Tool Bounds
stdout:
YES(?,O(n^1))
We consider the following Problem:
Strict Trs:
{ 5(1(4(5(3(3(3(x1))))))) -> 5(1(4(5(3(4(4(2(3(2(x1))))))))))
, 4(5(1(2(4(4(4(x1))))))) -> 4(1(1(4(5(3(0(1(0(4(x1))))))))))
, 4(1(4(2(4(0(1(x1))))))) -> 5(2(2(1(0(5(5(4(5(1(x1))))))))))
, 3(5(2(0(1(3(3(x1))))))) -> 3(4(3(2(3(2(4(4(5(5(x1))))))))))
, 1(1(3(3(5(3(1(x1))))))) -> 3(5(0(5(3(2(5(0(0(1(x1))))))))))
, 0(2(3(1(3(2(5(x1))))))) -> 0(4(3(1(2(3(2(3(2(0(x1))))))))))
, 0(1(3(5(2(2(3(x1))))))) -> 0(3(0(0(5(0(0(4(4(3(x1))))))))))
, 4(1(4(4(4(1(x1)))))) -> 4(1(0(3(3(5(5(5(4(1(x1))))))))))
, 4(0(1(3(4(0(x1)))))) -> 2(2(3(0(0(0(5(0(0(0(x1))))))))))
, 3(3(3(3(4(0(x1)))))) -> 3(0(0(2(1(0(5(3(5(4(x1))))))))))
, 0(3(3(1(4(3(x1)))))) -> 4(4(3(0(2(3(0(3(0(0(x1))))))))))
, 1(4(4(2(2(x1))))) -> 1(1(2(0(1(1(1(0(2(2(x1))))))))))
, 1(2(3(3(3(x1))))) -> 4(1(1(2(3(5(0(4(0(5(x1))))))))))
, 0(3(3(3(1(x1))))) -> 5(4(4(0(3(1(0(5(1(0(x1))))))))))
, 0(3(3(3(x1)))) -> 5(4(3(5(3(0(5(4(4(0(x1))))))))))
, 1(3(3(x1))) -> 3(5(3(2(5(0(2(4(5(4(x1))))))))))}
StartTerms: all
Strategy: none
Certificate: YES(?,O(n^1))
Proof:
The problem is match-bounded by 1.
The enriched problem is compatible with the following automaton:
{ 3_0(1) -> 1
, 3_1(1) -> 60
, 3_1(6) -> 5
, 3_1(10) -> 9
, 3_1(16) -> 15
, 3_1(28) -> 1
, 3_1(28) -> 27
, 3_1(28) -> 60
, 3_1(30) -> 29
, 3_1(32) -> 31
, 3_1(40) -> 39
, 3_1(46) -> 45
, 3_1(49) -> 48
, 3_1(51) -> 50
, 3_1(53) -> 44
, 3_1(62) -> 61
, 3_1(63) -> 62
, 3_1(69) -> 68
, 3_1(74) -> 87
, 3_1(81) -> 80
, 3_1(83) -> 82
, 3_1(86) -> 85
, 3_1(97) -> 96
, 3_1(104) -> 103
, 3_1(108) -> 101
, 3_1(110) -> 109
, 3_1(114) -> 37
, 1_0(1) -> 1
, 1_1(1) -> 27
, 1_1(3) -> 2
, 1_1(12) -> 11
, 1_1(13) -> 12
, 1_1(17) -> 92
, 1_1(18) -> 17
, 1_1(22) -> 21
, 1_1(47) -> 46
, 1_1(52) -> 107
, 1_1(78) -> 77
, 1_1(88) -> 1
, 1_1(88) -> 27
, 1_1(89) -> 88
, 1_1(92) -> 91
, 1_1(93) -> 92
, 1_1(94) -> 93
, 1_1(105) -> 104
, 4_0(1) -> 1
, 4_1(1) -> 19
, 4_1(4) -> 3
, 4_1(7) -> 6
, 4_1(8) -> 7
, 4_1(11) -> 1
, 4_1(11) -> 19
, 4_1(11) -> 25
, 4_1(11) -> 27
, 4_1(11) -> 52
, 4_1(11) -> 66
, 4_1(14) -> 13
, 4_1(19) -> 58
, 4_1(26) -> 25
