LTS Termination Proof

Input

Integer Transition System

Initial Location: l5, l7, l1, l3, l13, l18, l17, l2, l9, l14, l4, l6, l10, l8, l15, l16, l0, l12

Transitions: (pre-variables and post-variables)

l0	1	l1:	x1 = ___const_100HAT0 ∧ x2 = _i2HAT0 ∧ x3 = _iHAT0 ∧ x4 = _rHAT0 ∧ x1 = ___const_100HATpost ∧ x2 = _i2HATpost ∧ x3 = _iHATpost ∧ x4 = _rHATpost ∧ _rHAT0 = _rHATpost ∧ _i2HAT0 = _i2HATpost ∧ ___const_100HAT0 = ___const_100HATpost ∧ _iHATpost = 0 ∧ ___const_100HAT0 ≤ _iHAT0
l0	2	l2:	x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x4 = _x3 ∧ x1 = _x4 ∧ x2 = _x5 ∧ x3 = _x6 ∧ x4 = _x7 ∧ _x3 = _x7 ∧ _x2 = _x6 ∧ _x = _x4 ∧ _x5 = _x2 ∧ 1 + _x2 ≤ _x
l3	3	l1:	x1 = _x8 ∧ x2 = _x9 ∧ x3 = _x10 ∧ x4 = _x11 ∧ x1 = _x12 ∧ x2 = _x13 ∧ x3 = _x14 ∧ x4 = _x15 ∧ _x11 = _x15 ∧ _x9 = _x13 ∧ _x8 = _x12 ∧ _x14 = 1 + _x10
l4	4	l5:	x1 = _x16 ∧ x2 = _x17 ∧ x3 = _x18 ∧ x4 = _x19 ∧ x1 = _x20 ∧ x2 = _x21 ∧ x3 = _x22 ∧ x4 = _x23 ∧ _x19 = _x23 ∧ _x17 = _x21 ∧ _x18 = _x22 ∧ _x16 = _x20
l5	5	l3:	x1 = _x24 ∧ x2 = _x25 ∧ x3 = _x26 ∧ x4 = _x27 ∧ x1 = _x28 ∧ x2 = _x29 ∧ x3 = _x30 ∧ x4 = _x31 ∧ _x27 = _x31 ∧ _x25 = _x29 ∧ _x26 = _x30 ∧ _x24 = _x28
l6	6	l0:	x1 = _x32 ∧ x2 = _x33 ∧ x3 = _x34 ∧ x4 = _x35 ∧ x1 = _x36 ∧ x2 = _x37 ∧ x3 = _x38 ∧ x4 = _x39 ∧ _x35 = _x39 ∧ _x33 = _x37 ∧ _x34 = _x38 ∧ _x32 = _x36
l7	7	l4:	x1 = _x40 ∧ x2 = _x41 ∧ x3 = _x42 ∧ x4 = _x43 ∧ x1 = _x44 ∧ x2 = _x45 ∧ x3 = _x46 ∧ x4 = _x47 ∧ _x43 = _x47 ∧ _x41 = _x45 ∧ _x42 = _x46 ∧ _x40 = _x44
l7	8	l5:	x1 = _x48 ∧ x2 = _x49 ∧ x3 = _x50 ∧ x4 = _x51 ∧ x1 = _x52 ∧ x2 = _x53 ∧ x3 = _x54 ∧ x4 = _x55 ∧ _x51 = _x55 ∧ _x49 = _x53 ∧ _x50 = _x54 ∧ _x48 = _x52
l8	9	l7:	x1 = _x56 ∧ x2 = _x57 ∧ x3 = _x58 ∧ x4 = _x59 ∧ x1 = _x60 ∧ x2 = _x61 ∧ x3 = _x62 ∧ x4 = _x63 ∧ _x57 = _x61 ∧ _x58 = _x62 ∧ _x56 = _x60 ∧ _x63 = _x63
l9	10	l8:	x1 = _x64 ∧ x2 = _x65 ∧ x3 = _x66 ∧ x4 = _x67 ∧ x1 = _x68 ∧ x2 = _x69 ∧ x3 = _x70 ∧ x4 = _x71 ∧ _x67 = _x71 ∧ _x65 = _x69 ∧ _x66 = _x70 ∧ _x64 = _x68
l9	11	l3:	x1 = _x72 ∧ x2 = _x73 ∧ x3 = _x74 ∧ x4 = _x75 ∧ x1 = _x76 ∧ x2 = _x77 ∧ x3 = _x78 ∧ x4 = _x79 ∧ _x75 = _x79 ∧ _x73 = _x77 ∧ _x74 = _x78 ∧ _x72 = _x76
l10	12	l11:	x1 = _x80 ∧ x2 = _x81 ∧ x3 = _x82 ∧ x4 = _x83 ∧ x1 = _x84 ∧ x2 = _x85 ∧ x3 = _x86 ∧ x4 = _x87 ∧ _x83 = _x87 ∧ _x81 = _x85 ∧ _x82 = _x86 ∧ _x80 = _x84 ∧ _x80 ≤ _x82
