# LTS Termination Proof

by AProVE

## Input

Integer Transition System
• Initial Location: l1, l3, l0, l2
• Transitions: (pre-variables and post-variables)  l0 1 l1: x1 = _xHAT0 ∧ x2 = _yHAT0 ∧ x1 = _xHATpost ∧ x2 = _yHATpost ∧ _yHAT0 = _yHATpost ∧ _xHAT0 = _xHATpost ∧ 1 ≤ _yHAT0 ∧ 1 ≤ _xHAT0 l1 2 l0: x1 = _x ∧ x2 = _x1 ∧ x1 = _x2 ∧ x2 = _x3 ∧ _x = _x2 ∧ _x3 = −1 + _x1 l1 3 l0: x1 = _x4 ∧ x2 = _x5 ∧ x1 = _x6 ∧ x2 = _x7 ∧ _x5 = _x7 ∧ _x6 = −1 + _x4 l2 4 l1: x1 = _x8 ∧ x2 = _x9 ∧ x1 = _x10 ∧ x2 = _x11 ∧ _x9 = _x11 ∧ _x8 = _x10 ∧ 1 ≤ _x9 ∧ 1 ≤ _x8 l3 5 l2: x1 = _x12 ∧ x2 = _x13 ∧ x1 = _x14 ∧ x2 = _x15 ∧ _x13 = _x15 ∧ _x12 = _x14

## Proof

### 1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
 l1 l1 l1: x1 = x1 ∧ x2 = x2 l3 l3 l3: x1 = x1 ∧ x2 = x2 l0 l0 l0: x1 = x1 ∧ x2 = x2 l2 l2 l2: x1 = x1 ∧ x2 = x2
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, l0 }.

### 2.1.1 Transition Removal

We remove transition 1 using the following ranking functions, which are bounded by 0.

 l0: x1 + x2 l1: −1 + x1 + x2

### 2.1.2 Transition Removal

We remove transitions 3, 2 using the following ranking functions, which are bounded by 0.

 l1: 0 l0: −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)