LTS Termination Proof

by T2Cert

Input

Integer Transition System

Initial Location: 3

Transitions: (pre-variables and post-variables)

0	0	1:	0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ − arg1P ≤ 0 ∧ − arg2 ≤ 0 ∧ − arg3P ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ −1 + arg1P − arg2P ≤ 0 ∧ 1 − arg1P + arg2P ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − arg3P + arg3 ≤ 0 ∧ arg3P − arg3 ≤ 0
1	1	2:	1 + arg2 − arg3 ≤ 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
1	2	2:	0 ≤ 0 ∧ 0 ≤ 0 ∧ − arg2 + arg3 ≤ 0 ∧ − arg2P + arg3 ≤ 0 ∧ arg2P − arg3 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − arg3P + arg3P ≤ 0 ∧ arg3P − arg3P ≤ 0 ∧ − arg3 + arg3 ≤ 0 ∧ arg3 − arg3 ≤ 0 ∧ − arg1P + arg1P ≤ 0 ∧ arg1P − arg1P ≤ 0 ∧ − arg1 + arg1 ≤ 0 ∧ arg1 − arg1 ≤ 0
2	3	1:	0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ − arg1 ≤ 0 ∧ arg2 − arg3 ≤ 0 ∧ − arg2 + arg3 ≤ 0 ∧ −1 + arg1 − arg2P ≤ 0 ∧ 1 − arg1 + arg2P ≤ 0 ∧ 1 + arg2 − arg3P ≤ 0 ∧ −1 − arg2 + arg3P ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − arg3P + arg3 ≤ 0 ∧ arg3P − arg3 ≤ 0 ∧ − arg1P + arg1P ≤ 0 ∧ arg1P − arg1P ≤ 0 ∧ − arg1 + arg1 ≤ 0 ∧ arg1 − arg1 ≤ 0
3	4	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

Proof

1 Invariant Updates

The following invariants are asserted.

0:	TRUE
1:	− arg1P ≤ 0 ∧ − arg1 ≤ 0
2:	− arg1P ≤ 0 ∧ − arg1 ≤ 0
3:	TRUE

The invariants are proved as follows.

IMPACT Invariant Proof

nodes (location) invariant:

0	(0)	TRUE
1	(1)	− arg1P ≤ 0 ∧ − arg1 ≤ 0
2	(2)	− arg1P ≤ 0 ∧ − arg1 ≤ 0
3	(3)	TRUE

initial node: 3
cover edges:
transition edges:

0 0 1

1 1 2

1 2 2

2 3 1

3 4 0

2 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:

− 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, 4 using the following ranking functions, which are bounded by −11.

3:	0
0:	0
1:	0
2:	0
3:	−4
0:	−5
1:	−6
2:	−6
1_var_snapshot:	−6
1*:	−6

Hints:

6	lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]
1	lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]
2	lexWeak[ [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	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	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] ]

4 Location Addition

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

1* 8 1: − 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 6 1_var_snapshot: − 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 SCC Decomposition

We consider subproblems for each of the 1 SCC(s) of the program graph.

6.1 SCC Subproblem 1/1

Here we consider the SCC { 1, 2, 1_var_snapshot, 1* }.

6.1.1 Transition Removal

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

1:	0
2:	arg2 − arg3
1_var_snapshot:	0
1*:	0

Hints:

6	lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]
8	lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]
1	lexStrict[ [0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]
2	lexWeak[ [0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0] ]
3	lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

6.1.2 Splitting Cut-Point Transitions

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

6.1.2.1 Cut-Point Subproblem 1/1

Here we consider cut-point transition 5.

6.1.2.1.1 Fresh Variable Addition

The new variable __snapshot_1_arg3P is introduced. The transition formulas are extended as follows:

6:	__snapshot_1_arg3P ≤ arg3P ∧ arg3P ≤ __snapshot_1_arg3P
8:	__snapshot_1_arg3P ≤ __snapshot_1_arg3P ∧ __snapshot_1_arg3P ≤ __snapshot_1_arg3P
2:	__snapshot_1_arg3P ≤ __snapshot_1_arg3P ∧ __snapshot_1_arg3P ≤ __snapshot_1_arg3P
3:	__snapshot_1_arg3P ≤ __snapshot_1_arg3P ∧ __snapshot_1_arg3P ≤ __snapshot_1_arg3P

6.1.2.1.2 Fresh Variable Addition

The new variable __snapshot_1_arg3 is introduced. The transition formulas are extended as follows:

6:	__snapshot_1_arg3 ≤ arg3 ∧ arg3 ≤ __snapshot_1_arg3
8:	__snapshot_1_arg3 ≤ __snapshot_1_arg3 ∧ __snapshot_1_arg3 ≤ __snapshot_1_arg3
2:	__snapshot_1_arg3 ≤ __snapshot_1_arg3 ∧ __snapshot_1_arg3 ≤ __snapshot_1_arg3
3:	__snapshot_1_arg3 ≤ __snapshot_1_arg3 ∧ __snapshot_1_arg3 ≤ __snapshot_1_arg3

