1(3) -> 3
, s_1(3) -> 4
, s_1(4) -> 3
, half_0(2) -> 1
, half_1(2) -> 3
, -_0(2, 2) -> 1
, -_1(2, 2) -> 1
, if_0(2, 2, 2) -> 1}
Hurray, we answered YES(?,O(n^1))Tool pair3irc
Execution Time | Unknown |
---|
Answer | YES(?,O(n^1)) |
---|
Input | SK90 2.14 |
---|
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ double(0()) -> 0()
, double(s(x)) -> s(s(double(x)))
, half(0()) -> 0()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x))
, -(x, 0()) -> x
, -(s(x), s(y)) -> -(x, y)
, if(0(), y, z) -> y
, if(s(x), y, z) -> z
, half(double(x)) -> x}
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:
{ double(0()) -> 0()
, double(s(x)) -> s(s(double(x)))
, half(0()) -> 0()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x))
, -(x, 0()) -> x
, -(s(x), s(y)) -> -(x, y)
, if(0(), y, z) -> y
, if(s(x), y, z) -> z
, half(double(x)) -> x}
StartTerms: basic terms
Strategy: innermost
1) 'Fastest' proved the goal fastest:
'Fastest' proved the goal fastest:
'Bounds with minimal-enrichment and initial automaton 'match'' proved the goal fastest:
The problem is match-bounded by 1.
The enriched problem is compatible with the following automaton:
{ double_0(2) -> 1
, double_1(2) -> 4
, 0_0() -> 1
, 0_0() -> 2
, 0_1() -> 1
, 0_1() -> 3
, 0_1() -> 4
, s_0(2) -> 1
, s_0(2) -> 2
, s_1(3) -> 1
, s_1(3) -> 3
, s_1(3) -> 4
, s_1(4) -> 3
, half_0(2) -> 1
, half_1(2) -> 3
, -_0(2, 2) -> 1
, -_1(2, 2) -> 1
, if_0(2, 2, 2) -> 1}
Hurray, we answered YES(?,O(n^1))Tool pair3rc
Execution Time | Unknown |
---|
Answer | YES(?,O(n^1)) |
---|
Input | SK90 2.14 |
---|
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ double(0()) -> 0()
, double(s(x)) -> s(s(double(x)))
, half(0()) -> 0()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x))
, -(x, 0()) -> x
, -(s(x), s(y)) -> -(x, y)
, if(0(), y, z) -> y
, if(s(x), y, z) -> z
, half(double(x)) -> x}
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:
{ double(0()) -> 0()
, double(s(x)) -> s(s(double(x)))
, half(0()) -> 0()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x))
, -(x, 0()) -> x
, -(s(x), s(y)) -> -(x, y)
, if(0(), y, z) -> y
, if(s(x), y, z) -> z
, half(double(x)) -> x}
StartTerms: basic terms
Strategy: none
1) 'Fastest' proved the goal fastest:
'Fastest' proved the goal fastest:
'Bounds with minimal-enrichment and initial automaton 'match'' proved the goal fastest:
The problem is match-bounded by 1.
The enriched problem is compatible with the following automaton:
{ double_0(2) -> 1
, double_1(2) -> 4
, 0_0() -> 1
, 0_0() -> 2
, 0_1() -> 1
, 0_1() -> 3
, 0_1() -> 4
, s_0(2) -> 1
, s_0(2) -> 2
, s_1(3) -> 1
, s_1(3) -> 3
, s_1(3) -> 4
, s_1(4) -> 3
, half_0(2) -> 1
, half_1(2) -> 3
, -_0(2, 2) -> 1
, -_1(2, 2) -> 1
, if_0(2, 2, 2) -> 1}
Hurray, we answered YES(?,O(n^1))Tool rc
Execution Time | Unknown |
---|
Answer | YES(?,O(n^1)) |
---|
Input | SK90 2.14 |
---|
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ double(0()) -> 0()
, double(s(x)) -> s(s(double(x)))
, half(0()) -> 0()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x))
, -(x, 0()) -> x
, -(s(x), s(y)) -> -(x, y)
, if(0(), y, z) -> y
, if(s(x), y, z) -> z
, half(double(x)) -> x}
StartTerms: basic terms
Strategy: none
Certificate: YES(?,O(n^1))
Application of 'rc (timeout of 60.0 seconds)':
----------------------------------------------
'Fastest' proved the goal fastest:
'Fastest' proved the goal fastest:
'Bounds with minimal-enrichment and initial automaton 'match' (timeout of 100.0 seconds)' proved the goal fastest:
The problem is match-bounded by 1.
The enriched problem is compatible with the following automaton:
{ double_0(2) -> 1
, double_1(2) -> 4
, 0_0() -> 1
, 0_0() -> 2
, 0_1() -> 1
, 0_1() -> 3
, 0_1() -> 4
, s_0(2) -> 1
, s_0(2) -> 2
, s_1(3) -> 1
, s_1(3) -> 3
, s_1(3) -> 4
, s_1(4) -> 3
, half_0(2) -> 1
, half_1(2) -> 3
, -_0(2, 2) -> 1
, -_1(2, 2) -> 1
, if_0(2, 2, 2) -> 1}
Hurray, we answered YES(?,O(n^1))Tool tup3irc
Execution Time | 6.838322e-2ms |
---|
Answer | YES(?,O(n^1)) |
---|
Input | SK90 2.14 |
---|
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ double(0()) -> 0()
, double(s(x)) -> s(s(double(x)))
, half(0()) -> 0()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x))
, -(x, 0()) -> x
, -(s(x), s(y)) -> -(x, y)
, if(0(), y, z) -> y
, if(s(x), y, z) -> z
, half(double(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)))
, half(0()) -> 0()
, half(s(0())) -> 0()
, half(s(s(x))) -> s(half(x))
, -(x, 0()) -> x
, -(s(x), s(y)) -> -(x, y)
, if(0(), y, z) -> y
, if(s(x), y, z) -> z
, half(double(x)) -> x}
StartTerms: basic terms
Strategy: innermost
1) 'Fastest' proved the goal fastest:
'Fastest' proved the goal fastest:
'Bounds with minimal-enrichment and initial automaton 'match'' proved the goal fastest:
The problem is match-bounded by 1.
The enriched problem is compatible with the following automaton:
{ double_0(2) -> 1
, double_1(2) -> 4
, 0_0() -> 1
, 0_0() -> 2
, 0_1() -> 1
, 0_1() -> 3
, 0_1() -> 4
, s_0(2) -> 1
, s_0(2) -> 2
, s_1(3) -> 1
, s_1(3) -> 3
, s_1(3) -> 4
, s_1(4) -> 3
, half_0(2) -> 1
, half_1(2) -> 3
, -_0(2, 2) -> 1
, -_1(2, 2) -> 1
, if_0(2, 2, 2) -> 1}
Hurray, we answered YES(?,O(n^1))