, 4_1(27) -> 66
, 4_1(28) -> 19
, 4_1(29) -> 28
, 4_1(34) -> 33
, 4_1(35) -> 34
, 4_1(44) -> 19
, 4_1(45) -> 44
, 4_1(52) -> 113
, 4_1(59) -> 58
, 4_1(60) -> 59
, 4_1(68) -> 19
, 4_1(81) -> 118
, 4_1(82) -> 11
, 4_1(100) -> 99
, 4_1(101) -> 2
, 4_1(102) -> 101
, 4_1(113) -> 112
, 5_0(1) -> 1
, 5_1(1) -> 36
, 5_1(2) -> 1
, 5_1(2) -> 19
, 5_1(2) -> 26
, 5_1(2) -> 36
, 5_1(2) -> 52
, 5_1(2) -> 66
, 5_1(5) -> 4
, 5_1(15) -> 14
, 5_1(19) -> 81
, 5_1(24) -> 23
, 5_1(25) -> 24
, 5_1(27) -> 26
, 5_1(28) -> 36
, 5_1(35) -> 63
, 5_1(36) -> 35
, 5_1(37) -> 28
, 5_1(39) -> 38
, 5_1(42) -> 41
, 5_1(56) -> 55
, 5_1(64) -> 63
, 5_1(65) -> 64
, 5_1(66) -> 65
, 5_1(73) -> 72
, 5_1(80) -> 79
, 5_1(98) -> 97
, 5_1(107) -> 106
, 5_1(109) -> 108
, 5_1(112) -> 111
, 5_1(116) -> 115
, 2_0(1) -> 1
, 2_1(1) -> 10
, 2_1(9) -> 8
, 2_1(10) -> 95
, 2_1(20) -> 2
, 2_1(21) -> 20
, 2_1(28) -> 10
, 2_1(31) -> 30
, 2_1(33) -> 32
, 2_1(41) -> 40
, 2_1(48) -> 47
, 2_1(50) -> 49
, 2_1(52) -> 51
, 2_1(67) -> 1
, 2_1(67) -> 19
, 2_1(67) -> 113
, 2_1(68) -> 67
, 2_1(77) -> 76
, 2_1(85) -> 84
, 2_1(90) -> 89
, 2_1(96) -> 13
, 2_1(115) -> 114
, 2_1(118) -> 117
, 0_0(1) -> 1
, 0_1(1) -> 52
, 0_1(2) -> 52
, 0_1(10) -> 94
, 0_1(11) -> 18
, 0_1(17) -> 16
, 0_1(19) -> 18
, 0_1(23) -> 22
, 0_1(27) -> 43
, 0_1(28) -> 52
, 0_1(36) -> 100
, 0_1(38) -> 37
, 0_1(43) -> 42
, 0_1(44) -> 1
, 0_1(44) -> 43
, 0_1(44) -> 52
, 0_1(44) -> 94
, 0_1(52) -> 74
, 0_1(54) -> 53
, 0_1(55) -> 54
, 0_1(57) -> 56
, 0_1(58) -> 57
, 0_1(61) -> 12
, 0_1(70) -> 69
, 0_1(71) -> 70
, 0_1(72) -> 71
, 0_1(74) -> 73
, 0_1(75) -> 28
, 0_1(76) -> 75
, 0_1(79) -> 78
, 0_1(84) -> 83
, 0_1(87) -> 86
, 0_1(88) -> 52
, 0_1(91) -> 90
, 0_1(95) -> 94
, 0_1(99) -> 98
, 0_1(103) -> 102
, 0_1(106) -> 105
, 0_1(111) -> 110
, 0_1(117) -> 116}
Hurray, we answered YES(?,O(n^1))Tool CDI
stdout:
TIMEOUT
Statistics:
Number of monomials: 0
Last formula building started for bound 0
Last SAT solving started for bound 0Tool EDA
stdout:
TIMEOUT
We consider the following Problem:
Strict Trs:
{ 5(1(4(5(3(3(3(x1))))))) -> 5(1(4(5(3(4(4(2(3(2(x1))))))))))
