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 ∧ 1 − arg1 ≤ 0 ∧ − arg2 ≤ 0 ∧ − arg1P + arg2 ≤ 0 ∧ arg1P − arg2 ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg3P + arg3 ≤ 0 ∧ arg3P − arg3 ≤ 0 ∧ − arg2P + arg2P ≤ 0 ∧ arg2P − arg2P ≤ 0 ∧ − arg2 + arg2 ≤ 0 ∧ arg2 − arg2 ≤ 0
1	1	2:	0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ 2 − arg2P ≤ 0 ∧ −2 + arg2P ≤ 0 ∧ arg2 − arg3P ≤ 0 ∧ − 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
2	2	1:	0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 1 − arg2 ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ − arg1P + arg1 − arg2 ≤ 0 ∧ arg1P − arg1 + arg2 ≤ 0 ∧ −1 − arg2P + arg3 ≤ 0 ∧ 1 + arg2P − arg3 ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − arg3P + arg3 ≤ 0 ∧ arg3P − arg3 ≤ 0
2	3	2:	0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 1 − arg2 ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ −1 − arg2P + arg2 ≤ 0 ∧ 1 + arg2P − arg2 ≤ 0 ∧ −1 − arg3P + arg3 ≤ 0 ∧ 1 + arg3P − arg3 ≤ 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
2	4	2:	0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 1 − arg2 ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ −1 − arg2P + arg2 ≤ 0 ∧ 1 + arg2P − arg2 ≤ 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	5	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:	TRUE
2:	1 − arg1 ≤ 0
3:	TRUE

The invariants are proved as follows.

IMPACT Invariant Proof

nodes (location) invariant:

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

initial node: 3
cover edges:
transition edges:

0 0 1

1 1 2

2 2 1

2 3 2

2 4 2

3 5 0

2 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:

1	6	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
2	13	2:	− 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, 5 using the following ranking functions, which are bounded by −13.

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

Hints:

7	lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]
14	lexWeak[ [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, 0, 0] ]
2	lexWeak[ [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] ]
4	lexWeak[ [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] ]
5	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* 9 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 7 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 Location Addition

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

2* 16 2: − 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.

2 14 2_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

8 SCC Decomposition

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

8.1 SCC Subproblem 1/1

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

8.1.1 Transition Removal

We remove transitions 1, 2, 3, 4 using the following ranking functions, which are bounded by 11.

1:	10 + 9⋅arg1
2:	1 + 9⋅arg1 + 3⋅arg2
1_var_snapshot:	9 + 9⋅arg1
1*:	11 + 9⋅arg1
2_var_snapshot:	9⋅arg1 + 3⋅arg2
2*:	2 + 9⋅arg1 + 3⋅arg2

Hints:

7	lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 9, 0] ]
9	lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 9, 0] ]
14	lexWeak[ [0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 9, 0] ]
16	lexWeak[ [0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 9, 0] ]
1	lexStrict[ [0, 0, 0, 0, 0, 0, 3, 0, 0, 3, 0, 0, 0, 0, 0, 9, 0] , [0, 0, 0, 0, 9, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]
2	lexStrict[ [0, 0, 0, 0, 0, 0, 0, 12, 0, 0, 9, 0, 0, 9, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 3, 9, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]
3	lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 3, 0, 0, 0, 0, 0, 9, 0] , [9, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]
4	lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 3, 3, 0, 0, 0, 0, 0, 9, 0] , [0, 0, 0, 0, 0, 3, 9, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

8.1.2 Transition Removal

We remove transitions 7, 9, 14, 16 using the following ranking functions, which are bounded by −2.

1:	−1
2:	0
1_var_snapshot:	−2
1*:	0
2_var_snapshot:	−1
2*:	1

Hints:

7	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] ]
9	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] ]
14	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] ]
16	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] ]

8.1.3 Splitting Cut-Point Transitions

We consider 2 subproblems corresponding to sets of cut-point transitions as follows.

8.1.3.1 Cut-Point Subproblem 1/2

Here we consider cut-point transition 6.

8.1.3.1.1 Splitting Cut-Point Transitions

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

8.1.3.2 Cut-Point Subproblem 2/2

Here we consider cut-point transition 13.

8.1.3.2.1 Splitting Cut-Point Transitions

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

Tool configuration

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