# LTS Termination Proof

by AProVE

## Input

Integer Transition System
• Initial Location: l5, l7, l1, l3, l13, l18, l17, l2, l9, l14, l4, l6, l10, l8, l15, l16, l0, l12
• Transitions: (pre-variables and post-variables)  l0 1 l1: x1 = _i2HAT0 ∧ x2 = _iHAT0 ∧ x3 = _rHAT0 ∧ x1 = _i2HATpost ∧ x2 = _iHATpost ∧ x3 = _rHATpost ∧ _rHAT0 = _rHATpost ∧ _i2HAT0 = _i2HATpost ∧ _iHATpost = 0 ∧ 1 ≤ _iHAT0 l0 2 l2: x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x1 = _x3 ∧ x2 = _x4 ∧ x3 = _x5 ∧ _x2 = _x5 ∧ _x1 = _x4 ∧ _x3 = _x1 ∧ 1 + _x1 ≤ 1 l3 3 l1: x1 = _x6 ∧ x2 = _x7 ∧ x3 = _x8 ∧ x1 = _x9 ∧ x2 = _x10 ∧ x3 = _x11 ∧ _x8 = _x11 ∧ _x6 = _x9 ∧ _x10 = 1 + _x7 l4 4 l5: x1 = _x12 ∧ x2 = _x13 ∧ x3 = _x14 ∧ x1 = _x15 ∧ x2 = _x16 ∧ x3 = _x17 ∧ _x14 = _x17 ∧ _x12 = _x15 ∧ _x13 = _x16 l5 5 l3: x1 = _x18 ∧ x2 = _x19 ∧ x3 = _x20 ∧ x1 = _x21 ∧ x2 = _x22 ∧ x3 = _x23 ∧ _x20 = _x23 ∧ _x18 = _x21 ∧ _x19 = _x22 l6 6 l0: x1 = _x24 ∧ x2 = _x25 ∧ x3 = _x26 ∧ x1 = _x27 ∧ x2 = _x28 ∧ x3 = _x29 ∧ _x26 = _x29 ∧ _x24 = _x27 ∧ _x25 = _x28 l7 7 l4: x1 = _x30 ∧ x2 = _x31 ∧ x3 = _x32 ∧ x1 = _x33 ∧ x2 = _x34 ∧ x3 = _x35 ∧ _x32 = _x35 ∧ _x30 = _x33 ∧ _x31 = _x34 l7 8 l5: x1 = _x36 ∧ x2 = _x37 ∧ x3 = _x38 ∧ x1 = _x39 ∧ x2 = _x40 ∧ x3 = _x41 ∧ _x38 = _x41 ∧ _x36 = _x39 ∧ _x37 = _x40 l8 9 l7: x1 = _x42 ∧ x2 = _x43 ∧ x3 = _x44 ∧ x1 = _x45 ∧ x2 = _x46 ∧ x3 = _x47 ∧ _x42 = _x45 ∧ _x43 = _x46 ∧ _x47 = _x47 l9 10 l8: x1 = _x48 ∧ x2 = _x49 ∧ x3 = _x50 ∧ x1 = _x51 ∧ x2 = _x52 ∧ x3 = _x53 ∧ _x50 = _x53 ∧ _x48 = _x51 ∧ _x49 = _x52 l9 11 l3: x1 = _x54 ∧ x2 = _x55 ∧ x3 = _x56 ∧ x1 = _x57 ∧ x2 = _x58 ∧ x3 = _x59 ∧ _x56 = _x59 ∧ _x54 = _x57 ∧ _x55 = _x58 l10 12 l11: x1 = _x60 ∧ x2 = _x61 ∧ x3 = _x62 ∧ x1 = _x63 ∧ x2 = _x64 ∧ x3 = _x65 ∧ _x62 = _x65 ∧ _x60 = _x63 ∧ _x61 = _x64 ∧ 1 ≤ _x61 l10 13 l9: x1 = _x66 ∧ x2 = _x67 ∧ x3 = _x68 ∧ x1 = _x69 ∧ x2 = _x70 ∧ x3 = _x71 ∧ _x68 = _x71 ∧ _x66 = _x69 ∧ _x67 = _x70 ∧ 1 + _x67 ≤ 1 l12 14 l6: x1 = _x72 ∧ x2 = _x73 ∧ x3 = _x74 ∧ x1 = _x75 ∧ x2 = _x76 ∧ x3 = _x77 ∧ _x74 = _x77 ∧ _x72 = _x75 ∧ _x76 = 1 + _x73 l1 15 l10: x1 = _x78 ∧ x2 = _x79 ∧ x3 = _x80 ∧ x1 = _x81 ∧ x2 = _x82 ∧ x3 = _x83 ∧ _x80 = _x83 ∧ _x78 = _x81 ∧ _x79 = _x82 l13 16 l14: x1 = _x84 ∧ x2 = _x85 ∧ x3 = _x86 ∧ x1 = _x87 ∧ x2 = _x88 ∧ x3 = _x89 ∧ _x86 = _x89 ∧ _x84 = _x87 ∧ _x85 = _x88 l14 17 l12: x1 = _x90 ∧ x2 = _x91 ∧ x3 = _x92 ∧ x1 = _x93 ∧ x2 = _x94 ∧ x3 = _x95 ∧ _x92 = _x95 ∧ _x90 = _x93 ∧ _x91 = _x94 l15 18 l13: x1 = _x96 ∧ x2 = _x97 ∧ x3 = _x98 ∧ x1 = _x99 ∧ x2 = _x100 ∧ x3 = _x101 ∧ _x98 = _x101 ∧ _x96 = _x99 ∧ _x97 = _x100 l15 19 l14: x1 = _x102 ∧ x2 = _x103 ∧ x3 = _x104 ∧ x1 = _x105 ∧ x2 = _x106 ∧ x3 = _x107 ∧ _x104 = _x107 ∧ _x102 = _x105 ∧ _x103 = _x106 l16 20 l15: x1 = _x108 ∧ x2 = _x109 ∧ x3 = _x110 ∧ x1 = _x111 ∧ x2 = _x112 ∧ x3 = _x113 ∧ _x110 = _x113 ∧ _x108 = _x111 ∧ _x109 = _x112 l2 21 l16: x1 = _x114 ∧ x2 = _x115 ∧ x3 = _x116 ∧ x1 = _x117 ∧ x2 = _x118 ∧ x3 = _x119 ∧ _x116 = _x119 ∧ _x114 = _x117 ∧ _x115 = _x118 l2 22 l12: x1 = _x120 ∧ x2 = _x121 ∧ x3 = _x122 ∧ x1 = _x123 ∧ x2 = _x124 ∧ x3 = _x125 ∧ _x122 = _x125 ∧ _x120 = _x123 ∧ _x121 = _x124 l17 23 l6: x1 = _x126 ∧ x2 = _x127 ∧ x3 = _x128 ∧ x1 = _x129 ∧ x2 = _x130 ∧ x3 = _x131 ∧ _x132 = 0 ∧ _x130 = 0 ∧ _x126 = _x129 ∧ _x128 = _x131 l18 24 l17: x1 = _x133 ∧ x2 = _x134 ∧ x3 = _x135 ∧ x1 = _x136 ∧ x2 = _x137 ∧ x3 = _x138 ∧ _x135 = _x138 ∧ _x133 = _x136 ∧ _x134 = _x137

