# LTS Termination Proof

by AProVE

## Input

Integer Transition System
• Initial Location: l5, l6, l1, l3, l0, l2
• Transitions: (pre-variables and post-variables)  l0 1 l1: x1 = ___const_10HAT0 ∧ x2 = _a_20HAT0 ∧ x3 = _i_13HAT0 ∧ x4 = _i_21HAT0 ∧ x5 = _rt_11HAT0 ∧ x6 = _st_14HAT0 ∧ x1 = ___const_10HATpost ∧ x2 = _a_20HATpost ∧ x3 = _i_13HATpost ∧ x4 = _i_21HATpost ∧ x5 = _rt_11HATpost ∧ x6 = _st_14HATpost ∧ _st_14HAT0 = _st_14HATpost ∧ _rt_11HAT0 = _rt_11HATpost ∧ _i_21HAT0 = _i_21HATpost ∧ _a_20HAT0 = _a_20HATpost ∧ ___const_10HAT0 = ___const_10HATpost ∧ _i_13HATpost ≤ 0 ∧ 0 ≤ _i_13HATpost ∧ _i_13HATpost = 0 l2 2 l1: x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x4 = _x3 ∧ x5 = _x4 ∧ x6 = _x5 ∧ x1 = _x6 ∧ x2 = _x7 ∧ x3 = _x8 ∧ x4 = _x9 ∧ x5 = _x10 ∧ x6 = _x11 ∧ 1 + _x2 ≤ _x ∧ _x12 = 1 + _x2 ∧ _x8 = _x8 ∧ 2 ≤ _x8 ∧ _x8 ≤ 2 ∧ _x7 = _x7 ∧ _x7 ≤ _x8 ∧ _x8 ≤ _x7 ∧ _x = _x6 ∧ _x3 = _x9 ∧ _x4 = _x10 ∧ _x5 = _x11 l3 3 l1: x1 = _x13 ∧ x2 = _x14 ∧ x3 = _x15 ∧ x4 = _x16 ∧ x5 = _x17 ∧ x6 = _x18 ∧ x1 = _x19 ∧ x2 = _x20 ∧ x3 = _x21 ∧ x4 = _x22 ∧ x5 = _x23 ∧ x6 = _x24 ∧ _x18 = _x24 ∧ _x17 = _x23 ∧ _x16 = _x22 ∧ _x14 = _x20 ∧ _x13 = _x19 ∧ _x21 ≤ 1 ∧ 1 ≤ _x21 ∧ _x21 = 1 + _x15 ∧ 1 + _x15 ≤ _x13 l1 4 l4: x1 = _x25 ∧ x2 = _x26 ∧ x3 = _x27 ∧ x4 = _x28 ∧ x5 = _x29 ∧ x6 = _x30 ∧ x1 = _x31 ∧ x2 = _x32 ∧ x3 = _x33 ∧ x4 = _x34 ∧ x5 = _x35 ∧ x6 = _x36 ∧ _x30 = _x36 ∧ _x28 = _x34 ∧ _x27 = _x33 ∧ _x26 = _x32 ∧ _x25 = _x31 ∧ _x35 = _x30 ∧ _x25 ≤ _x27 l1 5 l5: x1 = _x37 ∧ x2 = _x38 ∧ x3 = _x39 ∧ x4 = _x40 ∧ x5 = _x41 ∧ x6 = _x42 ∧ x1 = _x43 ∧ x2 = _x44 ∧ x3 = _x45 ∧ x4 = _x46 ∧ x5 = _x47 ∧ x6 = _x48 ∧ _x42 = _x48 ∧ _x41 = _x47 ∧ _x38 = _x44 ∧ _x37 = _x43 ∧ 1 + _x46 ≤ _x37 ∧ 1 + _x46 ≤ _x45 ∧ _x45 ≤ 1 + _x46 ∧ _x45 = 1 + _x39 ∧ 1 + _x39 ≤ _x37 ∧ _x46 = _x46 l5 6 l1: x1 = _x49 ∧ x2 = _x50 ∧ x3 = _x51 ∧ x4 = _x52 ∧ x5 = _x53 ∧ x6 = _x54 ∧ x1 = _x55 ∧ x2 = _x56 ∧ x3 = _x57 ∧ x4 = _x58 ∧ x5 = _x59 ∧ x6 = _x60 ∧ _x54 = _x60 ∧ _x53 = _x59 ∧ _x52 = _x58 ∧ _x51 = _x57 ∧ _x50 = _x56 ∧ _x49 = _x55 l6 7 l0: x1 = _x61 ∧ x2 = _x62 ∧ x3 = _x63 ∧ x4 = _x64 ∧ x5 = _x65 ∧ x6 = _x66 ∧ x1 = _x67 ∧ x2 = _x68 ∧ x3 = _x69 ∧ x4 = _x70 ∧ x5 = _x71 ∧ x6 = _x72 ∧ _x66 = _x72 ∧ _x65 = _x71 ∧ _x64 = _x70 ∧ _x63 = _x69 ∧ _x62 = _x68 ∧ _x61 = _x67

## Proof

### 1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
 l5 l5 l5: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 l6 l6 l6: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 l1 l1 l1: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 l3 l3 l3: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 l0 l0 l0: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6 l2 l2 l2: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5 ∧ x6 = x6
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 { l5, l1 }.

### 2.1.1 Transition Removal

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

 l1: −1 + x1 − x3 l5: −1 + x1 − x3

### 2.1.2 Transition Removal

We remove transition 6 using the following ranking functions, which are bounded by 0.

 l5: 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 % (6 real / 0 unknown / 0 assumptions / 6 total proof steps)