LTS Termination Proof

by T2Cert

Input

Integer Transition System

Initial Location: 4

Transitions: (pre-variables and post-variables)

0	0	1:	0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ − arg1P ≤ 0 ∧ 1 − arg2 ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ 1 − arg3P ≤ 0 ∧ −1 + arg3P ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg3P + arg3 ≤ 0 ∧ arg3P − arg3 ≤ 0 ∧ − x8 + x8 ≤ 0 ∧ x8 − x8 ≤ 0 ∧ − x18 + x18 ≤ 0 ∧ x18 − x18 ≤ 0 ∧ − x14 + x14 ≤ 0 ∧ x14 − x14 ≤ 0 ∧ − arg2P + arg2P ≤ 0 ∧ arg2P − arg2P ≤ 0 ∧ − arg2 + arg2 ≤ 0 ∧ arg2 − arg2 ≤ 0
0	1	2:	0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ − x8 ≤ 0 ∧ 1 − arg2 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ 3 − arg2P ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − arg3P + arg3 ≤ 0 ∧ arg3P − arg3 ≤ 0 ∧ − x18 + x18 ≤ 0 ∧ x18 − x18 ≤ 0 ∧ − x14 + x14 ≤ 0 ∧ x14 − x14 ≤ 0
0	2	2:	0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ − x14 ≤ 0 ∧ 1 − arg2 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ −1 − arg1 + arg2P ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ 2 − arg2P ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − arg3P + arg3 ≤ 0 ∧ arg3P − arg3 ≤ 0 ∧ − x8 + x8 ≤ 0 ∧ x8 − x8 ≤ 0 ∧ − x18 + x18 ≤ 0 ∧ x18 − x18 ≤ 0
1	3	1:	0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ − arg1 ≤ 0 ∧ − arg2 ≤ 0 ∧ 1 − arg3 ≤ 0 ∧ 0 ≤ 0 ∧ 1 − arg2 + arg3 ≤ 0 ∧ −1 − arg1P + arg1 ≤ 0 ∧ 1 + arg1P − arg1 ≤ 0 ∧ 1 − arg3P + arg3 ≤ 0 ∧ −1 + arg3P − arg3 ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg3P + arg3 ≤ 0 ∧ arg3P − arg3 ≤ 0 ∧ − x8 + x8 ≤ 0 ∧ x8 − x8 ≤ 0 ∧ − x18 + x18 ≤ 0 ∧ x18 − x18 ≤ 0 ∧ − x14 + x14 ≤ 0 ∧ x14 − x14 ≤ 0 ∧ − arg2P + arg2P ≤ 0 ∧ arg2P − arg2P ≤ 0 ∧ − arg2 + arg2 ≤ 0 ∧ arg2 − arg2 ≤ 0
2	4	3:	0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 1 − x18 ≤ 0 ∧ 2 − arg3 ≤ 0 ∧ arg1P − arg2 ≤ 0 ∧ 1 + arg2P − arg2 ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ 1 − arg2 ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ − arg2P ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − x8 + x8 ≤ 0 ∧ x8 − x8 ≤ 0 ∧ − x14 + x14 ≤ 0 ∧ x14 − x14 ≤ 0 ∧ − arg3P + arg3P ≤ 0 ∧ arg3P − arg3P ≤ 0 ∧ − arg3 + arg3 ≤ 0 ∧ arg3 − arg3 ≤ 0
3	5	3:	0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 2 + arg1P − arg1 ≤ 0 ∧ 2 − arg3 ≤ 0 ∧ arg1P − arg2 ≤ 0 ∧ 3 − arg1 + arg2P ≤ 0 ∧ 1 + arg2P − arg2 ≤ 0 ∧ 3 − arg1 ≤ 0 ∧ 1 − arg2 ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ − arg2P ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − x8 + x8 ≤ 0 ∧ x8 − x8 ≤ 0 ∧ − x18 + x18 ≤ 0 ∧ x18 − x18 ≤ 0 ∧ − x14 + x14 ≤ 0 ∧ x14 − x14 ≤ 0 ∧ − arg3P + arg3P ≤ 0 ∧ arg3P − arg3P ≤ 0 ∧ − arg3 + arg3 ≤ 0 ∧ arg3 − arg3 ≤ 0
4	6	0:	0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − arg3P + arg3 ≤ 0 ∧ arg3P − arg3 ≤ 0 ∧ − x8 + x8 ≤ 0 ∧ x8 − x8 ≤ 0 ∧ − x18 + x18 ≤ 0 ∧ x18 − x18 ≤ 0 ∧ − x14 + x14 ≤ 0 ∧ x14 − x14 ≤ 0

Proof

1 Invariant Updates

The following invariants are asserted.