l10	13	l9:	x1 = _x88 ∧ x2 = _x89 ∧ x3 = _x90 ∧ x4 = _x91 ∧ x1 = _x92 ∧ x2 = _x93 ∧ x3 = _x94 ∧ x4 = _x95 ∧ _x91 = _x95 ∧ _x89 = _x93 ∧ _x90 = _x94 ∧ _x88 = _x92 ∧ 1 + _x90 ≤ _x88
l12	14	l6:	x1 = _x96 ∧ x2 = _x97 ∧ x3 = _x98 ∧ x4 = _x99 ∧ x1 = _x100 ∧ x2 = _x101 ∧ x3 = _x102 ∧ x4 = _x103 ∧ _x99 = _x103 ∧ _x97 = _x101 ∧ _x96 = _x100 ∧ _x102 = 1 + _x98
l1	15	l10:	x1 = _x104 ∧ x2 = _x105 ∧ x3 = _x106 ∧ x4 = _x107 ∧ x1 = _x108 ∧ x2 = _x109 ∧ x3 = _x110 ∧ x4 = _x111 ∧ _x107 = _x111 ∧ _x105 = _x109 ∧ _x106 = _x110 ∧ _x104 = _x108
l13	16	l14:	x1 = _x112 ∧ x2 = _x113 ∧ x3 = _x114 ∧ x4 = _x115 ∧ x1 = _x116 ∧ x2 = _x117 ∧ x3 = _x118 ∧ x4 = _x119 ∧ _x115 = _x119 ∧ _x113 = _x117 ∧ _x114 = _x118 ∧ _x112 = _x116
l14	17	l12:	x1 = _x120 ∧ x2 = _x121 ∧ x3 = _x122 ∧ x4 = _x123 ∧ x1 = _x124 ∧ x2 = _x125 ∧ x3 = _x126 ∧ x4 = _x127 ∧ _x123 = _x127 ∧ _x121 = _x125 ∧ _x122 = _x126 ∧ _x120 = _x124
l15	18	l13:	x1 = _x128 ∧ x2 = _x129 ∧ x3 = _x130 ∧ x4 = _x131 ∧ x1 = _x132 ∧ x2 = _x133 ∧ x3 = _x134 ∧ x4 = _x135 ∧ _x131 = _x135 ∧ _x129 = _x133 ∧ _x130 = _x134 ∧ _x128 = _x132
l15	19	l14:	x1 = _x136 ∧ x2 = _x137 ∧ x3 = _x138 ∧ x4 = _x139 ∧ x1 = _x140 ∧ x2 = _x141 ∧ x3 = _x142 ∧ x4 = _x143 ∧ _x139 = _x143 ∧ _x137 = _x141 ∧ _x138 = _x142 ∧ _x136 = _x140
l16	20	l15:	x1 = _x144 ∧ x2 = _x145 ∧ x3 = _x146 ∧ x4 = _x147 ∧ x1 = _x148 ∧ x2 = _x149 ∧ x3 = _x150 ∧ x4 = _x151 ∧ _x147 = _x151 ∧ _x145 = _x149 ∧ _x146 = _x150 ∧ _x144 = _x148
l2	21	l16:	x1 = _x152 ∧ x2 = _x153 ∧ x3 = _x154 ∧ x4 = _x155 ∧ x1 = _x156 ∧ x2 = _x157 ∧ x3 = _x158 ∧ x4 = _x159 ∧ _x155 = _x159 ∧ _x153 = _x157 ∧ _x154 = _x158 ∧ _x152 = _x156
l2	22	l12:	x1 = _x160 ∧ x2 = _x161 ∧ x3 = _x162 ∧ x4 = _x163 ∧ x1 = _x164 ∧ x2 = _x165 ∧ x3 = _x166 ∧ x4 = _x167 ∧ _x163 = _x167 ∧ _x161 = _x165 ∧ _x162 = _x166 ∧ _x160 = _x164
l17	23	l6:	x1 = _x168 ∧ x2 = _x169 ∧ x3 = _x170 ∧ x4 = _x171 ∧ x1 = _x172 ∧ x2 = _x173 ∧ x3 = _x174 ∧ x4 = _x175 ∧ _x176 = 0 ∧ _x174 = 0 ∧ _x168 = _x172 ∧ _x169 = _x173 ∧ _x171 = _x175
l18	24	l17:	x1 = _x177 ∧ x2 = _x178 ∧ x3 = _x179 ∧ x4 = _x180 ∧ x1 = _x181 ∧ x2 = _x182 ∧ x3 = _x183 ∧ x4 = _x184 ∧ _x180 = _x184 ∧ _x178 = _x182 ∧ _x179 = _x183 ∧ _x177 = _x181

