LTS Termination Proof

Input

Integer Transition System

Transitions: (pre-variables and post-variables)

f1_0_main_Load	1	f74_0_main_LE:	x1 = _arg1 ∧ x2 = _arg2 ∧ x1 = _arg1P ∧ x2 = _arg2P ∧ 0 ≤ _arg1 − 1 ∧ −1 ≤ _arg1P − 1 ∧ −1 ≤ _arg2 − 1
f74_0_main_LE	2	f74_0_main_LE':	x1 = _x ∧ x2 = _x1 ∧ x1 = _x2 ∧ x2 = _x3 ∧ _x − 2⋅_x4 = 0 ∧ 0 ≤ _x − 1 ∧ _x = _x2
f74_0_main_LE'	3	f74_0_main_LE:	x1 = _x6 ∧ x2 = _x8 ∧ x1 = _x9 ∧ x2 = _x10 ∧ 0 ≤ _x6 − 1 ∧ _x6 − 2⋅_x11 = 0 ∧ _x6 − 2⋅_x11 ≤ 1 ∧ 0 ≤ _x6 − 2⋅_x11 ∧ _x6 − 1 = _x9
__init	4	f1_0_main_Load:	x1 = _x12 ∧ x2 = _x13 ∧ x1 = _x14 ∧ x2 = _x15 ∧ 0 ≤ 0

We consider the following cutpoint-transitions:

and for every transition t, a duplicate t is considered.

We consider subproblems for each of the 1 SCC(s) of the program graph.

Here we consider the SCC { f74_0_main_LE, f74_0_main_LE' }.

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

f74_0_main_LE:	1 + 2⋅x1
f74_0_main_LE':	2⋅x1

There are no more "sharp" transitions in the cooperation program. Hence the cooperation termination is proved.

version: AProVE Commit ID: unknown
strategy: Statistics for single proof: 100.00 % (5 real / 0 unknown / 0 assumptions / 5 total proof steps)