LTS Termination Proof

Input

Integer Transition System

Initial Location: l5, l4, l7, l6, l11, l8, l1, l3, l0, l12, l2, l9

Transitions: (pre-variables and post-variables)

l0	1	l1:	x1 = _i4HAT0 ∧ x2 = _j5HAT0 ∧ x3 = _k6HAT0 ∧ x1 = _i4HATpost ∧ x2 = _j5HATpost ∧ x3 = _k6HATpost ∧ _k6HAT0 = _k6HATpost ∧ _j5HAT0 = _j5HATpost ∧ _i4HAT0 = _i4HATpost
l2	2	l3:	x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x1 = _x3 ∧ x2 = _x4 ∧ x3 = _x5 ∧ _x2 = _x5 ∧ _x1 = _x4 ∧ _x = _x3
l4	3	l5:	x1 = _x6 ∧ x2 = _x7 ∧ x3 = _x8 ∧ x1 = _x9 ∧ x2 = _x10 ∧ x3 = _x11 ∧ _x8 = _x11 ∧ _x7 = _x10 ∧ _x6 = _x9
l6	4	l7:	x1 = _x12 ∧ x2 = _x13 ∧ x3 = _x14 ∧ x1 = _x15 ∧ x2 = _x16 ∧ x3 = _x17 ∧ _x14 = _x17 ∧ _x13 = _x16 ∧ _x12 = _x15
l8	5	l6:	x1 = _x18 ∧ x2 = _x19 ∧ x3 = _x20 ∧ x1 = _x21 ∧ x2 = _x22 ∧ x3 = _x23 ∧ _x20 = _x23 ∧ _x18 = _x21 ∧ _x22 = 1 + _x19 ∧ 6 ≤ _x20
l8	6	l9:	x1 = _x24 ∧ x2 = _x25 ∧ x3 = _x26 ∧ x1 = _x27 ∧ x2 = _x28 ∧ x3 = _x29 ∧ _x25 = _x28 ∧ _x24 = _x27 ∧ _x29 = 1 + _x26 ∧ _x26 ≤ 5
l7	7	l4:	x1 = _x30 ∧ x2 = _x31 ∧ x3 = _x32 ∧ x1 = _x33 ∧ x2 = _x34 ∧ x3 = _x35 ∧ _x32 = _x35 ∧ _x31 = _x34 ∧ _x33 = 1 + _x30 ∧ 6 ≤ _x31
l7	8	l9:	x1 = _x36 ∧ x2 = _x37 ∧ x3 = _x38 ∧ x1 = _x39 ∧ x2 = _x40 ∧ x3 = _x41 ∧ _x37 = _x40 ∧ _x36 = _x39 ∧ _x41 = 1 ∧ _x37 ≤ 5
l5	9	l10:	x1 = _x42 ∧ x2 = _x43 ∧ x3 = _x44 ∧ x1 = _x45 ∧ x2 = _x46 ∧ x3 = _x47 ∧ _x44 = _x47 ∧ _x43 = _x46 ∧ _x42 = _x45 ∧ 6 ≤ _x42
l5	10	l6:	x1 = _x48 ∧ x2 = _x49 ∧ x3 = _x50 ∧ x1 = _x51 ∧ x2 = _x52 ∧ x3 = _x53 ∧ _x50 = _x53 ∧ _x48 = _x51 ∧ _x52 = 1 ∧ _x48 ≤ 5
l9	11	l8:	x1 = _x54 ∧ x2 = _x55 ∧ x3 = _x56 ∧ x1 = _x57 ∧ x2 = _x58 ∧ x3 = _x59 ∧ _x56 = _x59 ∧ _x55 = _x58 ∧ _x54 = _x57
l3	12	l0:	x1 = _x60 ∧ x2 = _x61 ∧ x3 = _x62 ∧ x1 = _x63 ∧ x2 = _x64 ∧ x3 = _x65 ∧ _x62 = _x65 ∧ _x61 = _x64 ∧ _x63 = 1 + _x60 ∧ 6 ≤ _x61
l3	13	l2:	x1 = _x66 ∧ x2 = _x67 ∧ x3 = _x68 ∧ x1 = _x69 ∧ x2 = _x70 ∧ x3 = _x71 ∧ _x68 = _x71 ∧ _x66 = _x69 ∧ _x70 = 1 + _x67 ∧ _x67 ≤ 5
l1	14	l4:	x1 = _x72 ∧ x2 = _x73 ∧ x3 = _x74 ∧ x1 = _x75 ∧ x2 = _x76 ∧ x3 = _x77 ∧ _x74 = _x77 ∧ _x73 = _x76 ∧ _x75 = 1 ∧ 6 ≤ _x72
l1	15	l2:	x1 = _x78 ∧ x2 = _x79 ∧ x3 = _x80 ∧ x1 = _x81 ∧ x2 = _x82 ∧ x3 = _x83 ∧ _x80 = _x83 ∧ _x78 = _x81 ∧ _x82 = 1 ∧ _x78 ≤ 5
l11	16	l0:	x1 = _x84 ∧ x2 = _x85 ∧ x3 = _x86 ∧ x1 = _x87 ∧ x2 = _x88 ∧ x3 = _x89 ∧ _x86 = _x89 ∧ _x85 = _x88 ∧ _x87 = 1
l12	17	l11:	x1 = _x90 ∧ x2 = _x91 ∧ x3 = _x92 ∧ x1 = _x93 ∧ x2 = _x94 ∧ x3 = _x95 ∧ _x92 = _x95 ∧ _x91 = _x94 ∧ _x90 = _x93

