# LTS Termination Proof

by AProVE

## 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 = _maxRetries_9HAT0 ∧ x5 = _retryCount_10HAT0 ∧ x6 = _selected_11HAT0 ∧ x7 = _x_5HAT0 ∧ x1 = _Result_4HATpost ∧ x2 = ____cil_tmp2_6HATpost ∧ x3 = ____cil_tmp6_12HATpost ∧ x4 = _maxRetries_9HATpost ∧ x5 = _retryCount_10HATpost ∧ x6 = _selected_11HATpost ∧ x7 = _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 ∧ _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 ∧ x1 = _x7 ∧ x2 = _x8 ∧ x3 = _x9 ∧ x4 = _x10 ∧ x5 = _x11 ∧ x6 = _x12 ∧ x7 = _x13 ∧ _x8 = _x6 ∧ _x14 = _x8 ∧ _x12 = _x14 ∧ _x7 = _x7 ∧ _x11 = 1 + _x4 ∧ _x2 = _x9 ∧ _x3 = _x10 ∧ _x6 = _x13 l3 3 l5: x1 = _x15 ∧ x2 = _x16 ∧ x3 = _x17 ∧ x4 = _x18 ∧ x5 = _x19 ∧ x6 = _x20 ∧ x7 = _x21 ∧ x1 = _x22 ∧ x2 = _x23 ∧ x3 = _x24 ∧ x4 = _x25 ∧ x5 = _x26 ∧ x6 = _x27 ∧ x7 = _x28 ∧ _x21 = _x28 ∧ _x20 = _x27 ∧ _x19 = _x26 ∧ _x18 = _x25 ∧ _x17 = _x24 ∧ _x16 = _x23 ∧ _x15 = _x22 ∧ 1 + _x20 ≤ 0 l3 4 l5: 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 ∧ 1 ≤ _x34 l5 5 l4: 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 ∧ _x49 = _x56 ∧ _x48 = _x55 ∧ _x47 = _x54 ∧ _x46 = _x53 ∧ _x44 = _x51 ∧ _x50 = _x52 ∧ _x52 = _x48 l3 6 l6: x1 = _x57 ∧ x2 = _x58 ∧ x3 = _x59 ∧ x4 = _x60 ∧ x5 = _x61 ∧ x6 = _x62 ∧ x7 = _x63 ∧ x1 = _x64 ∧ x2 = _x65 ∧ x3 = _x66 ∧ x4 = _x67 ∧ x5 = _x68 ∧ x6 = _x69 ∧ x7 = _x70 ∧ _x63 = _x70 ∧ _x62 = _x69 ∧ _x61 = _x68 ∧ _x60 = _x67 ∧ _x58 = _x65 ∧ _x64 = _x66 ∧ _x66 = _x62 ∧ _x60 − _x61 ≤ 0 ∧ 0 ≤ _x62 ∧ _x62 ≤ 0 l3 7 l1: x1 = _x71 ∧ x2 = _x72 ∧ x3 = _x73 ∧ x4 = _x74 ∧ x5 = _x75 ∧ x6 = _x76 ∧ x7 = _x77 ∧ x1 = _x78 ∧ x2 = _x79 ∧ x3 = _x80 ∧ x4 = _x81 ∧ x5 = _x82 ∧ x6 = _x83 ∧ x7 = _x84 ∧ _x76 ≤ 0 ∧ 0 ≤ _x76 ∧ 0 ≤ −1 + _x74 − _x75 ∧ _x79 = _x77 ∧ _x85 = _x79 ∧ _x83 = _x85 ∧ _x78 = _x78 ∧ _x73 = _x80 ∧ _x74 = _x81 ∧ _x75 = _x82 ∧ _x77 = _x84 l1 8 l3: 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 ∧ _x92 = _x99 ∧ _x91 = _x98 ∧ _x89 = _x96 ∧ _x88 = _x95 ∧ _x87 = _x94 ∧ _x86 = _x93 ∧ _x97 = 1 + _x90 l7 9 l3: x1 = _x100 ∧ x2 = _x101 ∧ x3 = _x102 ∧ x4 = _x103 ∧ x5 = _x104 ∧ x6 = _x105 ∧ x7 = _x106 ∧ x1 = _x107 ∧ x2 = _x108 ∧ x3 = _x109 ∧ x4 = _x110 ∧ x5 = _x111 ∧ x6 = _x112 ∧ x7 = _x113 ∧ _x106 = _x113 ∧ _x102 = _x109 ∧ _x101 = _x108 ∧ _x100 = _x107 ∧ _x112 = 0 ∧ _x111 = 0 ∧ _x110 = 400 l8 10 l7: x1 = _x114 ∧ x2 = _x115 ∧ x3 = _x116 ∧ x4 = _x117 ∧ x5 = _x118 ∧ x6 = _x119 ∧ x7 = _x120 ∧ x1 = _x121 ∧ x2 = _x122 ∧ x3 = _x123 ∧ x4 = _x124 ∧ x5 = _x125 ∧ x6 = _x126 ∧ x7 = _x127 ∧ _x120 = _x127 ∧ _x119 = _x126 ∧ _x118 = _x125 ∧ _x117 = _x124 ∧ _x116 = _x123 ∧ _x115 = _x122 ∧ _x114 = _x121

## 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 l7 l7 l7: 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 l8 l8 l8: 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 l2 l2 l2: 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 { l1, l3 }.

### 2.1.1 Transition Removal

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

 l1: −1 + x4 − x5 l3: x4 − x5

### 2.1.2 Transition Removal

We remove transition 8 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)