LTS Termination Proof

Input

Integer Transition System

Initial Location: l5, l4, l7, l6, l10, l11, l1, l3, l0, l12, l2, l9

Transitions: (pre-variables and post-variables)

l0	1	l1:	x1 = ___const_100HAT0 ∧ x2 = ___const_200HAT0 ∧ x3 = _iHAT0 ∧ x4 = _jHAT0 ∧ x5 = _y4HAT0 ∧ x6 = _y6HAT0 ∧ x7 = _y8HAT0 ∧ x1 = ___const_100HATpost ∧ x2 = ___const_200HATpost ∧ x3 = _iHATpost ∧ x4 = _jHATpost ∧ x5 = _y4HATpost ∧ x6 = _y6HATpost ∧ x7 = _y8HATpost ∧ _y8HAT0 = _y8HATpost ∧ _y6HAT0 = _y6HATpost ∧ _y4HAT0 = _y4HATpost ∧ _jHAT0 = _jHATpost ∧ _iHAT0 = _iHATpost ∧ ___const_200HAT0 = ___const_200HATpost ∧ ___const_100HAT0 = ___const_100HATpost
l2	2	l3:	x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x4 = _x3 ∧ x5 = _x4 ∧ x6 = _x5 ∧ x7 = _x6 ∧ x1 = _x7 ∧ x2 = _x8 ∧ x3 = _x9 ∧ x4 = _x10 ∧ x5 = _x11 ∧ x6 = _x12 ∧ x7 = _x13 ∧ _x6 = _x13 ∧ _x5 = _x12 ∧ _x4 = _x11 ∧ _x2 = _x9 ∧ _x1 = _x8 ∧ _x = _x7 ∧ _x10 = _x ∧ _x ≤ _x2
l2	3	l0:	x1 = _x14 ∧ x2 = _x15 ∧ x3 = _x16 ∧ x4 = _x17 ∧ x5 = _x18 ∧ x6 = _x19 ∧ x7 = _x20 ∧ x1 = _x21 ∧ x2 = _x22 ∧ x3 = _x23 ∧ x4 = _x24 ∧ x5 = _x25 ∧ x6 = _x26 ∧ x7 = _x27 ∧ _x20 = _x27 ∧ _x19 = _x26 ∧ _x17 = _x24 ∧ _x16 = _x23 ∧ _x15 = _x22 ∧ _x14 = _x21 ∧ _x25 = _x16 ∧ 1 + _x16 ≤ _x14
l4	4	l3:	x1 = _x28 ∧ x2 = _x29 ∧ x3 = _x30 ∧ x4 = _x31 ∧ x5 = _x32 ∧ x6 = _x33 ∧ x7 = _x34 ∧ x1 = _x35 ∧ x2 = _x36 ∧ x3 = _x37 ∧ x4 = _x38 ∧ x5 = _x39 ∧ x6 = _x40 ∧ x7 = _x41 ∧ _x34 = _x41 ∧ _x33 = _x40 ∧ _x32 = _x39 ∧ _x30 = _x37 ∧ _x29 = _x36 ∧ _x28 = _x35 ∧ _x38 = 1 + _x31
l5	5	l2:	x1 = _x42 ∧ x2 = _x43 ∧ x3 = _x44 ∧ x4 = _x45 ∧ x5 = _x46 ∧ x6 = _x47 ∧ x7 = _x48 ∧ x1 = _x49 ∧ x2 = _x50 ∧ x3 = _x51 ∧ x4 = _x52 ∧ x5 = _x53 ∧ x6 = _x54 ∧ x7 = _x55 ∧ _x48 = _x55 ∧ _x47 = _x54 ∧ _x46 = _x53 ∧ _x45 = _x52 ∧ _x44 = _x51 ∧ _x43 = _x50 ∧ _x42 = _x49
l6	6	l4:	x1 = _x56 ∧ x2 = _x57 ∧ x3 = _x58 ∧ x4 = _x59 ∧ x5 = _x60 ∧ x6 = _x61 ∧ x7 = _x62 ∧ x1 = _x63 ∧ x2 = _x64 ∧ x3 = _x65 ∧ x4 = _x66 ∧ x5 = _x67 ∧ x6 = _x68 ∧ x7 = _x69 ∧ _x62 = _x69 ∧ _x61 = _x68 ∧ _x60 = _x67 ∧ _x59 = _x66 ∧ _x58 = _x65 ∧ _x57 = _x64 ∧ _x56 = _x63
l7	7	l8:	x1 = _x70 ∧ x2 = _x71 ∧ x3 = _x72 ∧ x4 = _x73 ∧ x5 = _x74 ∧ x6 = _x75 ∧ x7 = _x76 ∧ x1 = _x77 ∧ x2 = _x78 ∧ x3 = _x79 ∧ x4 = _x80 ∧ x5 = _x81 ∧ x6 = _x82 ∧ x7 = _x83 ∧ _x76 = _x83 ∧ _x75 = _x82 ∧ _x74 = _x81 ∧ _x73 = _x80 ∧ _x72 = _x79 ∧ _x71 = _x78 ∧ _x70 = _x77 ∧ _x71 ≤ _x73
l7	8	l6:	x1 = _x84 ∧ x2 = _x85 ∧ x3 = _x86 ∧ x4 = _x87 ∧ x5 = _x88 ∧ x6 = _x89 ∧ x7 = _x90 ∧ x1 = _x91 ∧ x2 = _x92 ∧ x3 = _x93 ∧ x4 = _x94 ∧ x5 = _x95 ∧ x6 = _x96 ∧ x7 = _x97 ∧ _x89 = _x96 ∧ _x88 = _x95 ∧ _x87 = _x94 ∧ _x86 = _x93 ∧ _x85 = _x92 ∧ _x84 = _x91 ∧ _x97 = _x87 ∧ 1 + _x87 ≤ _x85
l3	9	l7:	x1 = _x98 ∧ x2 = _x99 ∧ x3 = _x100 ∧ x4 = _x101 ∧ x5 = _x102 ∧ x6 = _x103 ∧ x7 = _x104 ∧ x1 = _x105 ∧ x2 = _x106 ∧ x3 = _x107 ∧ x4 = _x108 ∧ x5 = _x109 ∧ x6 = _x110 ∧ x7 = _x111 ∧ _x104 = _x111 ∧ _x103 = _x110 ∧ _x102 = _x109 ∧ _x101 = _x108 ∧ _x100 = _x107 ∧ _x99 = _x106 ∧ _x98 = _x105
l9	10	l5:	x1 = _x112 ∧ x2 = _x113 ∧ x3 = _x114 ∧ x4 = _x115 ∧ x5 = _x116 ∧ x6 = _x117 ∧ x7 = _x118 ∧ x1 = _x119 ∧ x2 = _x120 ∧ x3 = _x121 ∧ x4 = _x122 ∧ x5 = _x123 ∧ x6 = _x124 ∧ x7 = _x125 ∧ _x118 = _x125 ∧ _x117 = _x124 ∧ _x116 = _x123 ∧ _x115 = _x122 ∧ _x113 = _x120 ∧ _x112 = _x119 ∧ _x121 = 1 + _x114
l10	11	l9:	x1 = _x126 ∧ x2 = _x127 ∧ x3 = _x128 ∧ x4 = _x129 ∧ x5 = _x130 ∧ x6 = _x131 ∧ x7 = _x132 ∧ x1 = _x133 ∧ x2 = _x134 ∧ x3 = _x135 ∧ x4 = _x136 ∧ x5 = _x137 ∧ x6 = _x138 ∧ x7 = _x139 ∧ _x132 = _x139 ∧ _x131 = _x138 ∧ _x130 = _x137 ∧ _x129 = _x136 ∧ _x128 = _x135 ∧ _x127 = _x134 ∧ _x126 = _x133
l1	12	l10:	x1 = _x140 ∧ x2 = _x141 ∧ x3 = _x142 ∧ x4 = _x143 ∧ x5 = _x144 ∧ x6 = _x145 ∧ x7 = _x146 ∧ x1 = _x147 ∧ x2 = _x148 ∧ x3 = _x149 ∧ x4 = _x150 ∧ x5 = _x151 ∧ x6 = _x152 ∧ x7 = _x153 ∧ _x146 = _x153 ∧ _x144 = _x151 ∧ _x143 = _x150 ∧ _x142 = _x149 ∧ _x141 = _x148 ∧ _x140 = _x147 ∧ _x152 = _x142
l11	13	l5:	x1 = _x154 ∧ x2 = _x155 ∧ x3 = _x156 ∧ x4 = _x157 ∧ x5 = _x158 ∧ x6 = _x159 ∧ x7 = _x160 ∧ x1 = _x161 ∧ x2 = _x162 ∧ x3 = _x163 ∧ x4 = _x164 ∧ x5 = _x165 ∧ x6 = _x166 ∧ x7 = _x167 ∧ _x160 = _x167 ∧ _x159 = _x166 ∧ _x158 = _x165 ∧ _x157 = _x164 ∧ _x155 = _x162 ∧ _x154 = _x161 ∧ _x163 = 0
l12	14	l11:	x1 = _x168 ∧ x2 = _x169 ∧ x3 = _x170 ∧ x4 = _x171 ∧ x5 = _x172 ∧ x6 = _x173 ∧ x7 = _x174 ∧ x1 = _x175 ∧ x2 = _x176 ∧ x3 = _x177 ∧ x4 = _x178 ∧ x5 = _x179 ∧ x6 = _x180 ∧ x7 = _x181 ∧ _x174 = _x181 ∧ _x173 = _x180 ∧ _x172 = _x179 ∧ _x171 = _x178 ∧ _x170 = _x177 ∧ _x169 = _x176 ∧ _x168 = _x175

