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 = _Result_4HAT0 ∧ x2 = ___disjvr_0HAT0 ∧ x3 = _tmp_7HAT0 ∧ x4 = _x_5HAT0 ∧ x5 = _y_6HAT0 ∧ x1 = _Result_4HATpost ∧ x2 = ___disjvr_0HATpost ∧ x3 = _tmp_7HATpost ∧ x4 = _x_5HATpost ∧ x5 = _y_6HATpost ∧ _y_6HAT0 = _y_6HATpost ∧ _x_5HAT0 = _x_5HATpost ∧ _tmp_7HAT0 = _tmp_7HATpost ∧ ___disjvr_0HAT0 = ___disjvr_0HATpost ∧ _Result_4HATpost = _Result_4HATpost ∧ − _x_5HAT0 + _y_6HAT0 ≤ 0
l0	2	l2:	x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x4 = _x3 ∧ x5 = _x4 ∧ x1 = _x5 ∧ x2 = _x6 ∧ x3 = _x7 ∧ x4 = _x8 ∧ x5 = _x9 ∧ _x3 = _x8 ∧ _x1 = _x6 ∧ _x = _x5 ∧ _x9 = −1 + _x4 ∧ 0 ≤ _x7 ∧ _x7 ≤ 0 ∧ _x7 = _x7 ∧ 0 ≤ −1 − _x3 + _x4
l2	3	l0:	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
l0	4	l4:	x1 = _x20 ∧ x2 = _x21 ∧ x3 = _x22 ∧ x4 = _x23 ∧ x5 = _x24 ∧ x1 = _x25 ∧ x2 = _x26 ∧ x3 = _x27 ∧ x4 = _x28 ∧ x5 = _x29 ∧ _x24 = _x29 ∧ _x23 = _x28 ∧ _x21 = _x26 ∧ _x20 = _x25 ∧ _x27 = _x27 ∧ 0 ≤ −1 − _x23 + _x24
l4	5	l5:	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 ∧ _x30 = _x35 ∧ _x36 = _x31
l5	6	l3:	x1 = _x40 ∧ x2 = _x41 ∧ x3 = _x42 ∧ x4 = _x43 ∧ x5 = _x44 ∧ x1 = _x45 ∧ x2 = _x46 ∧ x3 = _x47 ∧ x4 = _x48 ∧ x5 = _x49 ∧ _x44 = _x49 ∧ _x42 = _x47 ∧ _x41 = _x46 ∧ _x40 = _x45 ∧ _x48 = 1 + _x43
l3	7	l0:	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 ∧ _x50 = _x55
l6	8	l0:	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
l7	9	l6:	x1 = _x70 ∧ x2 = _x71 ∧ x3 = _x72 ∧ x4 = _x73 ∧ x5 = _x74 ∧ x1 = _x75 ∧ x2 = _x76 ∧ x3 = _x77 ∧ x4 = _x78 ∧ x5 = _x79 ∧ _x74 = _x79 ∧ _x73 = _x78 ∧ _x72 = _x77 ∧ _x71 = _x76 ∧ _x70 = _x75

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
l7	l7	l7:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5
l6	l6	l6:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5
l3	l3	l3:	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 { l5, l4, l3, l0, l2 }.

2.1.1 Transition Removal

We remove transitions 2, 4 using the following ranking functions, which are bounded by 0.

l0:	−1 − x4 + x5
l2:	−1 − x4 + x5
l3:	−1 − x4 + x5
l5:	−2 − x4 + x5
l4:	−2 − x4 + x5

2.1.2 Transition Removal

We remove transitions 7, 6, 5, 3 using the following ranking functions, which are bounded by 0.

l3:	0
l0:	−1
l5:	1
l4:	2
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 % (6 real / 0 unknown / 0 assumptions / 6 total proof steps)