LTS Termination Proof

Input

Integer Transition System

Initial Location: l5, l4, l7, l6, l1, l3, l0

Transitions: (pre-variables and post-variables)

l0	1	l1:	x1 = _Result_4HAT0 ∧ x2 = ___disjvr_0HAT0 ∧ x3 = _x_5HAT0 ∧ x4 = _y_6HAT0 ∧ x1 = _Result_4HATpost ∧ x2 = ___disjvr_0HATpost ∧ x3 = _x_5HATpost ∧ x4 = _y_6HATpost ∧ _y_6HAT0 = _y_6HATpost ∧ _x_5HAT0 = _x_5HATpost ∧ ___disjvr_0HAT0 = ___disjvr_0HATpost ∧ _Result_4HAT0 = _Result_4HATpost
l1	2	l3:	x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x4 = _x3 ∧ x1 = _x4 ∧ x2 = _x5 ∧ x3 = _x6 ∧ x4 = _x7 ∧ _x3 = _x7 ∧ _x2 = _x6 ∧ _x1 = _x5 ∧ _x = _x4 ∧ − _x2 + _x3 ≤ 0 ∧ − _x2 + _x3 ≤ 0
l3	3	l4:	x1 = _x8 ∧ x2 = _x9 ∧ x3 = _x10 ∧ x4 = _x11 ∧ x1 = _x12 ∧ x2 = _x13 ∧ x3 = _x14 ∧ x4 = _x15 ∧ _x11 = _x15 ∧ _x10 = _x14 ∧ _x9 = _x13 ∧ _x8 = _x12 ∧ _x13 = _x9
l4	4	l2:	x1 = _x16 ∧ x2 = _x17 ∧ x3 = _x18 ∧ x4 = _x19 ∧ x1 = _x20 ∧ x2 = _x21 ∧ x3 = _x22 ∧ x4 = _x23 ∧ _x19 = _x23 ∧ _x18 = _x22 ∧ _x17 = _x21 ∧ _x20 = _x20
l1	5	l5:	x1 = _x24 ∧ x2 = _x25 ∧ x3 = _x26 ∧ x4 = _x27 ∧ x1 = _x28 ∧ x2 = _x29 ∧ x3 = _x30 ∧ x4 = _x31 ∧ _x27 = _x31 ∧ _x25 = _x29 ∧ _x24 = _x28 ∧ _x30 = 1 + _x26 ∧ _x27 ≤ _x26 ∧ _x26 ≤ _x27 ∧ − _x26 + _x27 ≤ 0 ∧ − _x26 + _x27 ≤ 0
l5	6	l1:	x1 = _x32 ∧ x2 = _x33 ∧ x3 = _x34 ∧ x4 = _x35 ∧ x1 = _x36 ∧ x2 = _x37 ∧ x3 = _x38 ∧ x4 = _x39 ∧ _x35 = _x39 ∧ _x34 = _x38 ∧ _x33 = _x37 ∧ _x32 = _x36
l1	7	l6:	x1 = _x40 ∧ x2 = _x41 ∧ x3 = _x42 ∧ x4 = _x43 ∧ x1 = _x44 ∧ x2 = _x45 ∧ x3 = _x46 ∧ x4 = _x47 ∧ _x43 = _x47 ∧ _x41 = _x45 ∧ _x40 = _x44 ∧ _x46 = 1 + _x42 ∧ 0 ≤ −1 − _x42 + _x43
l6	8	l1:	x1 = _x48 ∧ x2 = _x49 ∧ x3 = _x50 ∧ x4 = _x51 ∧ x1 = _x52 ∧ x2 = _x53 ∧ x3 = _x54 ∧ x4 = _x55 ∧ _x51 = _x55 ∧ _x50 = _x54 ∧ _x49 = _x53 ∧ _x48 = _x52
l7	9	l0:	x1 = _x56 ∧ x2 = _x57 ∧ x3 = _x58 ∧ x4 = _x59 ∧ x1 = _x60 ∧ x2 = _x61 ∧ x3 = _x62 ∧ x4 = _x63 ∧ _x59 = _x63 ∧ _x58 = _x62 ∧ _x57 = _x61 ∧ _x56 = _x60

Proof

1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:

l5	l5	l5:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
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
l3	l3	l3:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4
l0	l0	l0:	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 { l5, l6, l1 }.

2.1.1 Transition Removal

We remove transitions 5, 7 using the following ranking functions, which are bounded by 0.

l1:	− x3 + x4
l5:	− x3 + x4
l6:	− x3 + x4

2.1.2 Transition Removal

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

l6:	0
l1:	−1
l5:	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)