LTS Termination Proof

by AProVE

Input

Integer Transition System
• Initial Location: f66_0_main_GE, f1_0_main_ConstantStackPush, __init
• Transitions: (pre-variables and post-variables)  f1_0_main_ConstantStackPush 1 f66_0_main_GE: x1 = _arg1 ∧ x2 = _arg2 ∧ x3 = _arg3 ∧ x1 = _arg1P ∧ x2 = _arg2P ∧ x3 = _arg3P ∧ _arg2 = _arg3P ∧ 0 = _arg2P ∧ 0 ≤ _arg1P − 1 ∧ 0 ≤ _arg1 − 1 ∧ −1 ≤ _arg2 − 1 ∧ _arg1P ≤ _arg1 f66_0_main_GE 2 f66_0_main_GE: x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x1 = _x3 ∧ x2 = _x4 ∧ x3 = _x5 ∧ _x2 = _x5 ∧ _x1 + 1 = _x4 ∧ 0 ≤ _x3 − 1 ∧ 0 ≤ _x − 1 ∧ _x3 ≤ _x ∧ −1 ≤ _x2 − 1 ∧ _x1 ≤ _x2 − 1 __init 3 f1_0_main_ConstantStackPush: x1 = _x6 ∧ x2 = _x7 ∧ x3 = _x8 ∧ x1 = _x9 ∧ x2 = _x10 ∧ x3 = _x11 ∧ 0 ≤ 0

Proof

1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
 f66_0_main_GE f66_0_main_GE f66_0_main_GE: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 f1_0_main_ConstantStackPush f1_0_main_ConstantStackPush f1_0_main_ConstantStackPush: x1 = x1 ∧ x2 = x2 ∧ x3 = x3 __init __init __init: x1 = x1 ∧ x2 = x2 ∧ x3 = x3
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 { f66_0_main_GE }.

2.1.1 Transition Removal

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

 f66_0_main_GE: − x2 + x3

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)