# LTS Termination Proof

by AProVE

## Input

Integer Transition System
• Initial Location: l4, l7, l6, l1, l8, l3, l0, l2
• Transitions: (pre-variables and post-variables)  l0 1 l1: x1 = _iHAT0 ∧ x1 = _iHATpost ∧ _iHAT0 = _iHATpost ∧ 3 ≤ _iHAT0 l0 2 l2: x1 = _x ∧ x1 = _x1 ∧ _x1 = 1 + _x ∧ 1 + _x ≤ 3 l3 3 l0: x1 = _x2 ∧ x1 = _x3 ∧ _x2 = _x3 l3 4 l1: x1 = _x4 ∧ x1 = _x5 ∧ _x4 = _x5 l2 5 l3: x1 = _x6 ∧ x1 = _x7 ∧ _x6 = _x7 l4 6 l5: x1 = _x8 ∧ x1 = _x9 ∧ _x8 = _x9 l1 7 l6: x1 = _x10 ∧ x1 = _x11 ∧ _x10 = _x11 l6 8 l4: x1 = _x12 ∧ x1 = _x13 ∧ _x12 = _x13 ∧ 1 + _x12 ≤ 2 l6 9 l4: x1 = _x14 ∧ x1 = _x15 ∧ _x14 = _x15 ∧ 2 ≤ _x14 l7 10 l2: x1 = _x16 ∧ x1 = _x17 ∧ _x17 = 0 l8 11 l7: x1 = _x18 ∧ x1 = _x19 ∧ _x18 = _x19

## Proof

### 1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
 l4 l4 l4: x1 = x1 l7 l7 l7: x1 = x1 l6 l6 l6: x1 = x1 l1 l1 l1: x1 = x1 l8 l8 l8: 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, l0, l2 }.

### 2.1.1 Transition Removal

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

 l0: 2 − x1 l2: 2 − x1 l3: 2 − x1

### 2.1.2 Transition Removal

We remove transitions 3, 5 using the following ranking functions, which are bounded by 0.

 l3: 0 l0: −1 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)