# LTS Termination Proof

by AProVE

## Input

Integer Transition System
• Initial Location: l5, l4, l6, l3, l0, l2
• Transitions: (pre-variables and post-variables)  l0 1 l1: x1 = ___const_20HAT0 ∧ x2 = _iHAT0 ∧ x1 = ___const_20HATpost ∧ x2 = _iHATpost ∧ _iHAT0 = _iHATpost ∧ ___const_20HAT0 = ___const_20HATpost l2 2 l0: x1 = _x ∧ x2 = _x1 ∧ x1 = _x2 ∧ x2 = _x3 ∧ _x1 = _x3 ∧ _x = _x2 ∧ _x ≤ _x1 l2 3 l3: x1 = _x4 ∧ x2 = _x5 ∧ x1 = _x6 ∧ x2 = _x7 ∧ _x4 = _x6 ∧ _x7 = 1 + _x5 ∧ 1 + _x5 ≤ _x4 l4 4 l0: x1 = _x8 ∧ x2 = _x9 ∧ x1 = _x10 ∧ x2 = _x11 ∧ _x9 = _x11 ∧ _x8 = _x10 ∧ _x9 ≤ 0 l4 5 l3: x1 = _x12 ∧ x2 = _x13 ∧ x1 = _x14 ∧ x2 = _x15 ∧ _x13 = _x15 ∧ _x12 = _x14 ∧ 1 ≤ _x13 l3 6 l2: x1 = _x16 ∧ x2 = _x17 ∧ x1 = _x18 ∧ x2 = _x19 ∧ _x17 = _x19 ∧ _x16 = _x18 l5 7 l4: x1 = _x20 ∧ x2 = _x21 ∧ x1 = _x22 ∧ x2 = _x23 ∧ _x24 = 0 ∧ _x23 = _x23 ∧ _x20 = _x22 l6 8 l5: x1 = _x25 ∧ x2 = _x26 ∧ x1 = _x27 ∧ x2 = _x28 ∧ _x26 = _x28 ∧ _x25 = _x27

## Proof

### 1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
 l5 l5 l5: x1 = x1 ∧ x2 = x2 l4 l4 l4: x1 = x1 ∧ x2 = x2 l6 l6 l6: 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 { l3, l2 }.

### 2.1.1 Transition Removal

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

 l2: −2⋅x2 + 2⋅x1 l3: 2⋅x1 − 2⋅x2 + 1

### 2.1.2 Transition Removal

We remove transition 6 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)