# LTS Termination Proof

by AProVE

## Input

Integer Transition System
• Initial Location: l5, l4, l7, l6, l10, l1, l3, l0, l2, l9
• Transitions: (pre-variables and post-variables)  l0 1 l1: x1 = _NHAT0 ∧ x2 = _iHAT0 ∧ x3 = _tmpHAT0 ∧ x4 = _xHAT0 ∧ x1 = _NHATpost ∧ x2 = _iHATpost ∧ x3 = _tmpHATpost ∧ x4 = _xHATpost ∧ _tmpHAT0 = _tmpHATpost ∧ _iHAT0 = _iHATpost ∧ _NHAT0 = _NHATpost ∧ _xHATpost = 2 + _xHAT0 l0 2 l1: x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x4 = _x3 ∧ x1 = _x4 ∧ x2 = _x5 ∧ x3 = _x6 ∧ x4 = _x7 ∧ _x3 = _x7 ∧ _x2 = _x6 ∧ _x1 = _x5 ∧ _x = _x4 l2 3 l3: x1 = _x8 ∧ x2 = _x9 ∧ x3 = _x10 ∧ x4 = _x11 ∧ x1 = _x12 ∧ x2 = _x13 ∧ x3 = _x14 ∧ x4 = _x15 ∧ _x11 = _x15 ∧ _x10 = _x14 ∧ _x9 = _x13 ∧ _x8 = _x12 ∧ _x8 ≤ _x9 l2 4 l0: x1 = _x16 ∧ x2 = _x17 ∧ x3 = _x18 ∧ x4 = _x19 ∧ x1 = _x20 ∧ x2 = _x21 ∧ x3 = _x22 ∧ x4 = _x23 ∧ _x19 = _x23 ∧ _x18 = _x22 ∧ _x17 = _x21 ∧ _x16 = _x20 ∧ 1 + _x17 ≤ _x16 l4 5 l2: x1 = _x24 ∧ x2 = _x25 ∧ x3 = _x26 ∧ x4 = _x27 ∧ x1 = _x28 ∧ x2 = _x29 ∧ x3 = _x30 ∧ x4 = _x31 ∧ _x27 = _x31 ∧ _x26 = _x30 ∧ _x25 = _x29 ∧ _x24 = _x28 l5 6 l6: x1 = _x32 ∧ x2 = _x33 ∧ x3 = _x34 ∧ x4 = _x35 ∧ x1 = _x36 ∧ x2 = _x37 ∧ x3 = _x38 ∧ x4 = _x39 ∧ _x35 = _x39 ∧ _x33 = _x37 ∧ _x32 = _x36 ∧ _x38 = 0 l7 7 l5: x1 = _x40 ∧ x2 = _x41 ∧ x3 = _x42 ∧ x4 = _x43 ∧ x1 = _x44 ∧ x2 = _x45 ∧ x3 = _x46 ∧ x4 = _x47 ∧ _x43 = _x47 ∧ _x42 = _x46 ∧ _x41 = _x45 ∧ _x40 = _x44 ∧ 2 + _x40 ≤ _x43 l7 8 l5: x1 = _x48 ∧ x2 = _x49 ∧ x3 = _x50 ∧ x4 = _x51 ∧ x1 = _x52 ∧ x2 = _x53 ∧ x3 = _x54 ∧ x4 = _x55 ∧ _x51 = _x55 ∧ _x50 = _x54 ∧ _x49 = _x53 ∧ _x48 = _x52 ∧ 1 + _x51 ≤ 1 + _x48 l7 9 l6: x1 = _x56 ∧ x2 = _x57 ∧ x3 = _x58 ∧ x4 = _x59 ∧ x1 = _x60 ∧ x2 = _x61 ∧ x3 = _x62 ∧ x4 = _x63 ∧ _x59 = _x63 ∧ _x57 = _x61 ∧ _x56 = _x60 ∧ _x62 = 1 ∧ 1 + _x56 ≤ _x59 ∧ _x59 ≤ 1 + _x56 l6 10 l8: x1 = _x64 ∧ x2 = _x65 ∧ x3 = _x66 ∧ x4 = _x67 ∧ x1 = _x68 ∧ x2 = _x69 ∧ x3 = _x70 ∧ x4 = _x71 ∧ _x67 = _x71 ∧ _x66 = _x70 ∧ _x65 = _x69 ∧ _x64 = _x68 l3 11 l7: x1 = _x72 ∧ x2 = _x73 ∧ x3 = _x74 ∧ x4 = _x75 ∧ x1 = _x76 ∧ x2 = _x77 ∧ x3 = _x78 ∧ x4 = _x79 ∧ _x75 = _x79 ∧ _x74 = _x78 ∧ _x73 = _x77 ∧ _x72 = _x76 ∧ 1 + _x72 ≤ _x75 l3 12 l7: x1 = _x80 ∧ x2 = _x81 ∧ x3 = _x82 ∧ x4 = _x83 ∧ x1 = _x84 ∧ x2 = _x85 ∧ x3 = _x86 ∧ x4 = _x87 ∧ _x83 = _x87 ∧ _x82 = _x86 ∧ _x81 = _x85 ∧ _x80 = _x84 ∧ 1 + _x83 ≤ _x80 l3 13 l6: x1 = _x88 ∧ x2 = _x89 ∧ x3 = _x90 ∧ x4 = _x91 ∧ x1 = _x92 ∧ x2 = _x93 ∧ x3 = _x94 ∧ x4 = _x95 ∧ _x91 = _x95 ∧ _x89 = _x93 ∧ _x88 = _x92 ∧ _x94 = 1 ∧ _x88 ≤ _x91 ∧ _x91 ≤ _x88 l1 14 l4: x1 = _x96 ∧ x2 = _x97 ∧ x3 = _x98 ∧ x4 = _x99 ∧ x1 = _x100 ∧ x2 = _x101 ∧ x3 = _x102 ∧ x4 = _x103 ∧ _x99 = _x103 ∧ _x98 = _x102 ∧ _x96 = _x100 ∧ _x101 = 1 + _x97 l9 15 l4: x1 = _x104 ∧ x2 = _x105 ∧ x3 = _x106 ∧ x4 = _x107 ∧ x1 = _x108 ∧ x2 = _x109 ∧ x3 = _x110 ∧ x4 = _x111 ∧ _x106 = _x110 ∧ _x104 = _x108 ∧ _x109 = 0 ∧ _x111 = 0 l10 16 l9: x1 = _x112 ∧ x2 = _x113 ∧ x3 = _x114 ∧ x4 = _x115 ∧ x1 = _x116 ∧ x2 = _x117 ∧ x3 = _x118 ∧ x4 = _x119 ∧ _x115 = _x119 ∧ _x114 = _x118 ∧ _x113 = _x117 ∧ _x112 = _x116

## Proof

### 1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
 l5 l5 l5: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 l4 l4 l4: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 l7 l7 l7: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 l6 l6 l6: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 l10 l10 l10: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 l1 l1 l1: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 l3 l3 l3: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 l0 l0 l0: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 l2 l2 l2: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 l9 l9 l9: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
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 { l4, l1, l0, l2 }.

### 2.1.1 Transition Removal

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

 l0: −2 + x1 − x2 l1: −2 + x1 − x2 l2: −1 + x1 − x2 l4: −1 + x1 − x2

### 2.1.2 Transition Removal

We remove transitions 1, 5, 14, 2 using the following ranking functions, which are bounded by 0.

 l0: 3 l1: 2 l4: 1 l2: 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)