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
l11	l11	l11:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3
l8	l8	l8:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3
l1	l1	l1:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3
l3	l3	l3:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3
l0	l0	l0:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3
l12	l12	l12:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3
l2	l2	l2:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3
l9	l9	l9:	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 2 SCC(s) of the program graph.

2.1 SCC Subproblem 1/2

Here we consider the SCC { l1, l3, l0, l2 }.

2.1.1 Transition Removal

We remove transition 15 using the following ranking functions, which are bounded by 0.

l0:	5 − x1
l1:	5 − x1
l3:	4 − x1
l2:	4 − x1

2.1.2 Transition Removal

We remove transitions 1, 12 using the following ranking functions, which are bounded by 0.

l0:	0
l1:	−1
l3:	1
l2:	1

2.1.3 Transition Removal

We remove transition 13 using the following ranking functions, which are bounded by 0.

l2:	5 − x2
l3:	5 − x2

2.1.4 Transition Removal

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

l2:	0
l3:	−1

2.1.5 Trivial Cooperation Program

There are no more "sharp" transitions in the cooperation program. Hence the cooperation termination is proved.

2.2 SCC Subproblem 2/2

Here we consider the SCC { l5, l4, l7, l6, l8, l9 }.

2.2.1 Transition Removal

We remove transition 10 using the following ranking functions, which are bounded by −14.

l4:	−3⋅x1 + 2
l5:	−3⋅x1 + 1
l7:	−3⋅x1
l6:	−3⋅x1
l8:	−3⋅x1
l9:	−3⋅x1

2.2.2 Transition Removal

We remove transition 3 using the following ranking functions, which are bounded by −1.

l4:	−1
l5:	−2
l7:	0
l6:	0
l8:	0
l9:	0

2.2.3 Transition Removal

We remove transition 8 using the following ranking functions, which are bounded by −15.

l7:	−3⋅x2
l4:	−3⋅x2 − 1
l6:	−3⋅x2 + 1
l8:	−3⋅x2 − 1
l9:	−3⋅x2 − 1

2.2.4 Transition Removal

We remove transition 6 using the following ranking functions, which are bounded by −10.

l7:	−2⋅x3 − 2
l4:	−2⋅x3 − 3
l6:	−2⋅x3 − 1
l8:	−2⋅x3
l9:	−2⋅x3 + 1

2.2.5 Transition Removal

We remove transitions 7, 4, 5, 11 using the following ranking functions, which are bounded by 0.

l7:	0
l4:	−1
l6:	1
l8:	2
l9:	3

2.2.6 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 % (15 real / 0 unknown / 0 assumptions / 15 total proof steps)