# LTS Termination Proof

by AProVE

## Input

Integer Transition System
• Initial Location: l5, l4, l3, l0, l2
• Transitions: (pre-variables and post-variables)  l0 1 l1: x1 = _Result_4HAT0 ∧ x2 = ___cil_tmp4_8HAT0 ∧ x3 = ___const_100HAT0 ∧ x4 = ___retres3_7HAT0 ∧ x5 = _i_5HAT0 ∧ x6 = _x_6HAT0 ∧ x1 = _Result_4HATpost ∧ x2 = ___cil_tmp4_8HATpost ∧ x3 = ___const_100HATpost ∧ x4 = ___retres3_7HATpost ∧ x5 = _i_5HATpost ∧ x6 = _x_6HATpost ∧ _x_6HAT0 = _x_6HATpost ∧ _i_5HAT0 = _i_5HATpost ∧ ___const_100HAT0 = ___const_100HATpost ∧ _Result_4HATpost = ___cil_tmp4_8HATpost ∧ ___cil_tmp4_8HATpost = ___retres3_7HATpost ∧ ___retres3_7HATpost = 0 ∧ 1 + ___const_100HAT0 − _i_5HAT0 ≤ 0 l0 2 l2: x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x4 = _x3 ∧ x5 = _x4 ∧ x6 = _x5 ∧ x1 = _x6 ∧ x2 = _x7 ∧ x3 = _x8 ∧ x4 = _x9 ∧ x5 = _x10 ∧ x6 = _x11 ∧ _x5 = _x11 ∧ _x3 = _x9 ∧ _x2 = _x8 ∧ _x1 = _x7 ∧ _x = _x6 ∧ _x10 = 1 + _x4 ∧ 0 ≤ −1 + _x5 ∧ 0 ≤ _x2 − _x4 l2 3 l0: x1 = _x12 ∧ x2 = _x13 ∧ x3 = _x14 ∧ x4 = _x15 ∧ x5 = _x16 ∧ x6 = _x17 ∧ x1 = _x18 ∧ x2 = _x19 ∧ x3 = _x20 ∧ x4 = _x21 ∧ x5 = _x22 ∧ x6 = _x23 ∧ _x17 = _x23 ∧ _x16 = _x22 ∧ _x15 = _x21 ∧ _x14 = _x20 ∧ _x13 = _x19 ∧ _x12 = _x18 l3 4 l4: x1 = _x24 ∧ x2 = _x25 ∧ x3 = _x26 ∧ x4 = _x27 ∧ x5 = _x28 ∧ x6 = _x29 ∧ x1 = _x30 ∧ x2 = _x31 ∧ x3 = _x32 ∧ x4 = _x33 ∧ x5 = _x34 ∧ x6 = _x35 ∧ _x29 = _x35 ∧ _x27 = _x33 ∧ _x26 = _x32 ∧ _x25 = _x31 ∧ _x24 = _x30 ∧ _x34 = 0 l4 5 l1: x1 = _x36 ∧ x2 = _x37 ∧ x3 = _x38 ∧ x4 = _x39 ∧ x5 = _x40 ∧ x6 = _x41 ∧ x1 = _x42 ∧ x2 = _x43 ∧ x3 = _x44 ∧ x4 = _x45 ∧ x5 = _x46 ∧ x6 = _x47 ∧ _x41 = _x47 ∧ _x40 = _x46 ∧ _x38 = _x44 ∧ _x42 = _x43 ∧ _x43 = _x45 ∧ _x45 = 0 ∧ _x41 ≤ 0 ∧ 0 ≤ _x38 − _x40 l4 6 l0: x1 = _x48 ∧ x2 = _x49 ∧ x3 = _x50 ∧ x4 = _x51 ∧ x5 = _x52 ∧ x6 = _x53 ∧ x1 = _x54 ∧ x2 = _x55 ∧ x3 = _x56 ∧ x4 = _x57 ∧ x5 = _x58 ∧ x6 = _x59 ∧ _x53 = _x59 ∧ _x51 = _x57 ∧ _x50 = _x56 ∧ _x49 = _x55 ∧ _x48 = _x54 ∧ _x58 = 1 + _x52 ∧ 0 ≤ −1 + _x53 ∧ 0 ≤ _x50 − _x52 l5 7 l3: x1 = _x60 ∧ x2 = _x61 ∧ x3 = _x62 ∧ x4 = _x63 ∧ x5 = _x64 ∧ x6 = _x65 ∧ x1 = _x66 ∧ x2 = _x67 ∧ x3 = _x68 ∧ x4 = _x69 ∧ x5 = _x70 ∧ x6 = _x71 ∧ _x65 = _x71 ∧ _x64 = _x70 ∧ _x63 = _x69 ∧ _x62 = _x68 ∧ _x61 = _x67 ∧ _x60 = _x66

## 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 l4 l4 l4: 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 { l0, l2 }.

### 2.1.1 Transition Removal

We remove transition 2 using the following ranking functions, which are bounded by 0.

 l0: −1 + x3 − x5 + x6 l2: −1 + x3 − x5 + x6

### 2.1.2 Transition Removal

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

 l2: 0 l0: −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)