LTS Termination Proof

Input

Integer Transition System

Transitions: (pre-variables and post-variables)

l0	1	l1:	x1 = _iHAT0 ∧ x2 = _jHAT0 ∧ x3 = _xHAT0 ∧ x1 = _iHATpost ∧ x2 = _jHATpost ∧ x3 = _xHATpost ∧ _xHAT0 = _xHATpost ∧ _jHAT0 = _jHATpost ∧ _iHAT0 = _iHATpost ∧ _xHAT0 ≤ _iHAT0
l0	2	l2:	x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x1 = _x3 ∧ x2 = _x4 ∧ x3 = _x5 ∧ _x2 = _x5 ∧ _x3 = 1 + _x ∧ _x4 = 2 + _x1 ∧ 1 + _x ≤ _x2
l3	3	l2:	x1 = _x6 ∧ x2 = _x7 ∧ x3 = _x8 ∧ x1 = _x9 ∧ x2 = _x10 ∧ x3 = _x11 ∧ _x8 = _x11 ∧ _x7 = _x10 ∧ _x9 = 0 ∧ 2 ≤ _x8
l3	4	l4:	x1 = _x12 ∧ x2 = _x13 ∧ x3 = _x14 ∧ x1 = _x15 ∧ x2 = _x16 ∧ x3 = _x17 ∧ _x14 = _x17 ∧ _x13 = _x16 ∧ _x12 = _x15 ∧ _x14 ≤ 1
l2	5	l0:	x1 = _x18 ∧ x2 = _x19 ∧ x3 = _x20 ∧ x1 = _x21 ∧ x2 = _x22 ∧ x3 = _x23 ∧ _x20 = _x23 ∧ _x19 = _x22 ∧ _x18 = _x21
l5	6	l4:	x1 = _x24 ∧ x2 = _x25 ∧ x3 = _x26 ∧ x1 = _x27 ∧ x2 = _x28 ∧ x3 = _x29 ∧ _x26 = _x29 ∧ _x25 = _x28 ∧ _x24 = _x27
l1	7	l5:	x1 = _x30 ∧ x2 = _x31 ∧ x3 = _x32 ∧ x1 = _x33 ∧ x2 = _x34 ∧ x3 = _x35 ∧ _x32 = _x35 ∧ _x31 = _x34 ∧ _x30 = _x33 ∧ 1 + _x31 ≤ 2⋅_x32
l1	8	l5:	x1 = _x36 ∧ x2 = _x37 ∧ x3 = _x38 ∧ x1 = _x39 ∧ x2 = _x40 ∧ x3 = _x41 ∧ _x38 = _x41 ∧ _x37 = _x40 ∧ _x36 = _x39 ∧ 2⋅_x38 ≤ _x37
l6	9	l3:	x1 = _x42 ∧ x2 = _x43 ∧ x3 = _x44 ∧ x1 = _x45 ∧ x2 = _x46 ∧ x3 = _x47 ∧ _x42 = _x45 ∧ _x47 = 10 ∧ _x46 = 0
l7	10	l6:	x1 = _x48 ∧ x2 = _x49 ∧ x3 = _x50 ∧ x1 = _x51 ∧ x2 = _x52 ∧ x3 = _x53 ∧ _x50 = _x53 ∧ _x49 = _x52 ∧ _x48 = _x51

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 { l0, l2 }.

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

l0:	−1 − x1 + x3
l2:	−1 − x1 + x3

We remove transition 5 using the following ranking functions, which are bounded by 0.

l2:	0
l0:	−1

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)