# LTS Termination Proof

by AProVE

## Input

Integer Transition System
• Initial Location: l0, l2
• Transitions: (pre-variables and post-variables)  l0 1 l1: x1 = _iHAT0 ∧ x2 = _tmp10HAT0 ∧ x3 = _tmp13HAT0 ∧ x4 = _tmp___011HAT0 ∧ x5 = _tmp___014HAT0 ∧ x1 = _iHATpost ∧ x2 = _tmp10HATpost ∧ x3 = _tmp13HATpost ∧ x4 = _tmp___011HATpost ∧ x5 = _tmp___014HATpost ∧ _iHAT1 = 0 ∧ _tmp10HATpost = _iHAT1 ∧ _iHAT2 = 1 + _iHAT1 ∧ _tmp___011HATpost = _iHAT2 ∧ _iHAT3 = 1 + _iHAT2 ∧ _tmp13HATpost = _iHAT3 ∧ _iHAT4 = 1 + _iHAT3 ∧ _tmp___014HATpost = _iHAT4 ∧ _iHATpost = 1 + _iHAT4 l2 2 l0: 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 ∧ _x1 = _x6 ∧ _x = _x5

## Proof

### 1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
 l0 l0 l0: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 l2 l2 l2: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5
and for every transition t, a duplicate t is considered.

### 2 SCC Decomposition

There exist no SCC in the program graph.

## Tool configuration

AProVE

• version: AProVE Commit ID: unknown
• strategy: Statistics for single proof: 100.00 % (2 real / 0 unknown / 0 assumptions / 2 total proof steps)