LTS Termination Proof

Input

Integer Transition System

Initial Location: l5, l7, l6, l1, l3, l0, l2

Transitions: (pre-variables and post-variables)

l0	1	l1:	x1 = ___const_10HAT0 ∧ x2 = _iHAT0 ∧ x3 = _jHAT0 ∧ x4 = _xHAT0 ∧ x1 = ___const_10HATpost ∧ x2 = _iHATpost ∧ x3 = _jHATpost ∧ x4 = _xHATpost ∧ _xHAT0 = _xHATpost ∧ _jHAT0 = _jHATpost ∧ _iHAT0 = _iHATpost ∧ ___const_10HAT0 = ___const_10HATpost ∧ _xHAT0 ≤ _iHAT0
l0	2	l2:	x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x4 = _x3 ∧ x1 = _x4 ∧ x2 = _x5 ∧ x3 = _x6 ∧ x4 = _x7 ∧ _x3 = _x7 ∧ _x = _x4 ∧ _x5 = 1 + _x1 ∧ _x6 = 2 + _x2 ∧ 1 + _x1 ≤ _x3
l3	3	l2:	x1 = _x8 ∧ x2 = _x9 ∧ x3 = _x10 ∧ x4 = _x11 ∧ x1 = _x12 ∧ x2 = _x13 ∧ x3 = _x14 ∧ x4 = _x15 ∧ _x11 = _x15 ∧ _x10 = _x14 ∧ _x8 = _x12 ∧ _x13 = 0 ∧ 2 ≤ _x11
l3	4	l4:	x1 = _x16 ∧ x2 = _x17 ∧ x3 = _x18 ∧ x4 = _x19 ∧ x1 = _x20 ∧ x2 = _x21 ∧ x3 = _x22 ∧ x4 = _x23 ∧ _x19 = _x23 ∧ _x18 = _x22 ∧ _x17 = _x21 ∧ _x16 = _x20 ∧ _x19 ≤ 1
l2	5	l0:	x1 = _x24 ∧ x2 = _x25 ∧ x3 = _x26 ∧ x4 = _x27 ∧ x1 = _x28 ∧ x2 = _x29 ∧ x3 = _x30 ∧ x4 = _x31 ∧ _x27 = _x31 ∧ _x26 = _x30 ∧ _x25 = _x29 ∧ _x24 = _x28
l5	6	l4:	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	l5:	x1 = _x40 ∧ x2 = _x41 ∧ x3 = _x42 ∧ x4 = _x43 ∧ x1 = _x44 ∧ x2 = _x45 ∧ x3 = _x46 ∧ x4 = _x47 ∧ _x43 = _x47 ∧ _x42 = _x46 ∧ _x41 = _x45 ∧ _x40 = _x44 ∧ 1 + _x42 ≤ 2⋅_x43
l1	8	l5:	x1 = _x48 ∧ x2 = _x49 ∧ x3 = _x50 ∧ x4 = _x51 ∧ x1 = _x52 ∧ x2 = _x53 ∧ x3 = _x54 ∧ x4 = _x55 ∧ _x51 = _x55 ∧ _x50 = _x54 ∧ _x49 = _x53 ∧ _x48 = _x52 ∧ 2⋅_x51 ≤ _x50
l6	9	l3:	x1 = _x56 ∧ x2 = _x57 ∧ x3 = _x58 ∧ x4 = _x59 ∧ x1 = _x60 ∧ x2 = _x61 ∧ x3 = _x62 ∧ x4 = _x63 ∧ _x57 = _x61 ∧ _x56 = _x60 ∧ _x63 = _x56 ∧ _x62 = 0
l7	10	l6:	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:

l5	l5	l5:	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
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 { l0, l2 }.

2.1.1 Transition Removal

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

l0:	−1 − x2 + x4
l2:	−1 − x2 + x4

2.1.2 Transition Removal

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

l2:	0
l0:	−1

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)