LTS Termination Proof

Input

Integer Transition System

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

Transitions: (pre-variables and post-variables)

l0	1	l1:	x1 = _iHAT0 ∧ x2 = _jHAT0 ∧ x3 = _y4HAT0 ∧ x4 = _y6HAT0 ∧ x5 = _y8HAT0 ∧ x1 = _iHATpost ∧ x2 = _jHATpost ∧ x3 = _y4HATpost ∧ x4 = _y6HATpost ∧ x5 = _y8HATpost ∧ _y8HAT0 = _y8HATpost ∧ _y6HAT0 = _y6HATpost ∧ _y4HAT0 = _y4HATpost ∧ _iHAT0 = _iHATpost ∧ _jHATpost = 100 ∧ 100 ≤ _iHAT0
l0	2	l2:	x1 = _x ∧ x2 = _x1 ∧ x3 = _x2 ∧ x4 = _x3 ∧ x5 = _x4 ∧ x1 = _x5 ∧ x2 = _x6 ∧ x3 = _x7 ∧ x4 = _x8 ∧ x5 = _x9 ∧ _x4 = _x9 ∧ _x1 = _x6 ∧ _x5 = 1 + _x ∧ _x8 = _x ∧ _x7 = _x ∧ 1 + _x ≤ 100
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
l3	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 ∧ _x22 = _x27 ∧ _x21 = _x26 ∧ _x20 = _x25 ∧ 200 ≤ _x21
l3	5	l1:	x1 = _x30 ∧ x2 = _x31 ∧ x3 = _x32 ∧ x4 = _x33 ∧ x5 = _x34 ∧ x1 = _x35 ∧ x2 = _x36 ∧ x3 = _x37 ∧ x4 = _x38 ∧ x5 = _x39 ∧ _x33 = _x38 ∧ _x32 = _x37 ∧ _x30 = _x35 ∧ _x36 = 1 + _x31 ∧ _x39 = _x31 ∧ 1 + _x31 ≤ 200
l1	6	l3:	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
l5	7	l2:	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 = 0
l6	8	l5:	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
l6	l6	l6:	x1 = x1 ∧ x2 = x2 ∧ x3 = x3 ∧ x4 = x4 ∧ x5 = x5
l1	l1	l1:	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 2 SCC(s) of the program graph.

2.1 SCC Subproblem 1/2

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:	99 − x1
l2:	99 − x1

2.1.2 Transition Removal

We remove transition 3 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.

2.2 SCC Subproblem 2/2

Here we consider the SCC { l1, l3 }.

2.2.1 Transition Removal

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

l1:	199 − x2
l3:	199 − x2

2.2.2 Transition Removal

We remove transition 6 using the following ranking functions, which are bounded by 0.

l1:	0
l3:	−1

2.2.3 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 % (10 real / 0 unknown / 0 assumptions / 10 total proof steps)