, 4(5(1(2(4(4(4(x1))))))) -> 4(1(1(4(5(3(0(1(0(4(x1))))))))))
, 4(1(4(2(4(0(1(x1))))))) -> 5(2(2(1(0(5(5(4(5(1(x1))))))))))
, 3(5(2(0(1(3(3(x1))))))) -> 3(4(3(2(3(2(4(4(5(5(x1))))))))))
, 1(1(3(3(5(3(1(x1))))))) -> 3(5(0(5(3(2(5(0(0(1(x1))))))))))
, 0(2(3(1(3(2(5(x1))))))) -> 0(4(3(1(2(3(2(3(2(0(x1))))))))))
, 0(1(3(5(2(2(3(x1))))))) -> 0(3(0(0(5(0(0(4(4(3(x1))))))))))
, 4(1(4(4(4(1(x1)))))) -> 4(1(0(3(3(5(5(5(4(1(x1))))))))))
, 4(0(1(3(4(0(x1)))))) -> 2(2(3(0(0(0(5(0(0(0(x1))))))))))
, 3(3(3(3(4(0(x1)))))) -> 3(0(0(2(1(0(5(3(5(4(x1))))))))))
, 0(3(3(1(4(3(x1)))))) -> 4(4(3(0(2(3(0(3(0(0(x1))))))))))
, 1(4(4(2(2(x1))))) -> 1(1(2(0(1(1(1(0(2(2(x1))))))))))
, 1(2(3(3(3(x1))))) -> 4(1(1(2(3(5(0(4(0(5(x1))))))))))
, 0(3(3(3(1(x1))))) -> 5(4(4(0(3(1(0(5(1(0(x1))))))))))
, 0(3(3(3(x1)))) -> 5(4(3(5(3(0(5(4(4(0(x1))))))))))
, 1(3(3(x1))) -> 3(5(3(2(5(0(2(4(5(4(x1))))))))))}
StartTerms: all
Strategy: none
Certificate: TIMEOUT
Proof:
Computation stopped due to timeout after 60.0 seconds.
Arrrr..Tool IDA
stdout:
TIMEOUT
We consider the following Problem:
Strict Trs:
{ 5(1(4(5(3(3(3(x1))))))) -> 5(1(4(5(3(4(4(2(3(2(x1))))))))))
, 4(5(1(2(4(4(4(x1))))))) -> 4(1(1(4(5(3(0(1(0(4(x1))))))))))
, 4(1(4(2(4(0(1(x1))))))) -> 5(2(2(1(0(5(5(4(5(1(x1))))))))))
, 3(5(2(0(1(3(3(x1))))))) -> 3(4(3(2(3(2(4(4(5(5(x1))))))))))
, 1(1(3(3(5(3(1(x1))))))) -> 3(5(0(5(3(2(5(0(0(1(x1))))))))))
, 0(2(3(1(3(2(5(x1))))))) -> 0(4(3(1(2(3(2(3(2(0(x1))))))))))
, 0(1(3(5(2(2(3(x1))))))) -> 0(3(0(0(5(0(0(4(4(3(x1))))))))))
, 4(1(4(4(4(1(x1)))))) -> 4(1(0(3(3(5(5(5(4(1(x1))))))))))
, 4(0(1(3(4(0(x1)))))) -> 2(2(3(0(0(0(5(0(0(0(x1))))))))))
, 3(3(3(3(4(0(x1)))))) -> 3(0(0(2(1(0(5(3(5(4(x1))))))))))
, 0(3(3(1(4(3(x1)))))) -> 4(4(3(0(2(3(0(3(0(0(x1))))))))))
, 1(4(4(2(2(x1))))) -> 1(1(2(0(1(1(1(0(2(2(x1))))))))))
, 1(2(3(3(3(x1))))) -> 4(1(1(2(3(5(0(4(0(5(x1))))))))))
, 0(3(3(3(1(x1))))) -> 5(4(4(0(3(1(0(5(1(0(x1))))))))))
, 0(3(3(3(x1)))) -> 5(4(3(5(3(0(5(4(4(0(x1))))))))))
, 1(3(3(x1))) -> 3(5(3(2(5(0(2(4(5(4(x1))))))))))}
StartTerms: all
Strategy: none
Certificate: TIMEOUT
Proof:
Computation stopped due to timeout after 60.0 seconds.
