LTS Termination Proof

Input

Integer Transition System

Initial Location: l5, l4, l6, l3, l0, l2

Transitions: (pre-variables and post-variables)

l0	1	l1:	x1 = _oldX0HAT0 ∧ x2 = _oldX1HAT0 ∧ x3 = _x0HAT0 ∧ x1 = _oldX0HATpost ∧ x2 = _oldX1HATpost ∧ x3 = _x0HATpost ∧ _x0HATpost = _oldX1HATpost ∧ _oldX1HATpost = _oldX1HATpost ∧ _oldX0HATpost = _x0HAT0
l0	2	l2:	x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x1 = _x3 ∧ x2 = _x4 ∧ x3 = _x5 ∧ _x1 = _x4 ∧ _x5 = −1 + _x3 ∧ _x3 = _x2
l3	3	l1:	x1 = _x6 ∧ x2 = _x7 ∧ x3 = _x8 ∧ x1 = _x9 ∧ x2 = _x10 ∧ x3 = _x11 ∧ _x11 = _x10 ∧ _x10 = _x10 ∧ _x9 = _x8
l4	4	l0:	x1 = _x12 ∧ x2 = _x13 ∧ x3 = _x14 ∧ x1 = _x15 ∧ x2 = _x16 ∧ x3 = _x17 ∧ _x13 = _x16 ∧ _x17 = _x15 ∧ 0 ≤ _x15 ∧ _x15 = _x14
l4	5	l3:	x1 = _x18 ∧ x2 = _x19 ∧ x3 = _x20 ∧ x1 = _x21 ∧ x2 = _x22 ∧ x3 = _x23 ∧ _x19 = _x22 ∧ _x23 = _x21 ∧ 1 + _x21 ≤ 0 ∧ _x21 = _x20
l2	6	l4:	x1 = _x24 ∧ x2 = _x25 ∧ x3 = _x26 ∧ x1 = _x27 ∧ x2 = _x28 ∧ x3 = _x29 ∧ _x25 = _x28 ∧ _x29 = _x27 ∧ _x27 = _x26
l5	7	l1:	x1 = _x30 ∧ x2 = _x31 ∧ x3 = _x32 ∧ x1 = _x33 ∧ x2 = _x34 ∧ x3 = _x35 ∧ _x32 = _x35 ∧ _x31 = _x34 ∧ _x30 = _x33
l5	8	l0:	x1 = _x36 ∧ x2 = _x37 ∧ x3 = _x38 ∧ x1 = _x39 ∧ x2 = _x40 ∧ x3 = _x41 ∧ _x38 = _x41 ∧ _x37 = _x40 ∧ _x36 = _x39
l5	9	l3:	x1 = _x42 ∧ x2 = _x43 ∧ x3 = _x44 ∧ x1 = _x45 ∧ x2 = _x46 ∧ x3 = _x47 ∧ _x44 = _x47 ∧ _x43 = _x46 ∧ _x42 = _x45
l5	10	l4:	x1 = _x48 ∧ x2 = _x49 ∧ x3 = _x50 ∧ x1 = _x51 ∧ x2 = _x52 ∧ x3 = _x53 ∧ _x50 = _x53 ∧ _x49 = _x52 ∧ _x48 = _x51
l5	11	l2:	x1 = _x54 ∧ x2 = _x55 ∧ x3 = _x56 ∧ x1 = _x57 ∧ x2 = _x58 ∧ x3 = _x59 ∧ _x56 = _x59 ∧ _x55 = _x58 ∧ _x54 = _x57
l6	12	l5:	x1 = _x60 ∧ x2 = _x61 ∧ x3 = _x62 ∧ x1 = _x63 ∧ x2 = _x64 ∧ x3 = _x65 ∧ _x62 = _x65 ∧ _x61 = _x64 ∧ _x60 = _x63

Proof

1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:

l5	l5	l5:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3
l4	l4	l4:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3
l6	l6	l6:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3
l3	l3	l3:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3
l0	l0	l0:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3
l2	l2	l2:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3

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 { l4, l0, l2 }.

2.1.1 Transition Removal

We remove transition 4 using the following ranking functions, which are bounded by 0.

l0:	−1 + x3
l2:	x3
l4:	x3

2.1.2 Transition Removal

We remove transitions 2, 6 using the following ranking functions, which are bounded by 0.

l0:	2
l2:	1
l4:	0

2.1.3 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 % (6 real / 0 unknown / 0 assumptions / 6 total proof steps)