LTS Termination Proof

Input

Integer Transition System

Initial Location: l5, l4, l7, l6, l1, l3, l0

Transitions: (pre-variables and post-variables)

l0	1	l1:	x1 = _nd_12HAT0 ∧ x2 = _rt_11HAT0 ∧ x3 = _rv_15HAT0 ∧ x4 = _st_14HAT0 ∧ x5 = _st_16HAT0 ∧ x6 = _x_13HAT0 ∧ x7 = _y_17HAT0 ∧ x1 = _nd_12HATpost ∧ x2 = _rt_11HATpost ∧ x3 = _rv_15HATpost ∧ x4 = _st_14HATpost ∧ x5 = _st_16HATpost ∧ x6 = _x_13HATpost ∧ x7 = _y_17HATpost ∧ _y_17HAT0 = _y_17HATpost ∧ _x_13HAT0 = _x_13HATpost ∧ _st_16HAT0 = _st_16HATpost ∧ _st_14HAT0 = _st_14HATpost ∧ _rv_15HAT0 = _rv_15HATpost ∧ _rt_11HAT0 = _rt_11HATpost ∧ _nd_12HAT0 = _nd_12HATpost
l1	2	l2:	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 ∧ _x3 = _x10 ∧ _x2 = _x9 ∧ _x = _x7 ∧ _x8 = _x3 ∧ _x5 ≤ 0
l1	3	l3:	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 ∧ 1 ≤ _x19 ∧ _x28 = _x28 ∧ _x23 = _x28 ∧ _x21 = _x21 ∧ 0 ≤ _x23 ∧ _x23 ≤ 0 ∧ _x27 = −1 + _x20 ∧ _x25 = _x25 ∧ 2 ≤ _x27 ∧ _x15 = _x22 ∧ _x17 = _x24 ∧ _x19 = _x26
l3	4	l1:	x1 = _x29 ∧ x2 = _x30 ∧ x3 = _x31 ∧ x4 = _x32 ∧ x5 = _x33 ∧ x6 = _x34 ∧ x7 = _x35 ∧ x1 = _x36 ∧ x2 = _x37 ∧ x3 = _x38 ∧ x4 = _x39 ∧ x5 = _x40 ∧ x6 = _x41 ∧ x7 = _x42 ∧ _x35 = _x42 ∧ _x34 = _x41 ∧ _x33 = _x40 ∧ _x32 = _x39 ∧ _x31 = _x38 ∧ _x30 = _x37 ∧ _x29 = _x36
l1	5	l5:	x1 = _x43 ∧ x2 = _x44 ∧ x3 = _x45 ∧ x4 = _x46 ∧ x5 = _x47 ∧ x6 = _x48 ∧ x7 = _x49 ∧ x1 = _x50 ∧ x2 = _x51 ∧ x3 = _x52 ∧ x4 = _x53 ∧ x5 = _x54 ∧ x6 = _x55 ∧ x7 = _x56 ∧ 1 ≤ _x48 ∧ _x57 = _x57 ∧ _x52 = _x57 ∧ _x50 = _x50 ∧ _x44 = _x51 ∧ _x46 = _x53 ∧ _x47 = _x54 ∧ _x48 = _x55 ∧ _x49 = _x56
l5	6	l6:	x1 = _x58 ∧ x2 = _x59 ∧ x3 = _x60 ∧ x4 = _x61 ∧ x5 = _x62 ∧ x6 = _x63 ∧ x7 = _x64 ∧ x1 = _x65 ∧ x2 = _x66 ∧ x3 = _x67 ∧ x4 = _x68 ∧ x5 = _x69 ∧ x6 = _x70 ∧ x7 = _x71 ∧ _x64 = _x71 ∧ _x63 = _x70 ∧ _x62 = _x69 ∧ _x61 = _x68 ∧ _x60 = _x67 ∧ _x59 = _x66 ∧ _x58 = _x65 ∧ 1 ≤ _x60
l5	7	l6:	x1 = _x72 ∧ x2 = _x73 ∧ x3 = _x74 ∧ x4 = _x75 ∧ x5 = _x76 ∧ x6 = _x77 ∧ x7 = _x78 ∧ x1 = _x79 ∧ x2 = _x80 ∧ x3 = _x81 ∧ x4 = _x82 ∧ x5 = _x83 ∧ x6 = _x84 ∧ x7 = _x85 ∧ _x78 = _x85 ∧ _x77 = _x84 ∧ _x76 = _x83 ∧ _x75 = _x82 ∧ _x74 = _x81 ∧ _x73 = _x80 ∧ _x72 = _x79 ∧ 1 + _x74 ≤ 0
l6	8	l4:	x1 = _x86 ∧ x2 = _x87 ∧ x3 = _x88 ∧ x4 = _x89 ∧ x5 = _x90 ∧ x6 = _x91 ∧ x7 = _x92 ∧ x1 = _x93 ∧ x2 = _x94 ∧ x3 = _x95 ∧ x4 = _x96 ∧ x5 = _x97 ∧ x6 = _x98 ∧ x7 = _x99 ∧ _x98 = −1 + _x91 ∧ _x100 = _x100 ∧ _x99 = _x100 ∧ _x93 = _x93 ∧ _x87 = _x94 ∧ _x88 = _x95 ∧ _x89 = _x96 ∧ _x90 = _x97
l4	9	l1:	x1 = _x101 ∧ x2 = _x102 ∧ x3 = _x103 ∧ x4 = _x104 ∧ x5 = _x105 ∧ x6 = _x106 ∧ x7 = _x107 ∧ x1 = _x108 ∧ x2 = _x109 ∧ x3 = _x110 ∧ x4 = _x111 ∧ x5 = _x112 ∧ x6 = _x113 ∧ x7 = _x114 ∧ _x107 = _x114 ∧ _x106 = _x113 ∧ _x105 = _x112 ∧ _x104 = _x111 ∧ _x103 = _x110 ∧ _x102 = _x109 ∧ _x101 = _x108
l7	10	l0:	x1 = _x115 ∧ x2 = _x116 ∧ x3 = _x117 ∧ x4 = _x118 ∧ x5 = _x119 ∧ x6 = _x120 ∧ x7 = _x121 ∧ x1 = _x122 ∧ x2 = _x123 ∧ x3 = _x124 ∧ x4 = _x125 ∧ x5 = _x126 ∧ x6 = _x127 ∧ x7 = _x128 ∧ _x121 = _x128 ∧ _x120 = _x127 ∧ _x119 = _x126 ∧ _x118 = _x125 ∧ _x117 = _x124 ∧ _x116 = _x123 ∧ _x115 = _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
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
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

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 { l5, l4, l6, l1, l3 }.

2.1.1 Transition Removal

We remove transition 5 using the following ranking functions, which are bounded by 0.

l1:	−1 + x6
l3:	−1 + x6
l4:	−1 + x6
l6:	−2 + x6
l5:	−2 + x6

2.1.2 Transition Removal

We remove transitions 9, 8, 7, 6 using the following ranking functions, which are bounded by 0.

l1:	−1
l3:	−1
l4:	0
l6:	1
l5:	2

2.1.3 Transition Removal

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

l1:	−1 + x6 + x7
l3:	−1 + x6 + x7

2.1.4 Transition Removal

We remove transition 4 using the following ranking functions, which are bounded by 0.

l3:	0
l1:	−1

2.1.5 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 % (8 real / 0 unknown / 0 assumptions / 8 total proof steps)