# LTS Termination Proof

by AProVE

## Input

Integer Transition System
• Initial Location: l5, l4, l7, l6, l8, l3, l0, l2
• Transitions: (pre-variables and post-variables)  l0 1 l1: x1 = _cHAT0 ∧ x2 = _oxHAT0 ∧ x3 = _oyHAT0 ∧ x4 = _xHAT0 ∧ x5 = _yHAT0 ∧ x1 = _cHATpost ∧ x2 = _oxHATpost ∧ x3 = _oyHATpost ∧ x4 = _xHATpost ∧ x5 = _yHATpost ∧ _yHAT0 = _yHATpost ∧ _xHAT0 = _xHATpost ∧ _oyHAT0 = _oyHATpost ∧ _oxHAT0 = _oxHATpost ∧ _cHAT0 = _cHATpost ∧ 1 + _yHAT0 ≤ 0 l0 2 l1: x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x4 = _x3 ∧ x5 = _x4 ∧ x1 = _x5 ∧ x2 = _x6 ∧ x3 = _x7 ∧ x4 = _x8 ∧ x5 = _x9 ∧ _x4 = _x9 ∧ _x3 = _x8 ∧ _x2 = _x7 ∧ _x1 = _x6 ∧ _x = _x5 ∧ _x2 ≤ _x4 l2 3 l0: x1 = _x10 ∧ x2 = _x11 ∧ x3 = _x12 ∧ x4 = _x13 ∧ x5 = _x14 ∧ x1 = _x15 ∧ x2 = _x16 ∧ x3 = _x17 ∧ x4 = _x18 ∧ x5 = _x19 ∧ _x14 = _x19 ∧ _x13 = _x18 ∧ _x12 = _x17 ∧ _x11 = _x16 ∧ _x10 = _x15 ∧ 1 + _x13 ≤ 0 l2 4 l0: x1 = _x20 ∧ x2 = _x21 ∧ x3 = _x22 ∧ x4 = _x23 ∧ x5 = _x24 ∧ x1 = _x25 ∧ x2 = _x26 ∧ x3 = _x27 ∧ x4 = _x28 ∧ x5 = _x29 ∧ _x24 = _x29 ∧ _x23 = _x28 ∧ _x22 = _x27 ∧ _x21 = _x26 ∧ _x20 = _x25 ∧ _x21 ≤ _x23 l2 5 l3: x1 = _x30 ∧ x2 = _x31 ∧ x3 = _x32 ∧ x4 = _x33 ∧ x5 = _x34 ∧ x1 = _x35 ∧ x2 = _x36 ∧ x3 = _x37 ∧ x4 = _x38 ∧ x5 = _x39 ∧ _x34 = _x39 ∧ _x32 = _x37 ∧ _x31 = _x36 ∧ _x30 = _x35 ∧ _x38 = _x33 l4 6 l3: x1 = _x40 ∧ x2 = _x41 ∧ x3 = _x42 ∧ x4 = _x43 ∧ x5 = _x44 ∧ x1 = _x45 ∧ x2 = _x46 ∧ x3 = _x47 ∧ x4 = _x48 ∧ x5 = _x49 ∧ _x44 = _x49 ∧ _x43 = _x48 ∧ _x42 = _x47 ∧ _x41 = _x46 ∧ _x40 = _x45 ∧ _x40 ≤ 0 l4 7 l3: x1 = _x50 ∧ x2 = _x51 ∧ x3 = _x52 ∧ x4 = _x53 ∧ x5 = _x54 ∧ x1 = _x55 ∧ x2 = _x56 ∧ x3 = _x57 ∧ x4 = _x58 ∧ x5 = _x59 ∧ _x54 = _x59 ∧ _x53 = _x58 ∧ _x55 = 1 ∧ _x57 = _x54 ∧ _x56 = _x53 l5 8 l4: x1 = _x60 ∧ x2 = _x61 ∧ x3 = _x62 ∧ x4 = _x63 ∧ x5 = _x64 ∧ x1 = _x65 ∧ x2 = _x66 ∧ x3 = _x67 ∧ x4 = _x68 ∧ x5 = _x69 ∧ _x64 = _x69 ∧ _x63 = _x68 ∧ _x62 = _x67 ∧ _x61 = _x66 ∧ _x60 = _x65 ∧ _x60 ≤ 0 l5 9 l2: x1 = _x70 ∧ x2 = _x71 ∧ x3 = _x72 ∧ x4 = _x73 ∧ x5 = _x74 ∧ x1 = _x75 ∧ x2 = _x76 ∧ x3 = _x77 ∧ x4 = _x78 ∧ x5 = _x79 ∧ _x74 = _x79 ∧ _x73 = _x78 ∧ _x72 = _x77 ∧ _x71 = _x76 ∧ _x70 = _x75 ∧ 1 ≤ _x70 l6 10 l5: x1 = _x80 ∧ x2 = _x81 ∧ x3 = _x82 ∧ x4 = _x83 ∧ x5 = _x84 ∧ x1 = _x85 ∧ x2 = _x86 ∧ x3 = _x87 ∧ x4 = _x88 ∧ x5 = _x89 ∧ _x83 = _x88 ∧ _x82 = _x87 ∧ _x81 = _x86 ∧ _x80 = _x85 ∧ _x89 = −1 + _x84 l6 11 l5: x1 = _x90 ∧ x2 = _x91 ∧ x3 = _x92 ∧ x4 = _x93 ∧ x5 = _x94 ∧ x1 = _x95 ∧ x2 = _x96 ∧ x3 = _x97 ∧ x4 = _x98 ∧ x5 = _x99 ∧ _x94 = _x99 ∧ _x92 = _x97 ∧ _x91 = _x96 ∧ _x90 = _x95 ∧ _x98 = −1 + _x93 l3 12 l6: x1 = _x100 ∧ x2 = _x101 ∧ x3 = _x102 ∧ x4 = _x103 ∧ x5 = _x104 ∧ x1 = _x105 ∧ x2 = _x106 ∧ x3 = _x107 ∧ x4 = _x108 ∧ x5 = _x109 ∧ _x104 = _x109 ∧ _x103 = _x108 ∧ _x102 = _x107 ∧ _x101 = _x106 ∧ _x100 = _x105 ∧ 1 ≤ _x104 ∧ 1 ≤ _x103 l7 13 l3: x1 = _x110 ∧ x2 = _x111 ∧ x3 = _x112 ∧ x4 = _x113 ∧ x5 = _x114 ∧ x1 = _x115 ∧ x2 = _x116 ∧ x3 = _x117 ∧ x4 = _x118 ∧ x5 = _x119 ∧ _x114 = _x119 ∧ _x113 = _x118 ∧ _x112 = _x117 ∧ _x111 = _x116 ∧ _x110 = _x115 ∧ _x110 ≤ 0 l8 14 l7: x1 = _x120 ∧ x2 = _x121 ∧ x3 = _x122 ∧ x4 = _x123 ∧ x5 = _x124 ∧ x1 = _x125 ∧ x2 = _x126 ∧ x3 = _x127 ∧ x4 = _x128 ∧ x5 = _x129 ∧ _x124 = _x129 ∧ _x123 = _x128 ∧ _x122 = _x127 ∧ _x121 = _x126 ∧ _x120 = _x125

## 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 l4 l4 l4: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 l7 l7 l7: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 l6 l6 l6: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 l8 l8 l8: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 l3 l3 l3: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 l0 l0 l0: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 l2 l2 l2: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5
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, l3, l2 }.

### 2.1.1 Transition Removal

We remove transition 12 using the following ranking functions, which are bounded by 0.

 l2: −1 + x4 + x5 l3: −1 + x4 + x5 l5: −1 + x4 + x5 l6: −2 + x4 + x5 l4: −1 + x4 + x5

### 2.1.2 Transition Removal

We remove transitions 5, 9, 10, 7, 6, 8, 11 using the following ranking functions, which are bounded by 0.

 l2: 0 l3: −1 l5: 1 l6: 2 l4: 0

### 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 % (5 real / 0 unknown / 0 assumptions / 5 total proof steps)