LTS Termination Proof

Input

Integer Transition System

Initial Location: l5, l7, l1, l8, l3, l0, l2

Transitions: (pre-variables and post-variables)

l0	1	l1:	x1 = _Result_4HAT0 ∧ x2 = ___cil_tmp2_6HAT0 ∧ x3 = ___cil_tmp6_12HAT0 ∧ x4 = ___disjvr_0HAT0 ∧ x5 = _maxRetries_9HAT0 ∧ x6 = _retryCount_10HAT0 ∧ x7 = _selected_11HAT0 ∧ x8 = _x_5HAT0 ∧ x1 = _Result_4HATpost ∧ x2 = ___cil_tmp2_6HATpost ∧ x3 = ___cil_tmp6_12HATpost ∧ x4 = ___disjvr_0HATpost ∧ x5 = _maxRetries_9HATpost ∧ x6 = _retryCount_10HATpost ∧ x7 = _selected_11HATpost ∧ x8 = _x_5HATpost ∧ ___cil_tmp2_6HATpost = _x_5HAT0 ∧ _Result_4HAT1 = ___cil_tmp2_6HATpost ∧ _selected_11HATpost = _Result_4HAT1 ∧ _Result_4HATpost = _Result_4HATpost ∧ ___cil_tmp6_12HAT0 = ___cil_tmp6_12HATpost ∧ ___disjvr_0HAT0 = ___disjvr_0HATpost ∧ _maxRetries_9HAT0 = _maxRetries_9HATpost ∧ _retryCount_10HAT0 = _retryCount_10HATpost ∧ _x_5HAT0 = _x_5HATpost
l2	2	l3:	x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x4 = _x3 ∧ x5 = _x4 ∧ x6 = _x5 ∧ x7 = _x6 ∧ x8 = _x7 ∧ x1 = _x8 ∧ x2 = _x9 ∧ x3 = _x10 ∧ x4 = _x11 ∧ x5 = _x12 ∧ x6 = _x13 ∧ x7 = _x14 ∧ x8 = _x15 ∧ _x9 = _x7 ∧ _x16 = _x9 ∧ _x14 = _x16 ∧ _x8 = _x8 ∧ _x13 = 1 + _x5 ∧ _x2 = _x10 ∧ _x3 = _x11 ∧ _x4 = _x12 ∧ _x7 = _x15
l3	3	l5:	x1 = _x17 ∧ x2 = _x18 ∧ x3 = _x19 ∧ x4 = _x20 ∧ x5 = _x21 ∧ x6 = _x22 ∧ x7 = _x23 ∧ x8 = _x24 ∧ x1 = _x25 ∧ x2 = _x26 ∧ x3 = _x27 ∧ x4 = _x28 ∧ x5 = _x29 ∧ x6 = _x30 ∧ x7 = _x31 ∧ x8 = _x32 ∧ _x24 = _x32 ∧ _x23 = _x31 ∧ _x22 = _x30 ∧ _x21 = _x29 ∧ _x20 = _x28 ∧ _x19 = _x27 ∧ _x18 = _x26 ∧ _x17 = _x25 ∧ _x28 = _x20
l5	4	l4:	x1 = _x33 ∧ x2 = _x34 ∧ x3 = _x35 ∧ x4 = _x36 ∧ x5 = _x37 ∧ x6 = _x38 ∧ x7 = _x39 ∧ x8 = _x40 ∧ x1 = _x41 ∧ x2 = _x42 ∧ x3 = _x43 ∧ x4 = _x44 ∧ x5 = _x45 ∧ x6 = _x46 ∧ x7 = _x47 ∧ x8 = _x48 ∧ _x40 = _x48 ∧ _x39 = _x47 ∧ _x38 = _x46 ∧ _x37 = _x45 ∧ _x36 = _x44 ∧ _x34 = _x42 ∧ _x41 = _x43 ∧ _x43 = _x39
l3	5	l6:	x1 = _x49 ∧ x2 = _x50 ∧ x3 = _x51 ∧ x4 = _x52 ∧ x5 = _x53 ∧ x6 = _x54 ∧ x7 = _x55 ∧ x8 = _x56 ∧ x1 = _x57 ∧ x2 = _x58 ∧ x3 = _x59 ∧ x4 = _x60 ∧ x5 = _x61 ∧ x6 = _x62 ∧ x7 = _x63 ∧ x8 = _x64 ∧ _x56 = _x64 ∧ _x55 = _x63 ∧ _x54 = _x62 ∧ _x53 = _x61 ∧ _x52 = _x60 ∧ _x50 = _x58 ∧ _x57 = _x59 ∧ _x59 = _x55 ∧ _x53 − _x54 ≤ 0 ∧ 0 ≤ _x55 ∧ _x55 ≤ 0