6.1.2.1.3 Fresh Variable Addition

The new variable __snapshot_1_arg2P is introduced. The transition formulas are extended as follows:

6:	__snapshot_1_arg2P ≤ arg2P ∧ arg2P ≤ __snapshot_1_arg2P
8:	__snapshot_1_arg2P ≤ __snapshot_1_arg2P ∧ __snapshot_1_arg2P ≤ __snapshot_1_arg2P
2:	__snapshot_1_arg2P ≤ __snapshot_1_arg2P ∧ __snapshot_1_arg2P ≤ __snapshot_1_arg2P
3:	__snapshot_1_arg2P ≤ __snapshot_1_arg2P ∧ __snapshot_1_arg2P ≤ __snapshot_1_arg2P

6.1.2.1.4 Fresh Variable Addition

The new variable __snapshot_1_arg2 is introduced. The transition formulas are extended as follows:

6:	__snapshot_1_arg2 ≤ arg2 ∧ arg2 ≤ __snapshot_1_arg2
8:	__snapshot_1_arg2 ≤ __snapshot_1_arg2 ∧ __snapshot_1_arg2 ≤ __snapshot_1_arg2
2:	__snapshot_1_arg2 ≤ __snapshot_1_arg2 ∧ __snapshot_1_arg2 ≤ __snapshot_1_arg2
3:	__snapshot_1_arg2 ≤ __snapshot_1_arg2 ∧ __snapshot_1_arg2 ≤ __snapshot_1_arg2

6.1.2.1.5 Fresh Variable Addition

The new variable __snapshot_1_arg1P is introduced. The transition formulas are extended as follows:

6:	__snapshot_1_arg1P ≤ arg1P ∧ arg1P ≤ __snapshot_1_arg1P
8:	__snapshot_1_arg1P ≤ __snapshot_1_arg1P ∧ __snapshot_1_arg1P ≤ __snapshot_1_arg1P
2:	__snapshot_1_arg1P ≤ __snapshot_1_arg1P ∧ __snapshot_1_arg1P ≤ __snapshot_1_arg1P
3:	__snapshot_1_arg1P ≤ __snapshot_1_arg1P ∧ __snapshot_1_arg1P ≤ __snapshot_1_arg1P

6.1.2.1.6 Fresh Variable Addition

The new variable __snapshot_1_arg1 is introduced. The transition formulas are extended as follows:

6:	__snapshot_1_arg1 ≤ arg1 ∧ arg1 ≤ __snapshot_1_arg1
8:	__snapshot_1_arg1 ≤ __snapshot_1_arg1 ∧ __snapshot_1_arg1 ≤ __snapshot_1_arg1
2:	__snapshot_1_arg1 ≤ __snapshot_1_arg1 ∧ __snapshot_1_arg1 ≤ __snapshot_1_arg1
3:	__snapshot_1_arg1 ≤ __snapshot_1_arg1 ∧ __snapshot_1_arg1 ≤ __snapshot_1_arg1

6.1.2.1.7 Invariant Updates

The following invariants are asserted.

0:	TRUE
1:	arg2 − arg2P ≤ 0 ∧ −1 + arg1 − arg2P ≤ 0 ∧ − arg1P ≤ 0 ∧ − arg1 ≤ 0
2:	− arg1P ≤ 0 ∧ − arg1 ≤ 0
3:	TRUE
1:	arg2 − arg2P ≤ 0 ∧ −1 + arg1 − arg2P ≤ 0 ∧ − arg1P ≤ 0 ∧ − arg1 ≤ 0 ∨ − __snapshot_1_arg2P + __snapshot_1_arg3 ≤ 0 ∧ arg2 − arg2P ≤ 0 ∧ 1 − __snapshot_1_arg2P + __snapshot_1_arg3 + arg2P − arg3 ≤ 0 ∧ −1 + arg1 − arg2P ≤ 0 ∧ − arg1P ≤ 0 ∧ − arg1 ≤ 0
2:	− __snapshot_1_arg2P + __snapshot_1_arg3 ≤ 0 ∧ −1 − __snapshot_1_arg2P + __snapshot_1_arg3 + arg1 − arg2 ≤ 0 ∧ − arg1P ≤ 0 ∧ − arg1 ≤ 0
1_var_snapshot:	−1 − __snapshot_1_arg2P + __snapshot_1_arg3 + arg1 − arg3 ≤ 0 ∧ − __snapshot_1_arg2P + __snapshot_1_arg3 + arg2 − arg3 ≤ 0 ∧ − arg1P ≤ 0 ∧ − arg1 ≤ 0
1*:	− __snapshot_1_arg2P + __snapshot_1_arg3 ≤ 0 ∧ arg2 − arg2P ≤ 0 ∧ 1 − __snapshot_1_arg2P + __snapshot_1_arg3 + arg2P − arg3 ≤ 0 ∧ −1 + arg1 − arg2P ≤ 0 ∧ − arg1P ≤ 0 ∧ − arg1 ≤ 0

The invariants are proved as follows.

