LTS Termination Proof

Input

Integer Transition System

Initial Location: l5, l4, l7, l6, l3, l0, l2

Transitions: (pre-variables and post-variables)

l0	1	l1:	x1 = _ResultHAT0 ∧ x2 = _xHAT0 ∧ x3 = _yHAT0 ∧ x1 = _ResultHATpost ∧ x2 = _xHATpost ∧ x3 = _yHATpost ∧ _yHAT0 = _yHATpost ∧ _xHAT0 = _xHATpost ∧ _ResultHATpost = _ResultHATpost ∧ − _xHAT0 ≤ 0
l0	2	l3:	x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x1 = _x3 ∧ x2 = _x4 ∧ x3 = _x5 ∧ _x2 = _x5 ∧ _x1 = _x4 ∧ _x = _x3 ∧ 0 ≤ −1 − _x1
l3	3	l4:	x1 = _x6 ∧ x2 = _x7 ∧ x3 = _x8 ∧ x1 = _x9 ∧ x2 = _x10 ∧ x3 = _x11 ∧ _x8 = _x11 ∧ _x7 = _x10 ∧ _x6 = _x9 ∧ 1 + _x8 ≤ −1
l3	4	l4:	x1 = _x12 ∧ x2 = _x13 ∧ x3 = _x14 ∧ x1 = _x15 ∧ x2 = _x16 ∧ x3 = _x17 ∧ _x14 = _x17 ∧ _x13 = _x16 ∧ _x12 = _x15 ∧ 0 ≤ _x14
l4	5	l2:	x1 = _x18 ∧ x2 = _x19 ∧ x3 = _x20 ∧ x1 = _x21 ∧ x2 = _x22 ∧ x3 = _x23 ∧ _x18 = _x21 ∧ _x23 = 1 + _x20 ∧ _x22 = 1 + _x19
l2	6	l0:	x1 = _x24 ∧ x2 = _x25 ∧ x3 = _x26 ∧ x1 = _x27 ∧ x2 = _x28 ∧ x3 = _x29 ∧ _x26 = _x29 ∧ _x25 = _x28 ∧ _x24 = _x27
l0	7	l5:	x1 = _x30 ∧ x2 = _x31 ∧ x3 = _x32 ∧ x1 = _x33 ∧ x2 = _x34 ∧ x3 = _x35 ∧ _x30 = _x33 ∧ _x34 = −99 + _x31 ∧ _x35 = 1 + _x32 ∧ −1 ≤ _x32 ∧ _x32 ≤ −1 ∧ 0 ≤ −1 − _x31
l5	8	l0:	x1 = _x36 ∧ x2 = _x37 ∧ x3 = _x38 ∧ x1 = _x39 ∧ x2 = _x40 ∧ x3 = _x41 ∧ _x38 = _x41 ∧ _x37 = _x40 ∧ _x36 = _x39
l6	9	l0:	x1 = _x42 ∧ x2 = _x43 ∧ x3 = _x44 ∧ x1 = _x45 ∧ x2 = _x46 ∧ x3 = _x47 ∧ _x44 = _x47 ∧ _x43 = _x46 ∧ _x42 = _x45
l7	10	l6:	x1 = _x48 ∧ x2 = _x49 ∧ x3 = _x50 ∧ x1 = _x51 ∧ x2 = _x52 ∧ x3 = _x53 ∧ _x50 = _x53 ∧ _x49 = _x52 ∧ _x48 = _x51

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
l7	l7	l7:	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 { l5, l4, l3, l0, l2 }.

2.1.1 Transition Removal

We remove transitions 7, 3 using the following ranking functions, which are bounded by 0.

l0:	−4⋅x3 + 1
l3:	−4⋅x3
l5:	−4⋅x3 + 2
l2:	−4⋅x3 + 2
l4:	−4⋅x3 − 1

2.1.2 Transition Removal

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

l0:	−4⋅x2
l3:	−4⋅x2 − 1
l5:	−4⋅x2 + 1
l2:	−4⋅x2 + 1
l4:	−4⋅x2 − 2

2.1.3 Transition Removal

We remove transitions 8, 6, 5, 4 using the following ranking functions, which are bounded by −1.

l5:	−1
l0:	−2
l2:	−1
l4:	0
l3:	1

2.1.4 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)