l3	6	l1:	x1 = _x65 ∧ x2 = _x66 ∧ x3 = _x67 ∧ x4 = _x68 ∧ x5 = _x69 ∧ x6 = _x70 ∧ x7 = _x71 ∧ x8 = _x72 ∧ x1 = _x73 ∧ x2 = _x74 ∧ x3 = _x75 ∧ x4 = _x76 ∧ x5 = _x77 ∧ x6 = _x78 ∧ x7 = _x79 ∧ x8 = _x80 ∧ _x71 ≤ 0 ∧ 0 ≤ _x71 ∧ 0 ≤ −1 + _x69 − _x70 ∧ _x74 = _x72 ∧ _x81 = _x74 ∧ _x79 = _x81 ∧ _x73 = _x73 ∧ _x67 = _x75 ∧ _x68 = _x76 ∧ _x69 = _x77 ∧ _x70 = _x78 ∧ _x72 = _x80
l1	7	l3:	x1 = _x82 ∧ x2 = _x83 ∧ x3 = _x84 ∧ x4 = _x85 ∧ x5 = _x86 ∧ x6 = _x87 ∧ x7 = _x88 ∧ x8 = _x89 ∧ x1 = _x90 ∧ x2 = _x91 ∧ x3 = _x92 ∧ x4 = _x93 ∧ x5 = _x94 ∧ x6 = _x95 ∧ x7 = _x96 ∧ x8 = _x97 ∧ _x89 = _x97 ∧ _x88 = _x96 ∧ _x86 = _x94 ∧ _x85 = _x93 ∧ _x84 = _x92 ∧ _x83 = _x91 ∧ _x82 = _x90 ∧ _x95 = 1 + _x87
l7	8	l3:	x1 = _x98 ∧ x2 = _x99 ∧ x3 = _x100 ∧ x4 = _x101 ∧ x5 = _x102 ∧ x6 = _x103 ∧ x7 = _x104 ∧ x8 = _x105 ∧ x1 = _x106 ∧ x2 = _x107 ∧ x3 = _x108 ∧ x4 = _x109 ∧ x5 = _x110 ∧ x6 = _x111 ∧ x7 = _x112 ∧ x8 = _x113 ∧ _x105 = _x113 ∧ _x101 = _x109 ∧ _x100 = _x108 ∧ _x99 = _x107 ∧ _x98 = _x106 ∧ _x112 = 0 ∧ _x111 = 0 ∧ _x110 = 4
l8	9	l7:	x1 = _x114 ∧ x2 = _x115 ∧ x3 = _x116 ∧ x4 = _x117 ∧ x5 = _x118 ∧ x6 = _x119 ∧ x7 = _x120 ∧ x8 = _x121 ∧ x1 = _x122 ∧ x2 = _x123 ∧ x3 = _x124 ∧ x4 = _x125 ∧ x5 = _x126 ∧ x6 = _x127 ∧ x7 = _x128 ∧ x8 = _x129 ∧ _x121 = _x129 ∧ _x120 = _x128 ∧ _x119 = _x127 ∧ _x118 = _x126 ∧ _x117 = _x125 ∧ _x116 = _x124 ∧ _x115 = _x123 ∧ _x114 = _x122

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 ∧ x8 = x8
l7	l7	l7:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8
l1	l1	l1:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8
l8	l8	l8:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8
l3	l3	l3:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8
l0	l0	l0:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8
l2	l2	l2:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 ∧ x7 = x7 ∧ x8 = x8

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

2 SCC Decomposition

We consider subproblems for each of the 1 SCC(s) of the program graph.

2.1 SCC Subproblem 1/1

Here we consider the SCC { l1, l3 }.

2.1.1 Transition Removal

We remove transition 6 using the following ranking functions, which are bounded by 0.

l1:	−1 + x5 − x6
l3:	x5 − x6

2.1.2 Transition Removal

We remove transition 7 using the following ranking functions, which are bounded by 0.

l1:	0
l3:	−1

2.1.3 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 % (6 real / 0 unknown / 0 assumptions / 6 total proof steps)