IMPACT Invariant Proof

nodes (location) invariant:

0	(3)	TRUE
1	(0)	TRUE
2	(1)	arg2 − arg2P ≤ 0 ∧ −1 + arg1 − arg2P ≤ 0 ∧ − arg1P ≤ 0 ∧ − arg1 ≤ 0
3	(2)	− arg1P ≤ 0 ∧ − arg1 ≤ 0
4	(2)	− arg1P ≤ 0 ∧ − arg1 ≤ 0
5	(1)	arg2 − arg2P ≤ 0 ∧ −1 + arg1 − arg2P ≤ 0 ∧ − arg1P ≤ 0 ∧ − arg1 ≤ 0
6	(1_var_snapshot)	−1 − __snapshot_1_arg2P + __snapshot_1_arg3 + arg1 − arg3 ≤ 0 ∧ − __snapshot_1_arg2P + __snapshot_1_arg3 + arg2 − arg3 ≤ 0 ∧ − arg1P ≤ 0 ∧ − arg1 ≤ 0
11	(1)	arg2 − arg2P ≤ 0 ∧ −1 + arg1 − arg2P ≤ 0 ∧ − arg1P ≤ 0 ∧ − arg1 ≤ 0
15	(2)	− __snapshot_1_arg2P + __snapshot_1_arg3 ≤ 0 ∧ −1 − __snapshot_1_arg2P + __snapshot_1_arg3 + arg1 − arg2 ≤ 0 ∧ − arg1P ≤ 0 ∧ − arg1 ≤ 0
16	(1*)	− __snapshot_1_arg2P + __snapshot_1_arg3 ≤ 0 ∧ arg2 − arg2P ≤ 0 ∧ 1 − __snapshot_1_arg2P + __snapshot_1_arg3 + arg2P − arg3 ≤ 0 ∧ −1 + arg1 − arg2P ≤ 0 ∧ − arg1P ≤ 0 ∧ − arg1 ≤ 0
17	(1)	− __snapshot_1_arg2P + __snapshot_1_arg3 ≤ 0 ∧ arg2 − arg2P ≤ 0 ∧ 1 − __snapshot_1_arg2P + __snapshot_1_arg3 + arg2P − arg3 ≤ 0 ∧ −1 + arg1 − arg2P ≤ 0 ∧ − arg1P ≤ 0 ∧ − arg1 ≤ 0
18	(1_var_snapshot)	−1 − __snapshot_1_arg2P + __snapshot_1_arg3 + arg1 − arg3 ≤ 0 ∧ − __snapshot_1_arg2P + __snapshot_1_arg3 + arg2 − arg3 ≤ 0 ∧ − arg1P ≤ 0 ∧ − arg1 ≤ 0

initial node: 0

cover edges:

→ 4

Hint: distribute conclusion

	[1, 0]
	[0, 1]

→ 2

Hint: distribute conclusion

	[1, 0, 0, 0]
	[0, 1, 0, 0]
	[0, 0, 1, 0]
	[0, 0, 0, 1]

→ 6

Hint: distribute conclusion

	[1, 0, 0, 0]
	[0, 1, 0, 0]
	[0, 0, 1, 0]
	[0, 0, 0, 1]

transition edges:

4 1

Hint: auto

0 2

Hint: distribute conclusion

	[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, 1, 0, 1, 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, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0]

1 3

Hint: distribute conclusion

	[0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0]
	[0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1]

2 4

Hint: distribute conclusion

	[0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0]
	[0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1]

5 5

Hint: distribute conclusion

	[1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0]
	[0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0]
	[0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0]
	[0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1]

3 11

Hint: distribute conclusion

	[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, 1, 0, 0, 0, 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, 1, 0, 0]
	[0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1]

6 6

Hint: distribute conclusion

	[0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0]
	[1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0]
	[0, 0, 1, 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, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

2 15

Hint: distribute conclusion

	[0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0]
	[1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0]
	[0, 0, 1, 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, 1, 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]

3 16

Hint: distribute conclusion

	[1, 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, 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, 1, 0, 0, 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, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 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, 1, 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, 0, 1, 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, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

8 17

Hint: distribute conclusion

	[1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0]
	[0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 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, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0]
	[0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 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, 1, 0, 0, 0, 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, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

6 18

Hint: distribute conclusion

	[0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0]
	[0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0]
	[0, 0, 0, 0, 1, 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, 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]

6.1.2.1.8 Transition Removal

We remove transition 8 using the following ranking functions, which are bounded by −2.

1:	arg2P − arg3
2:	__snapshot_1_arg2P − __snapshot_1_arg3
1_var_snapshot:	__snapshot_1_arg2P − __snapshot_1_arg3
1*:	__snapshot_1_arg2P − __snapshot_1_arg3

Hints:

distribute assertion

	lexWeak[ [0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0] ]
	lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0] ]

lexStrict[ [0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 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, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0] ]

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, 1, 1, 0, 0, 0, 0, 0, 0, 0] ]

6.1.2.1.9 Transition Removal

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

1:	−1
1_var_snapshot:	−2
2:	−3
1*:	−4

Hints:

distribute assertion

	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] ]
	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] ]

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] ]

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, 0] ]

6.1.2.1.10 Splitting Cut-Point Transitions

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

Tool configuration

T2Cert

version: 1.0