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

Proof

1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:

1	5	1:	− 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	12	2:	− 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.

2 Transition Removal

We remove transitions 0, 4 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:

6	lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0] ]
13	lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0] ]
1	lexWeak[ [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] ]
3	lexWeak[ [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] ]
4	lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0] ]

3 Location Addition

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

1* 8 1: − arg2P + arg2P ≤ 0 ∧ arg2P − arg2P ≤ 0 ∧ − arg2 + arg2 ≤ 0 ∧ arg2 − arg2 ≤ 0 ∧ − arg1P + arg1P ≤ 0 ∧ arg1P − arg1P ≤ 0 ∧ − arg1 + arg1 ≤ 0 ∧ arg1 − arg1 ≤ 0

4 Location Addition

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

1 6 1_var_snapshot: − 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.

2* 15 2: − 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 13 2_var_snapshot: − 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 SCC Decomposition

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

7.1 SCC Subproblem 1/1

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

7.1.1 Transition Removal

We remove transitions 1, 3 using the following ranking functions, which are bounded by 8.

1:	16 + 9⋅arg1
2:	9⋅arg1
1_var_snapshot:	16 + 9⋅arg1
1*:	17 + 9⋅arg1
2_var_snapshot:	9⋅arg1
2*:	9⋅arg1

Hints:

6	lexWeak[ [0, 0, 0, 0, 0, 0, 9, 0] ]
8	lexWeak[ [0, 0, 0, 0, 0, 0, 9, 0] ]
13	lexWeak[ [0, 0, 0, 0, 0, 0, 9, 0] ]
15	lexWeak[ [0, 0, 0, 0, 0, 0, 9, 0] ]
1	lexStrict[ [0, 0, 0, 0, 0, 0, 9, 0, 0, 9, 0, 0, 0] , [0, 0, 0, 0, 9, 0, 0, 0, 0, 0, 0, 0, 0] ]
2	lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 9, 0] ]
3	lexStrict[ [0, 0, 0, 0, 0, 0, 0, 9, 9, 0, 0, 0] , [0, 0, 0, 0, 9, 0, 0, 0, 0, 0, 0, 0] ]

7.1.2 Transition Removal

We remove transitions 6, 8, 2 using the following ranking functions, which are bounded by −2.

1:	−1
2:	2⋅arg1 − 2⋅arg2
1_var_snapshot:	−2
1*:	0
2_var_snapshot:	2⋅arg1 − 2⋅arg2
2*:	1 + 2⋅arg1 − 2⋅arg2

Hints:

6	lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0] ]
8	lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0] ]
13	lexWeak[ [0, 0, 0, 2, 0, 0, 2, 0] ]
15	lexWeak[ [0, 0, 0, 2, 0, 0, 2, 0] ]
2	lexStrict[ [0, 0, 0, 2, 0, 0, 2, 0, 0, 2, 0] , [0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0] ]

7.1.3 Transition Removal

We remove transitions 13, 15 using the following ranking functions, which are bounded by −1.

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

Hints:

13	lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0] ]
15	lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0] ]

7.1.4 Splitting Cut-Point Transitions

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

7.1.4.1 Cut-Point Subproblem 1/2

Here we consider cut-point transition 5.

7.1.4.1.1 Splitting Cut-Point Transitions

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

7.1.4.2 Cut-Point Subproblem 2/2

Here we consider cut-point transition 12.

7.1.4.2.1 Splitting Cut-Point Transitions

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

Tool configuration

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