# LTS Termination Proof

by AProVE

## Input

Integer Transition System
• Initial Location: l5, l4, l7, l6, l3, l0, l2
• Transitions: (pre-variables and post-variables)  l0 1 l1: x1 = _copiedHAT0 ∧ x2 = _eHAT0 ∧ x3 = _nHAT0 ∧ x4 = _oldeHAT0 ∧ x5 = _oldnHAT0 ∧ x1 = _copiedHATpost ∧ x2 = _eHATpost ∧ x3 = _nHATpost ∧ x4 = _oldeHATpost ∧ x5 = _oldnHATpost ∧ _oldnHAT0 = _oldnHATpost ∧ _oldeHAT0 = _oldeHATpost ∧ _nHAT0 = _nHATpost ∧ _eHAT0 = _eHATpost ∧ _copiedHAT0 = _copiedHATpost ∧ _nHAT0 ≤ _oldnHAT0 ∧ _oldeHAT0 ≤ _eHAT0 ∧ 1 ≤ _copiedHAT0 l0 2 l2: x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x4 = _x3 ∧ x5 = _x4 ∧ x1 = _x5 ∧ x2 = _x6 ∧ x3 = _x7 ∧ x4 = _x8 ∧ x5 = _x9 ∧ _x6 = 1 + _x1 ∧ _x7 = 11 + _x2 ∧ _x2 ≤ 100 ∧ 1 ≤ _x1 ∧ _x8 = _x1 ∧ _x9 = _x2 ∧ _x5 = 1 ∧ _x ≤ 0 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 l0 4 l3: x1 = _x20 ∧ x2 = _x21 ∧ x3 = _x22 ∧ x4 = _x23 ∧ x5 = _x24 ∧ x1 = _x25 ∧ x2 = _x26 ∧ x3 = _x27 ∧ x4 = _x28 ∧ x5 = _x29 ∧ _x26 = −1 + _x21 ∧ _x27 = −10 + _x22 ∧ 101 ≤ _x22 ∧ 1 ≤ _x21 ∧ _x28 = _x21 ∧ _x29 = _x22 ∧ _x25 = 1 ∧ _x20 ≤ 0 l3 5 l0: x1 = _x30 ∧ x2 = _x31 ∧ x3 = _x32 ∧ x4 = _x33 ∧ x5 = _x34 ∧ x1 = _x35 ∧ x2 = _x36 ∧ x3 = _x37 ∧ x4 = _x38 ∧ x5 = _x39 ∧ _x34 = _x39 ∧ _x33 = _x38 ∧ _x32 = _x37 ∧ _x31 = _x36 ∧ _x30 = _x35 l0 6 l4: x1 = _x40 ∧ x2 = _x41 ∧ x3 = _x42 ∧ x4 = _x43 ∧ x5 = _x44 ∧ x1 = _x45 ∧ x2 = _x46 ∧ x3 = _x47 ∧ x4 = _x48 ∧ x5 = _x49 ∧ _x44 = _x49 ∧ _x43 = _x48 ∧ _x40 = _x45 ∧ _x46 = 1 + _x41 ∧ _x47 = 11 + _x42 ∧ _x42 ≤ 100 ∧ 1 ≤ _x41 l4 7 l0: x1 = _x50 ∧ x2 = _x51 ∧ x3 = _x52 ∧ x4 = _x53 ∧ x5 = _x54 ∧ x1 = _x55 ∧ x2 = _x56 ∧ x3 = _x57 ∧ x4 = _x58 ∧ x5 = _x59 ∧ _x54 = _x59 ∧ _x53 = _x58 ∧ _x52 = _x57 ∧ _x51 = _x56 ∧ _x50 = _x55 l0 8 l5: x1 = _x60 ∧ x2 = _x61 ∧ x3 = _x62 ∧ x4 = _x63 ∧ x5 = _x64 ∧ x1 = _x65 ∧ x2 = _x66 ∧ x3 = _x67 ∧ x4 = _x68 ∧ x5 = _x69 ∧ _x64 = _x69 ∧ _x63 = _x68 ∧ _x60 = _x65 ∧ _x66 = −1 + _x61 ∧ _x67 = −10 + _x62 ∧ 101 ≤ _x62 ∧ 1 ≤ _x61 l5 9 l0: 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 l6 10 l0: x1 = _x80 ∧ x2 = _x81 ∧ x3 = _x82 ∧ x4 = _x83 ∧ x5 = _x84 ∧ x1 = _x85 ∧ x2 = _x86 ∧ x3 = _x87 ∧ x4 = _x88 ∧ x5 = _x89 ∧ _x84 = _x89 ∧ _x83 = _x88 ∧ _x85 = 0 ∧ _x86 = 1 ∧ _x87 = _x87 l7 11 l6: x1 = _x90 ∧ x2 = _x91 ∧ x3 = _x92 ∧ x4 = _x93 ∧ x5 = _x94 ∧ x1 = _x95 ∧ x2 = _x96 ∧ x3 = _x97 ∧ x4 = _x98 ∧ x5 = _x99 ∧ _x94 = _x99 ∧ _x93 = _x98 ∧ _x92 = _x97 ∧ _x91 = _x96 ∧ _x90 = _x95

## 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 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, l3, l0, l2 }.

### 2.1.1 Transition Removal

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

 l0: 90 − x1 + 10⋅x2 − x3 l2: 90 − x1 + 10⋅x2 − x3 l3: 91 − x1 + 10⋅x2 − x3 l4: 91 − x1 + 10⋅x2 − x3 l5: 90 − x1 + 10⋅x2 − x3

### 2.1.2 Transition Removal

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

 l3: −1 l0: −1 l4: −1 l5: −1 l2: 0

### 2.1.3 Transition Removal

We remove transition 6 using the following ranking functions, which are bounded by 0.

 l3: 90 + 10⋅x2 − x3 l0: 90 + 10⋅x2 − x3 l4: 90 + 10⋅x2 − x3 l5: 90 + 10⋅x2 − x3

### 2.1.4 Transition Removal

We remove transitions 4, 8 using the following ranking functions, which are bounded by 0.

 l3: −1 + x2 + x3 l0: −1 + x2 + x3 l4: −1 + x2 + x3 l5: −1 + x2 + x3

### 2.1.5 Transition Removal

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

 l3: 0 l0: −1 l4: 0 l5: 0

### 2.1.6 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)