# 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 = _Result_4HAT0 ∧ x2 = _b_7HAT0 ∧ x3 = _x_5HAT0 ∧ x4 = _y_6HAT0 ∧ x1 = _Result_4HATpost ∧ x2 = _b_7HATpost ∧ x3 = _x_5HATpost ∧ x4 = _y_6HATpost ∧ _x_5HAT0 = _x_5HATpost ∧ _Result_4HAT0 = _Result_4HATpost ∧ _y_6HATpost = −1 + _y_6HAT0 ∧ _b_7HATpost = 0 ∧ 1 − _b_7HAT0 ≤ 0 ∧ 0 ≤ −1 − _x_5HAT0 + _y_6HAT0 l0 2 l2: x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x4 = _x3 ∧ x1 = _x4 ∧ x2 = _x5 ∧ x3 = _x6 ∧ x4 = _x7 ∧ _x3 = _x7 ∧ _x2 = _x6 ∧ _x1 = _x5 ∧ _x4 = _x4 ∧ − _x2 + _x3 ≤ 0 l3 3 l1: x1 = _x8 ∧ x2 = _x9 ∧ x3 = _x10 ∧ x4 = _x11 ∧ x1 = _x12 ∧ x2 = _x13 ∧ x3 = _x14 ∧ x4 = _x15 ∧ _x11 = _x15 ∧ _x10 = _x14 ∧ _x8 = _x12 ∧ _x13 = 0 l1 4 l0: x1 = _x16 ∧ x2 = _x17 ∧ x3 = _x18 ∧ x4 = _x19 ∧ x1 = _x20 ∧ x2 = _x21 ∧ x3 = _x22 ∧ x4 = _x23 ∧ _x19 = _x23 ∧ _x16 = _x20 ∧ _x22 = 1 + _x18 ∧ _x21 = 1 ∧ 0 ≤ − _x17 ∧ 0 ≤ −1 − _x18 + _x19 l1 5 l2: x1 = _x24 ∧ x2 = _x25 ∧ x3 = _x26 ∧ x4 = _x27 ∧ x1 = _x28 ∧ x2 = _x29 ∧ x3 = _x30 ∧ x4 = _x31 ∧ _x27 = _x31 ∧ _x26 = _x30 ∧ _x25 = _x29 ∧ _x28 = _x28 ∧ − _x26 + _x27 ≤ 0 l4 6 l3: x1 = _x32 ∧ x2 = _x33 ∧ x3 = _x34 ∧ x4 = _x35 ∧ x1 = _x36 ∧ x2 = _x37 ∧ x3 = _x38 ∧ x4 = _x39 ∧ _x35 = _x39 ∧ _x34 = _x38 ∧ _x33 = _x37 ∧ _x32 = _x36

## Proof

### 1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
 l4 l4 l4: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 l1 l1 l1: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 l3 l3 l3: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 l0 l0 l0: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
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 transitions 1, 4 using the following ranking functions, which are bounded by 0.

 l0: − x3 + x4 l1: − x3 + x4

### 2.1.2 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 % (4 real / 0 unknown / 0 assumptions / 4 total proof steps)