# LTS Termination Proof

by AProVE

## Input

Integer Transition System
• Initial Location: l4, l1, l3, l0, l2
• Transitions: (pre-variables and post-variables)  l0 1 l1: x1 = _xHAT0 ∧ x2 = _yHAT0 ∧ x3 = _zHAT0 ∧ x1 = _xHATpost ∧ x2 = _yHATpost ∧ x3 = _zHATpost ∧ _zHAT0 = _zHATpost ∧ _yHAT0 = _yHATpost ∧ _xHAT0 = _xHATpost l2 2 l1: x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x1 = _x3 ∧ x2 = _x4 ∧ x3 = _x5 ∧ _x2 = _x5 ∧ _x1 = _x4 ∧ _x3 = 1 + _x ∧ _x2 ≤ _x1 l2 3 l3: x1 = _x6 ∧ x2 = _x7 ∧ x3 = _x8 ∧ x1 = _x9 ∧ x2 = _x10 ∧ x3 = _x11 ∧ _x8 = _x11 ∧ _x6 = _x9 ∧ _x10 = 1 + _x7 ∧ 1 + _x7 ≤ _x8 l3 4 l2: x1 = _x12 ∧ x2 = _x13 ∧ x3 = _x14 ∧ x1 = _x15 ∧ x2 = _x16 ∧ x3 = _x17 ∧ _x14 = _x17 ∧ _x13 = _x16 ∧ _x12 = _x15 l1 5 l2: x1 = _x18 ∧ x2 = _x19 ∧ x3 = _x20 ∧ x1 = _x21 ∧ x2 = _x22 ∧ x3 = _x23 ∧ _x20 = _x23 ∧ _x19 = _x22 ∧ _x18 = _x21 ∧ 1 + _x18 ≤ _x19 l4 6 l0: x1 = _x24 ∧ x2 = _x25 ∧ x3 = _x26 ∧ x1 = _x27 ∧ x2 = _x28 ∧ x3 = _x29 ∧ _x26 = _x29 ∧ _x25 = _x28 ∧ _x24 = _x27

## Proof

### 1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
 l4 l4 l4: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 l1 l1 l1: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 l3 l3 l3: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 l0 l0 l0: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 l2 l2 l2: 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 1 SCC(s) of the program graph.

### 2.1 SCC Subproblem 1/1

Here we consider the SCC { l1, l3, l2 }.

### 2.1.1 Transition Removal

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

 l1: −2⋅x2 + 2⋅x3 + 1 l2: −2⋅x2 + 2⋅x3 + 1 l3: −2⋅x2 + 2⋅x3 + 2

### 2.1.2 Transition Removal

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

 l1: −1 l2: −1 l3: 0

### 2.1.3 Transition Removal

We remove transition 5 using the following ranking functions, which are bounded by 0.

 l1: −2⋅x1 + 2⋅x2 l2: −2⋅x1 + 2⋅x2 − 1

### 2.1.4 Transition Removal

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

 l2: 0 l1: −1

### 2.1.5 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 % (7 real / 0 unknown / 0 assumptions / 7 total proof steps)