LTS Termination Proof

Input

Integer Transition System

Initial Location: l4, l7, l6, l1, l8, l3, l0, l2

Transitions: (pre-variables and post-variables)

l0	1	l1:	x1 = _jHAT0 ∧ x2 = _tmp3HAT0 ∧ x3 = _x1HAT0 ∧ x4 = _y2HAT0 ∧ x1 = _jHATpost ∧ x2 = _tmp3HATpost ∧ x3 = _x1HATpost ∧ x4 = _y2HATpost ∧ _jHAT0 = _jHATpost ∧ _tmp3HATpost = _tmp3HATpost ∧ _y2HATpost = 1 + _jHAT0 ∧ _x1HATpost = _jHAT0
l0	2	l1:	x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x4 = _x3 ∧ x1 = _x4 ∧ x2 = _x5 ∧ x3 = _x6 ∧ x4 = _x7 ∧ _x3 = _x7 ∧ _x2 = _x6 ∧ _x1 = _x5 ∧ _x = _x4
l2	3	l3:	x1 = _x8 ∧ x2 = _x9 ∧ x3 = _x10 ∧ x4 = _x11 ∧ x1 = _x12 ∧ x2 = _x13 ∧ x3 = _x14 ∧ x4 = _x15 ∧ _x11 = _x15 ∧ _x10 = _x14 ∧ _x9 = _x13 ∧ _x8 = _x12 ∧ 4 ≤ _x8
l2	4	l0:	x1 = _x16 ∧ x2 = _x17 ∧ x3 = _x18 ∧ x4 = _x19 ∧ x1 = _x20 ∧ x2 = _x21 ∧ x3 = _x22 ∧ x4 = _x23 ∧ _x19 = _x23 ∧ _x18 = _x22 ∧ _x17 = _x21 ∧ _x16 = _x20 ∧ 1 + _x16 ≤ 4
l4	5	l5:	x1 = _x24 ∧ x2 = _x25 ∧ x3 = _x26 ∧ x4 = _x27 ∧ x1 = _x28 ∧ x2 = _x29 ∧ x3 = _x30 ∧ x4 = _x31 ∧ _x27 = _x31 ∧ _x26 = _x30 ∧ _x25 = _x29 ∧ _x24 = _x28
l6	6	l2:	x1 = _x32 ∧ x2 = _x33 ∧ x3 = _x34 ∧ x4 = _x35 ∧ x1 = _x36 ∧ x2 = _x37 ∧ x3 = _x38 ∧ x4 = _x39 ∧ _x35 = _x39 ∧ _x34 = _x38 ∧ _x33 = _x37 ∧ _x32 = _x36
l3	7	l4:	x1 = _x40 ∧ x2 = _x41 ∧ x3 = _x42 ∧ x4 = _x43 ∧ x1 = _x44 ∧ x2 = _x45 ∧ x3 = _x46 ∧ x4 = _x47 ∧ _x43 = _x47 ∧ _x42 = _x46 ∧ _x41 = _x45 ∧ _x40 = _x44
l1	8	l6:	x1 = _x48 ∧ x2 = _x49 ∧ x3 = _x50 ∧ x4 = _x51 ∧ x1 = _x52 ∧ x2 = _x53 ∧ x3 = _x54 ∧ x4 = _x55 ∧ _x51 = _x55 ∧ _x50 = _x54 ∧ _x49 = _x53 ∧ _x52 = 1 + _x48
l7	9	l6:	x1 = _x56 ∧ x2 = _x57 ∧ x3 = _x58 ∧ x4 = _x59 ∧ x1 = _x60 ∧ x2 = _x61 ∧ x3 = _x62 ∧ x4 = _x63 ∧ _x59 = _x63 ∧ _x58 = _x62 ∧ _x57 = _x61 ∧ _x60 = 0
l8	10	l7:	x1 = _x64 ∧ x2 = _x65 ∧ x3 = _x66 ∧ x4 = _x67 ∧ x1 = _x68 ∧ x2 = _x69 ∧ x3 = _x70 ∧ x4 = _x71 ∧ _x67 = _x71 ∧ _x66 = _x70 ∧ _x65 = _x69 ∧ _x64 = _x68

Proof

1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:

l4	l4	l4:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
l7	l7	l7:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
l6	l6	l6:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
l1	l1	l1:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
l8	l8	l8:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
l3	l3	l3:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
l0	l0	l0:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
l2	l2	l2:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4

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 { l6, l1, l0, l2 }.

2.1.1 Transition Removal

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

l0:	2 − x1
l1:	2 − x1
l2:	3 − x1
l6:	3 − x1

2.1.2 Transition Removal

We remove transitions 1, 6, 8, 2 using the following ranking functions, which are bounded by 0.

l0:	3
l1:	2
l6:	1
l2:	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 % (5 real / 0 unknown / 0 assumptions / 5 total proof steps)