Tool CaT
stdout:
YES(?,O(n^1))
Problem:
0(1(2(3(4(x1))))) -> 0(2(3(1(4(x1)))))
0(5(1(2(3(4(x1)))))) -> 0(1(2(5(3(4(x1))))))
0(5(1(2(3(4(x1)))))) -> 0(5(2(1(3(4(x1))))))
0(5(1(2(3(4(x1)))))) -> 5(0(2(3(1(4(x1))))))
0(5(2(3(1(4(x1)))))) -> 0(1(5(2(3(4(x1))))))
Proof:
Bounds Processor:
bound: 1
enrichment: match
automaton:
final states: {6}
transitions:
01(11) -> 12*
11(45) -> 46*
11(43) -> 44*
11(8) -> 9*
51(55) -> 56*
51(47) -> 48*
51(41) -> 42*
51(58) -> 59*
21(10) -> 11*
21(57) -> 58*
21(42) -> 43*
21(46) -> 47*
31(40) -> 41*
31(9) -> 10*
41(7) -> 8*
41(29) -> 30*
41(31) -> 32*
41(21) -> 22*
41(23) -> 24*
00(5) -> 6*
00(2) -> 6*
00(4) -> 6*
00(1) -> 6*
00(3) -> 6*
10(5) -> 1*
10(2) -> 1*
10(4) -> 1*
10(1) -> 1*
10(3) -> 1*
20(5) -> 2*
20(2) -> 2*
20(4) -> 2*
20(1) -> 2*
20(3) -> 2*
30(5) -> 3*
30(2) -> 3*
30(4) -> 3*
30(1) -> 3*
30(3) -> 3*
40(5) -> 4*
40(2) -> 4*
40(4) -> 4*
40(1) -> 4*
40(3) -> 4*
50(5) -> 5*
50(2) -> 5*
50(4) -> 5*
50(1) -> 5*
50(3) -> 5*
1 -> 29*
2 -> 21*
3 -> 31*
4 -> 23*
5 -> 7*
8 -> 40*
12 -> 55,6
22 -> 8*
24 -> 8*
30 -> 8*
32 -> 8*
41 -> 57,45
44 -> 11*
48 -> 11*
56 -> 6*
59 -> 43*
problem:
QedTool IRC1
stdout:
YES(?,O(1))
Tool IRC2
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(1(2(3(4(x1))))) -> 0(2(3(1(4(x1)))))
, 0(5(1(2(3(4(x1)))))) -> 0(1(2(5(3(4(x1))))))
, 0(5(1(2(3(4(x1)))))) -> 0(5(2(1(3(4(x1))))))
, 0(5(1(2(3(4(x1)))))) -> 5(0(2(3(1(4(x1))))))
, 0(5(2(3(1(4(x1)))))) -> 0(1(5(2(3(4(x1))))))}
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:
{ 0(1(2(3(4(x1))))) -> 0(2(3(1(4(x1)))))
, 0(5(1(2(3(4(x1)))))) -> 0(1(2(5(3(4(x1))))))
, 0(5(1(2(3(4(x1)))))) -> 0(5(2(1(3(4(x1))))))
, 0(5(1(2(3(4(x1)))))) -> 5(0(2(3(1(4(x1))))))
, 0(5(2(3(1(4(x1)))))) -> 0(1(5(2(3(4(x1))))))}
Proof Output:
The problem is match-bounded by 1.
The enriched problem is compatible with the following automaton:
{ 0_0(2) -> 1
, 0_1(3) -> 1
, 1_0(2) -> 2
, 1_1(6) -> 5
, 1_1(7) -> 3
, 1_1(9) -> 11
, 2_0(2) -> 2
, 2_1(4) -> 3
, 2_1(8) -> 7
, 2_1(9) -> 12
, 2_1(11) -> 10
, 3_0(2) -> 2
, 3_1(5) -> 4
, 3_1(6) -> 9
, 4_0(2) -> 2
, 4_1(2) -> 6
, 5_0(2) -> 2
, 5_1(1) -> 1
, 5_1(9) -> 8
, 5_1(10) -> 3
, 5_1(12) -> 7}Tool RC1
stdout:
YES(?,O(1))
Tool RC2
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(1(2(3(4(x1))))) -> 0(2(3(1(4(x1)))))
, 0(5(1(2(3(4(x1)))))) -> 0(1(2(5(3(4(x1))))))
, 0(5(1(2(3(4(x1)))))) -> 0(5(2(1(3(4(x1))))))
, 0(5(1(2(3(4(x1)))))) -> 5(0(2(3(1(4(x1))))))
, 0(5(2(3(1(4(x1)))))) -> 0(1(5(2(3(4(x1))))))}
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:
{ 0(1(2(3(4(x1))))) -> 0(2(3(1(4(x1)))))
, 0(5(1(2(3(4(x1)))))) -> 0(1(2(5(3(4(x1))))))
, 0(5(1(2(3(4(x1)))))) -> 0(5(2(1(3(4(x1))))))
, 0(5(1(2(3(4(x1)))))) -> 5(0(2(3(1(4(x1))))))
, 0(5(2(3(1(4(x1)))))) -> 0(1(5(2(3(4(x1))))))}
Proof Output:
The problem is match-bounded by 1.
The enriched problem is compatible with the following automaton:
{ 0_0(2) -> 1
, 0_1(3) -> 1
, 1_0(2) -> 2
, 1_1(6) -> 5
, 1_1(7) -> 3
, 1_1(9) -> 11
, 2_0(2) -> 2
, 2_1(4) -> 3
, 2_1(8) -> 7
, 2_1(9) -> 12
, 2_1(11) -> 10
, 3_0(2) -> 2
, 3_1(5) -> 4
, 3_1(6) -> 9
, 4_0(2) -> 2
, 4_1(2) -> 6
, 5_0(2) -> 2
, 5_1(1) -> 1
, 5_1(9) -> 8
, 5_1(10) -> 3
, 5_1(12) -> 7}