## Proof

### 1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
 l5 l5 l5: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 l7 l7 l7: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 l1 l1 l1: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 l3 l3 l3: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 l13 l13 l13: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 l18 l18 l18: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 l17 l17 l17: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 l2 l2 l2: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 l9 l9 l9: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 l14 l14 l14: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 l4 l4 l4: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 l6 l6 l6: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 l10 l10 l10: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 l8 l8 l8: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 l15 l15 l15: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 l16 l16 l16: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 l0 l0 l0: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 l12 l12 l12: x1 = x1 ∧ x2 = x2 ∧ x3 = x3
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 { l6, l13, l15, l16, l0, l12, l2, l14 }.

### 2.1.1 Transition Removal

We remove transition 2 using the following ranking functions, which are bounded by 0.

 l0: − x2 l2: −1 − x2 l6: − x2 l12: −1 − x2 l14: −1 − x2 l15: −1 − x2 l13: −1 − x2 l16: −1 − x2

### 2.1.2 Transition Removal

We remove transitions 6, 14, 22, 17, 19, 16, 18, 20, 21 using the following ranking functions, which are bounded by −6.

 l6: −6 l0: −7 l12: −5 l2: 0 l14: −4 l15: −2 l13: −3 l16: −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 { l5, l4, l7, l10, l1, l8, l3, l9 }.

### 2.2.1 Transition Removal

We remove transition 13 using the following ranking functions, which are bounded by 0.

 l1: − x2 l10: − x2 l3: −1 − x2 l9: −1 − x2 l5: −1 − x2 l7: −1 − x2 l4: −1 − x2 l8: −1 − x2

### 2.2.2 Transition Removal

We remove transitions 15, 3, 11, 5, 8, 4, 7, 9, 10 using the following ranking functions, which are bounded by 0.

 l1: 0 l10: −1 l3: 1 l9: 6 l5: 2 l7: 4 l4: 3 l8: 5

### 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)