0:	TRUE
1:	1 − arg2 ≤ 0
2:	1 − arg1P ≤ 0 ∧ 1 − arg2P ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ 1 − arg2 ≤ 0
3:	1 − arg1P ≤ 0 ∧ − arg2P ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ − arg2 ≤ 0 ∧ 2 − arg3 ≤ 0 ∧ 1 − x18 ≤ 0
4:	TRUE

The invariants are proved as follows.

IMPACT Invariant Proof

nodes (location) invariant:

0	(0)	TRUE
1	(1)	1 − arg2 ≤ 0
2	(2)	1 − arg1P ≤ 0 ∧ 1 − arg2P ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ 1 − arg2 ≤ 0
3	(3)	1 − arg1P ≤ 0 ∧ − arg2P ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ − arg2 ≤ 0 ∧ 2 − arg3 ≤ 0 ∧ 1 − x18 ≤ 0
4	(4)	TRUE

initial node: 4
cover edges:
transition edges:

0 0 1

0 1 2

0 2 2

1 3 1

2 4 3

3 5 3

4 6 0

2 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:

1	7	1:	− x8 + x8 ≤ 0 ∧ x8 − x8 ≤ 0 ∧ − x18 + x18 ≤ 0 ∧ x18 − x18 ≤ 0 ∧ − x14 + x14 ≤ 0 ∧ x14 − x14 ≤ 0 ∧ − arg3P + arg3P ≤ 0 ∧ arg3P − arg3P ≤ 0 ∧ − arg3 + arg3 ≤ 0 ∧ arg3 − arg3 ≤ 0 ∧ − arg2P + arg2P ≤ 0 ∧ arg2P − arg2P ≤ 0 ∧ − arg2 + arg2 ≤ 0 ∧ arg2 − arg2 ≤ 0 ∧ − arg1P + arg1P ≤ 0 ∧ arg1P − arg1P ≤ 0 ∧ − arg1 + arg1 ≤ 0 ∧ arg1 − arg1 ≤ 0
3	14	3:	− x8 + x8 ≤ 0 ∧ x8 − x8 ≤ 0 ∧ − x18 + x18 ≤ 0 ∧ x18 − x18 ≤ 0 ∧ − x14 + x14 ≤ 0 ∧ x14 − x14 ≤ 0 ∧ − arg3P + arg3P ≤ 0 ∧ arg3P − arg3P ≤ 0 ∧ − arg3 + arg3 ≤ 0 ∧ arg3 − arg3 ≤ 0 ∧ − arg2P + arg2P ≤ 0 ∧ arg2P − arg2P ≤ 0 ∧ − arg2 + arg2 ≤ 0 ∧ arg2 − arg2 ≤ 0 ∧ − arg1P + arg1P ≤ 0 ∧ arg1P − arg1P ≤ 0 ∧ − arg1 + arg1 ≤ 0 ∧ arg1 − arg1 ≤ 0

and for every transition t, a duplicate t is considered.

3 Transition Removal

We remove transitions 0, 1, 2, 4, 6 using the following ranking functions, which are bounded by −17.

4:	0
0:	0
1:	0
2:	0
3:	0
4:	−6
0:	−7
1:	−8
1_var_snapshot:	−8
1*:	−8
2:	−11
3:	−12
3_var_snapshot:	−12
3*:	−12

Hints:

8	lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]
15	lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]
3	lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]
5	lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]
0	lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]
1	lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]
2	lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]
4	lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]
6	lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

4 Location Addition

The following skip-transition is inserted and corresponding redirections w.r.t. the old location are performed.

1* 10 1: − x8 + x8 ≤ 0 ∧ x8 − x8 ≤ 0 ∧ − x18 + x18 ≤ 0 ∧ x18 − x18 ≤ 0 ∧ − x14 + x14 ≤ 0 ∧ x14 − x14 ≤ 0 ∧ − arg3P + arg3P ≤ 0 ∧ arg3P − arg3P ≤ 0 ∧ − arg3 + arg3 ≤ 0 ∧ arg3 − arg3 ≤ 0 ∧ − arg2P + arg2P ≤ 0 ∧ arg2P − arg2P ≤ 0 ∧ − arg2 + arg2 ≤ 0 ∧ arg2 − arg2 ≤ 0 ∧ − arg1P + arg1P ≤ 0 ∧ arg1P − arg1P ≤ 0 ∧ − arg1 + arg1 ≤ 0 ∧ arg1 − arg1 ≤ 0

5 Location Addition

The following skip-transition is inserted and corresponding redirections w.r.t. the old location are performed.

