LTS Termination Proof

Input

Integer Transition System

Initial Location: l5, l4, l1, l0, l2

Transitions: (pre-variables and post-variables)

l0	1	l2:	x1 = _Result_4HAT0 ∧ x2 = _k_7HAT0 ∧ x3 = _w_8HAT0 ∧ x4 = _x_5HAT0 ∧ x5 = _y_6HAT0 ∧ x1 = _Result_4HATpost ∧ x2 = _k_7HATpost ∧ x3 = _w_8HATpost ∧ x4 = _x_5HATpost ∧ x5 = _y_6HATpost ∧ _w_8HAT0 = _w_8HATpost ∧ _k_7HAT0 = _k_7HATpost ∧ _Result_4HAT0 = _Result_4HATpost ∧ _y_6HATpost = 1 + _y_6HAT0 ∧ _x_5HATpost = 1 + _x_5HAT0 ∧ 0 ≤ −2 − _x_5HAT0
l2	2	l1:	x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x4 = _x3 ∧ x5 = _x4 ∧ x1 = _x5 ∧ x2 = _x6 ∧ x3 = _x7 ∧ x4 = _x8 ∧ x5 = _x9 ∧ _x4 = _x9 ∧ _x3 = _x8 ∧ _x2 = _x7 ∧ _x1 = _x6 ∧ _x = _x5 ∧ 1 + _x4 ≤ 0
l2	3	l1:	x1 = _x10 ∧ x2 = _x11 ∧ x3 = _x12 ∧ x4 = _x13 ∧ x5 = _x14 ∧ x1 = _x15 ∧ x2 = _x16 ∧ x3 = _x17 ∧ x4 = _x18 ∧ x5 = _x19 ∧ _x14 = _x19 ∧ _x13 = _x18 ∧ _x12 = _x17 ∧ _x11 = _x16 ∧ _x10 = _x15 ∧ 1 ≤ _x14
l1	4	l0:	x1 = _x20 ∧ x2 = _x21 ∧ x3 = _x22 ∧ x4 = _x23 ∧ x5 = _x24 ∧ x1 = _x25 ∧ x2 = _x26 ∧ x3 = _x27 ∧ x4 = _x28 ∧ x5 = _x29 ∧ _x24 = _x29 ∧ _x23 = _x28 ∧ _x22 = _x27 ∧ _x21 = _x26 ∧ _x20 = _x25
l0	5	l3:	x1 = _x30 ∧ x2 = _x31 ∧ x3 = _x32 ∧ x4 = _x33 ∧ x5 = _x34 ∧ x1 = _x35 ∧ x2 = _x36 ∧ x3 = _x37 ∧ x4 = _x38 ∧ x5 = _x39 ∧ _x34 = _x39 ∧ _x33 = _x38 ∧ _x32 = _x37 ∧ _x31 = _x36 ∧ _x35 = _x35 ∧ −1 − _x33 ≤ 0
l4	6	l0:	x1 = _x40 ∧ x2 = _x41 ∧ x3 = _x42 ∧ x4 = _x43 ∧ x5 = _x44 ∧ x1 = _x45 ∧ x2 = _x46 ∧ x3 = _x47 ∧ x4 = _x48 ∧ x5 = _x49 ∧ _x44 = _x49 ∧ _x43 = _x48 ∧ _x42 = _x47 ∧ _x41 = _x46 ∧ _x40 = _x45 ∧ 0 ≤ −2 + _x42 ∧ 0 ≤ −2 + _x41
l4	7	l3:	x1 = _x50 ∧ x2 = _x51 ∧ x3 = _x52 ∧ x4 = _x53 ∧ x5 = _x54 ∧ x1 = _x55 ∧ x2 = _x56 ∧ x3 = _x57 ∧ x4 = _x58 ∧ x5 = _x59 ∧ _x54 = _x59 ∧ _x53 = _x58 ∧ _x52 = _x57 ∧ _x51 = _x56 ∧ _x55 = _x55 ∧ −1 + _x51 ≤ 0
l5	8	l4:	x1 = _x60 ∧ x2 = _x61 ∧ x3 = _x62 ∧ x4 = _x63 ∧ x5 = _x64 ∧ x1 = _x65 ∧ x2 = _x66 ∧ x3 = _x67 ∧ x4 = _x68 ∧ x5 = _x69 ∧ _x64 = _x69 ∧ _x63 = _x68 ∧ _x62 = _x67 ∧ _x61 = _x66 ∧ _x60 = _x65

Proof

1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:

l5	l5	l5:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5
l4	l4	l4:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5
l1	l1	l1:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5
l0	l0	l0:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5
l2	l2	l2:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5

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 { l1, l0, l2 }.

2.1.1 Transition Removal

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

l0:	−1 − x5
l2:	− x5
l1:	−1 − x5

2.1.2 Transition Removal

We remove transition 1 using the following ranking functions, which are bounded by 0.

l0:	−3⋅x4
l2:	−3⋅x4 + 2
l1:	−3⋅x4 + 1

2.1.3 Transition Removal

We remove transitions 4, 3 using the following ranking functions, which are bounded by 0.

l1:	0
l0:	−1
l2:	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)