# LTS Termination Proof

by AProVE

## Input

Integer Transition System
• Initial Location: l4, l1, l3, l0
• Transitions: (pre-variables and post-variables)  l0 1 l1: x1 = _nondetEXCL13HAT0 ∧ x2 = _resultEXCL12HAT0 ∧ x3 = _temp0EXCL15HAT0 ∧ x4 = _xEXCL14HAT0 ∧ x5 = _xEXCL20HAT0 ∧ x1 = _nondetEXCL13HATpost ∧ x2 = _resultEXCL12HATpost ∧ x3 = _temp0EXCL15HATpost ∧ x4 = _xEXCL14HATpost ∧ x5 = _xEXCL20HATpost ∧ _nondetEXCL13HAT1 = _nondetEXCL13HAT1 ∧ _xEXCL14HATpost = _nondetEXCL13HAT1 ∧ _nondetEXCL13HATpost = _nondetEXCL13HATpost ∧ _resultEXCL12HAT0 = _resultEXCL12HATpost ∧ _temp0EXCL15HAT0 = _temp0EXCL15HATpost ∧ _xEXCL20HAT0 = _xEXCL20HATpost l1 2 l2: x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x4 = _x3 ∧ x5 = _x4 ∧ x1 = _x5 ∧ x2 = _x6 ∧ x3 = _x7 ∧ x4 = _x8 ∧ x5 = _x9 ∧ _x4 = _x9 ∧ _x3 = _x8 ∧ _x2 = _x7 ∧ _x = _x5 ∧ _x6 = _x2 ∧ _x3 ≤ 0 l1 3 l3: x1 = _x10 ∧ x2 = _x11 ∧ x3 = _x12 ∧ x4 = _x13 ∧ x5 = _x14 ∧ x1 = _x15 ∧ x2 = _x16 ∧ x3 = _x17 ∧ x4 = _x18 ∧ x5 = _x19 ∧ _x14 = _x19 ∧ _x12 = _x17 ∧ _x11 = _x16 ∧ _x10 = _x15 ∧ 1 ≤ _x14 ∧ −1 + _x14 ≤ _x18 ∧ _x18 ≤ −1 + _x14 ∧ _x18 = −1 + _x13 ∧ 1 ≤ _x13 l3 4 l1: x1 = _x20 ∧ x2 = _x21 ∧ x3 = _x22 ∧ x4 = _x23 ∧ x5 = _x24 ∧ x1 = _x25 ∧ x2 = _x26 ∧ x3 = _x27 ∧ x4 = _x28 ∧ x5 = _x29 ∧ _x24 = _x29 ∧ _x23 = _x28 ∧ _x22 = _x27 ∧ _x21 = _x26 ∧ _x20 = _x25 l4 5 l0: x1 = _x30 ∧ x2 = _x31 ∧ x3 = _x32 ∧ x4 = _x33 ∧ x5 = _x34 ∧ x1 = _x35 ∧ x2 = _x36 ∧ x3 = _x37 ∧ x4 = _x38 ∧ x5 = _x39 ∧ _x34 = _x39 ∧ _x33 = _x38 ∧ _x32 = _x37 ∧ _x31 = _x36 ∧ _x30 = _x35

## Proof

### 1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
 l4 l4 l4: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 l1 l1 l1: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 l3 l3 l3: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 l0 l0 l0: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5
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 }.

### 2.1.1 Transition Removal

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

 l1: −1 + x4 + x5 l3: −1 + x4 + x5

### 2.1.2 Transition Removal

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

 l3: 0 l1: −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)