LTS Termination Proof

by AProVE

Input

Integer Transition System
• Initial Location: f1_0_main_New, f176_0_iter_NULL, __init
• Transitions: (pre-variables and post-variables)  f1_0_main_New 1 f176_0_iter_NULL: x1 = _arg1 ∧ x1 = _arg1P ∧ 3 ≤ _arg1P − 1 f176_0_iter_NULL 2 f176_0_iter_NULL: x1 = _x ∧ x1 = _x1 ∧ −1 ≤ _x1 − 1 ∧ 0 ≤ _x − 1 ∧ _x1 + 1 ≤ _x __init 3 f1_0_main_New: x1 = _x2 ∧ x1 = _x3 ∧ 0 ≤ 0

Proof

1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
 f1_0_main_New f1_0_main_New f1_0_main_New: x1 = x1 f176_0_iter_NULL f176_0_iter_NULL f176_0_iter_NULL: x1 = x1 __init __init __init: x1 = x1
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 { f176_0_iter_NULL }.

2.1.1 Transition Removal

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

 f176_0_iter_NULL: x1

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)