Proof

1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:

l5	l5	l5:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
l7	l7	l7:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
l1	l1	l1:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
l3	l3	l3:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
l13	l13	l13:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
l18	l18	l18:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
l17	l17	l17:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
l2	l2	l2:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
l9	l9	l9:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
l14	l14	l14:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
l4	l4	l4:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
l6	l6	l6:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
l10	l10	l10:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
l8	l8	l8:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
l15	l15	l15:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
l16	l16	l16:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
l0	l0	l0:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
l12	l12	l12:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4

and for every transition t, a duplicate t is considered.

2 SCC Decomposition

We consider subproblems for each of the 2 SCC(s) of the program graph.

2.1 SCC Subproblem 1/2

Here we consider the SCC { l6, l13, l15, l16, l0, l12, l2, l14 }.

2.1.1 Transition Removal

We remove transition 2 using the following ranking functions, which are bounded by 0.

l0:	−1 + x1 − x3
l2:	−2 + x1 − x3
l6:	−1 + x1 − x3
l12:	−2 + x1 − x3
l14:	−2 + x1 − x3
l15:	−2 + x1 − x3
l13:	−2 + x1 − x3
l16:	−2 + x1 − x3

2.1.2 Transition Removal

We remove transitions 6, 14, 22, 17, 19, 16, 18, 20, 21 using the following ranking functions, which are bounded by 0.

l6:	0
l0:	−1
l12:	1
l2:	6
l14:	2
l15:	4
l13:	3
l16:	5

2.1.3 Trivial Cooperation Program

There are no more "sharp" transitions in the cooperation program. Hence the cooperation termination is proved.

2.2 SCC Subproblem 2/2

Here we consider the SCC { l5, l4, l7, l10, l1, l8, l3, l9 }.

2.2.1 Transition Removal

We remove transition 13 using the following ranking functions, which are bounded by 0.

l1:	8⋅x1 − 8⋅x3 + 3
l10:	8⋅x1 − 8⋅x3 + 2
l3:	8⋅x1 − 8⋅x3 − 4
l9:	8⋅x1 − 8⋅x3 + 1
l5:	8⋅x1 − 8⋅x3 − 3
l7:	8⋅x1 − 8⋅x3 − 1
l4:	8⋅x1 − 8⋅x3 − 2
l8:	8⋅x1 − 8⋅x3

2.2.2 Transition Removal

We remove transitions 15, 3, 11, 5, 8, 4, 7, 9, 10 using the following ranking functions, which are bounded by 0.

l1:	0
l10:	−1
l3:	1
l9:	6
l5:	2
l7:	4
l4:	3
l8:	5

2.2.3 Trivial Cooperation Program

There are no more "sharp" transitions in the cooperation program. Hence the cooperation termination is proved.

Tool configuration

AProVE

version: AProVE Commit ID: unknown
strategy: Statistics for single proof: 100.00 % (10 real / 0 unknown / 0 assumptions / 10 total proof steps)