# LTS Termination Proof

by AProVE

## Input

Integer Transition System
• Initial Location: l5, l6, l1, l3, l0, l2
• Transitions: (pre-variables and post-variables)  l0 1 l1: x1 = _jHAT0 ∧ x1 = _jHATpost ∧ _jHAT0 = _jHATpost l2 2 l0: x1 = _x ∧ x1 = _x1 ∧ _x = _x1 ∧ 2 ≤ _x l2 3 l3: x1 = _x2 ∧ x1 = _x3 ∧ _x3 = 1 + _x2 ∧ 1 + _x2 ≤ 2 l3 4 l2: x1 = _x4 ∧ x1 = _x5 ∧ _x4 = _x5 l1 5 l4: x1 = _x6 ∧ x1 = _x7 ∧ _x6 = _x7 l5 6 l3: x1 = _x8 ∧ x1 = _x9 ∧ _x9 = 0 l6 7 l5: x1 = _x10 ∧ x1 = _x11 ∧ _x10 = _x11

## Proof

### 1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
 l5 l5 l5: x1 = x1 l6 l6 l6: x1 = x1 l1 l1 l1: x1 = x1 l3 l3 l3: x1 = x1 l0 l0 l0: x1 = x1 l2 l2 l2: x1 = x1
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 { l3, l2 }.

### 2.1.1 Transition Removal

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

 l2: 1 − x1 l3: 1 − x1

### 2.1.2 Transition Removal

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

 l3: 0 l2: −1

### 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)