Arrrr..Tool TRI
stdout:
TIMEOUT
We consider the following Problem:
Strict Trs:
{ 5(1(4(5(3(3(3(x1))))))) -> 5(1(4(5(3(4(4(2(3(2(x1))))))))))
, 4(5(1(2(4(4(4(x1))))))) -> 4(1(1(4(5(3(0(1(0(4(x1))))))))))
, 4(1(4(2(4(0(1(x1))))))) -> 5(2(2(1(0(5(5(4(5(1(x1))))))))))
, 3(5(2(0(1(3(3(x1))))))) -> 3(4(3(2(3(2(4(4(5(5(x1))))))))))
, 1(1(3(3(5(3(1(x1))))))) -> 3(5(0(5(3(2(5(0(0(1(x1))))))))))
, 0(2(3(1(3(2(5(x1))))))) -> 0(4(3(1(2(3(2(3(2(0(x1))))))))))
, 0(1(3(5(2(2(3(x1))))))) -> 0(3(0(0(5(0(0(4(4(3(x1))))))))))
, 4(1(4(4(4(1(x1)))))) -> 4(1(0(3(3(5(5(5(4(1(x1))))))))))
, 4(0(1(3(4(0(x1)))))) -> 2(2(3(0(0(0(5(0(0(0(x1))))))))))
, 3(3(3(3(4(0(x1)))))) -> 3(0(0(2(1(0(5(3(5(4(x1))))))))))
, 0(3(3(1(4(3(x1)))))) -> 4(4(3(0(2(3(0(3(0(0(x1))))))))))
, 1(4(4(2(2(x1))))) -> 1(1(2(0(1(1(1(0(2(2(x1))))))))))
, 1(2(3(3(3(x1))))) -> 4(1(1(2(3(5(0(4(0(5(x1))))))))))
, 0(3(3(3(1(x1))))) -> 5(4(4(0(3(1(0(5(1(0(x1))))))))))
, 0(3(3(3(x1)))) -> 5(4(3(5(3(0(5(4(4(0(x1))))))))))
, 1(3(3(x1))) -> 3(5(3(2(5(0(2(4(5(4(x1))))))))))}
StartTerms: all
Strategy: none
Certificate: TIMEOUT
Proof:
Computation stopped due to timeout after 60.0 seconds.
Arrrr..Tool TRI2
stdout:
YES(?,O(n^2))
We consider the following Problem:
Strict Trs:
{ 5(1(4(5(3(3(3(x1))))))) -> 5(1(4(5(3(4(4(2(3(2(x1))))))))))
, 4(5(1(2(4(4(4(x1))))))) -> 4(1(1(4(5(3(0(1(0(4(x1))))))))))
, 4(1(4(2(4(0(1(x1))))))) -> 5(2(2(1(0(5(5(4(5(1(x1))))))))))
, 3(5(2(0(1(3(3(x1))))))) -> 3(4(3(2(3(2(4(4(5(5(x1))))))))))
, 1(1(3(3(5(3(1(x1))))))) -> 3(5(0(5(3(2(5(0(0(1(x1))))))))))
, 0(2(3(1(3(2(5(x1))))))) -> 0(4(3(1(2(3(2(3(2(0(x1))))))))))
, 0(1(3(5(2(2(3(x1))))))) -> 0(3(0(0(5(0(0(4(4(3(x1))))))))))
, 4(1(4(4(4(1(x1)))))) -> 4(1(0(3(3(5(5(5(4(1(x1))))))))))
, 4(0(1(3(4(0(x1)))))) -> 2(2(3(0(0(0(5(0(0(0(x1))))))))))
, 3(3(3(3(4(0(x1)))))) -> 3(0(0(2(1(0(5(3(5(4(x1))))))))))
, 0(3(3(1(4(3(x1)))))) -> 4(4(3(0(2(3(0(3(0(0(x1))))))))))
, 1(4(4(2(2(x1))))) -> 1(1(2(0(1(1(1(0(2(2(x1))))))))))
, 1(2(3(3(3(x1))))) -> 4(1(1(2(3(5(0(4(0(5(x1))))))))))
, 0(3(3(3(1(x1))))) -> 5(4(4(0(3(1(0(5(1(0(x1))))))))))
, 0(3(3(3(x1)))) -> 5(4(3(5(3(0(5(4(4(0(x1))))))))))
, 1(3(3(x1))) -> 3(5(3(2(5(0(2(4(5(4(x1))))))))))}
StartTerms: all
Strategy: none
Certificate: YES(?,O(n^2))
Proof:
We have the following triangular matrix interpretation:
Interpretation Functions:
3(x1) = [1 1] x1 + [0]
[0 1] [3]
1(x1) = [1 2] x1 + [0]
[0 1] [0]
4(x1) = [1 0] x1 + [2]
[0 0] [0]
5(x1) = [1 0] x1 + [0]
[0 0] [0]
2(x1) = [1 0] x1 + [0]
[0 0] [0]
0(x1) = [1 0] x1 + [0]
[0 0] [0]
Hurray, we answered YES(?,O(n^2))