1 8 1_var_snapshot: − x8 + x8 ≤ 0 ∧ x8 − x8 ≤ 0 ∧ − x18 + x18 ≤ 0 ∧ x18 − x18 ≤ 0 ∧ − x14 + x14 ≤ 0 ∧ x14 − x14 ≤ 0 ∧ − arg3P + arg3P ≤ 0 ∧ arg3P − arg3P ≤ 0 ∧ − arg3 + arg3 ≤ 0 ∧ arg3 − arg3 ≤ 0 ∧ − arg2P + arg2P ≤ 0 ∧ arg2P − arg2P ≤ 0 ∧ − arg2 + arg2 ≤ 0 ∧ arg2 − arg2 ≤ 0 ∧ − arg1P + arg1P ≤ 0 ∧ arg1P − arg1P ≤ 0 ∧ − arg1 + arg1 ≤ 0 ∧ arg1 − arg1 ≤ 0

6 Location Addition

The following skip-transition is inserted and corresponding redirections w.r.t. the old location are performed.

3* 17 3: − x8 + x8 ≤ 0 ∧ x8 − x8 ≤ 0 ∧ − x18 + x18 ≤ 0 ∧ x18 − x18 ≤ 0 ∧ − x14 + x14 ≤ 0 ∧ x14 − x14 ≤ 0 ∧ − arg3P + arg3P ≤ 0 ∧ arg3P − arg3P ≤ 0 ∧ − arg3 + arg3 ≤ 0 ∧ arg3 − arg3 ≤ 0 ∧ − arg2P + arg2P ≤ 0 ∧ arg2P − arg2P ≤ 0 ∧ − arg2 + arg2 ≤ 0 ∧ arg2 − arg2 ≤ 0 ∧ − arg1P + arg1P ≤ 0 ∧ arg1P − arg1P ≤ 0 ∧ − arg1 + arg1 ≤ 0 ∧ arg1 − arg1 ≤ 0

7 Location Addition

The following skip-transition is inserted and corresponding redirections w.r.t. the old location are performed.

3 15 3_var_snapshot: − x8 + x8 ≤ 0 ∧ x8 − x8 ≤ 0 ∧ − x18 + x18 ≤ 0 ∧ x18 − x18 ≤ 0 ∧ − x14 + x14 ≤ 0 ∧ x14 − x14 ≤ 0 ∧ − arg3P + arg3P ≤ 0 ∧ arg3P − arg3P ≤ 0 ∧ − arg3 + arg3 ≤ 0 ∧ arg3 − arg3 ≤ 0 ∧ − arg2P + arg2P ≤ 0 ∧ arg2P − arg2P ≤ 0 ∧ − arg2 + arg2 ≤ 0 ∧ arg2 − arg2 ≤ 0 ∧ − arg1P + arg1P ≤ 0 ∧ arg1P − arg1P ≤ 0 ∧ − arg1 + arg1 ≤ 0 ∧ arg1 − arg1 ≤ 0

8 SCC Decomposition

We consider subproblems for each of the 2 SCC(s) of the program graph.

8.1 SCC Subproblem 1/2

Here we consider the SCC { 1, 1_var_snapshot, 1* }.

8.1.1 Transition Removal

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

1:	arg1 + arg2 − arg3
1_var_snapshot:	arg1 + arg2 − arg3
1*:	1 + arg1 + arg2 − arg3

Hints:

8	lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0] ]
10	lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0] ]
3	lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0] , [0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

8.1.2 Transition Removal

We remove transitions 8, 10 using the following ranking functions, which are bounded by −1.

1:	0
1_var_snapshot:	− arg2
1*:	arg2

Hints:

8	lexStrict[ [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]
10	lexStrict[ [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] , [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

8.1.3 Splitting Cut-Point Transitions

We consider 1 subproblems corresponding to sets of cut-point transitions as follows.

8.1.3.1 Cut-Point Subproblem 1/1

Here we consider cut-point transition 7.

8.1.3.1.1 Splitting Cut-Point Transitions

There remain no cut-point transition to consider. Hence the cooperation termination is trivial.

8.2 SCC Subproblem 2/2

Here we consider the SCC { 3, 3_var_snapshot, 3* }.

8.2.1 Transition Removal

We remove transitions 15, 17, 5 using the following ranking functions, which are bounded by −1.

3:	1 + 3⋅arg2
3_var_snapshot:	3⋅arg2
3*:	2 + 3⋅arg2

Hints:

15	lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0] , [0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]
17	lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0] , [0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]
5	lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] , [0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

8.2.2 Splitting Cut-Point Transitions

We consider 1 subproblems corresponding to sets of cut-point transitions as follows.

8.2.2.1 Cut-Point Subproblem 1/1

Here we consider cut-point transition 14.

8.2.2.1.1 Splitting Cut-Point Transitions

There remain no cut-point transition to consider. Hence the cooperation termination is trivial.

Tool configuration

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