LTS Termination Proof

by AProVE

Input

Integer Transition System
• Initial Location: f55_0_main_GE, f1_0_main_ConstantStackPush, __init
• Transitions: (pre-variables and post-variables)  f1_0_main_ConstantStackPush 1 f55_0_main_GE: x1 = _arg1 ∧ x1 = _arg1P ∧ 0 = _arg1P f55_0_main_GE 2 f55_0_main_GE: x1 = _x ∧ x1 = _x1 ∧ _x + 1 = _x1 ∧ _x ≤ 19 ∧ 9 ≤ _x − 1 f55_0_main_GE 3 f55_0_main_GE: x1 = _x2 ∧ x1 = _x3 ∧ _x2 + 1 = _x3 ∧ _x2 ≤ 19 ∧ _x2 ≤ 9 __init 4 f1_0_main_ConstantStackPush: x1 = _x4 ∧ x1 = _x5 ∧ 0 ≤ 0

Proof

1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
 f55_0_main_GE f55_0_main_GE f55_0_main_GE: x1 = x1 f1_0_main_ConstantStackPush f1_0_main_ConstantStackPush f1_0_main_ConstantStackPush: 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 { f55_0_main_GE }.

2.1.1 Transition Removal

We remove transitions 2, 3 using the following ranking functions, which are bounded by 0.

 f55_0_main_GE: 19 − 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)