Proof

1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:

l5	l5	l5:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l4	l4	l4:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l7	l7	l7:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l6	l6	l6:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l10	l10	l10:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l11	l11	l11:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l1	l1	l1:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l3	l3	l3:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l0	l0	l0:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l12	l12	l12:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l2	l2	l2:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7
l9	l9	l9:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7

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 { l5, l10, l1, l0, l2, l9 }.

2.1.1 Transition Removal

We remove transition 3 using the following ranking functions, which are bounded by 0.

l0:	6⋅x1 − 6⋅x3 − 1
l1:	6⋅x1 − 6⋅x3 − 2
l2:	−6⋅x3 + 6⋅x1
l5:	6⋅x1 − 6⋅x3 + 1
l9:	6⋅x1 − 6⋅x3 − 4
l10:	6⋅x1 − 6⋅x3 − 3

2.1.2 Transition Removal

We remove transitions 1, 5, 10, 11, 12 using the following ranking functions, which are bounded by −3.

l0:	1
l1:	0
l5:	−3
l2:	−4
l9:	−2
l10:	−1

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 { l4, l7, l6, l3 }.

2.2.1 Transition Removal

We remove transition 8 using the following ranking functions, which are bounded by 0.

l3:	4⋅x2 − 4⋅x4 + 1
l7:	4⋅x2 − 4⋅x4
l4:	4⋅x2 − 4⋅x4 − 2
l6:	4⋅x2 − 4⋅x4 − 1

2.2.2 Transition Removal

We remove transitions 9, 4, 6 using the following ranking functions, which are bounded by 0.

l3:	0
l7:	−1
l4:	1
l6:	2

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)