LTS Termination Proof

Input

Integer Transition System

Transitions: (pre-variables and post-variables)

f1_0_main_Load	1	f168_0_main_EQ:	x1 = _arg1 ∧ x2 = _arg2 ∧ x1 = _arg1P ∧ x2 = _arg2P ∧ 0 ≤ _arg1 − 1 ∧ 0 ≤ _arg2P − 1 ∧ 0 ≤ _arg1P − 1 ∧ −1 ≤ _arg2 − 1
f168_0_main_EQ	2	f168_0_main_EQ:	x1 = _x ∧ x2 = _x1 ∧ x1 = _x2 ∧ x2 = _x3 ∧ _x1 − _x = _x3 ∧ _x = _x2 ∧ _x ≤ _x1 − 1 ∧ 0 ≤ _x − 1 ∧ 0 ≤ _x1 − 1
f168_0_main_EQ	3	f168_0_main_EQ:	x1 = _x4 ∧ x2 = _x5 ∧ x1 = _x6 ∧ x2 = _x7 ∧ _x5 = _x7 ∧ _x4 − _x5 = _x6 ∧ _x5 ≤ _x4 − 1 ∧ 0 ≤ _x5 − 1 ∧ 0 ≤ _x4 − 1
__init	4	f1_0_main_Load:	x1 = _x8 ∧ x2 = _x9 ∧ x1 = _x10 ∧ x2 = _x11 ∧ 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 { f168_0_main_EQ }.

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

f168_0_main_EQ:

x1 + x2

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 % (4 real / 0 unknown / 0 assumptions / 4 total proof steps)