LTS Termination Proof

by T2Cert

Input

Integer Transition System

Proof

1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
1 7 1: arg3P + arg3P ≤ 0arg3Parg3P ≤ 0arg3 + arg3 ≤ 0arg3arg3 ≤ 0arg2P + arg2P ≤ 0arg2Parg2P ≤ 0arg2 + arg2 ≤ 0arg2arg2 ≤ 0arg1P + arg1P ≤ 0arg1Parg1P ≤ 0arg1 + arg1 ≤ 0arg1arg1 ≤ 0
2 14 2: arg3P + arg3P ≤ 0arg3Parg3P ≤ 0arg3 + arg3 ≤ 0arg3arg3 ≤ 0arg2P + arg2P ≤ 0arg2Parg2P ≤ 0arg2 + arg2 ≤ 0arg2arg2 ≤ 0arg1P + arg1P ≤ 0arg1Parg1P ≤ 0arg1 + arg1 ≤ 0arg1arg1 ≤ 0
3 21 3: arg3P + arg3P ≤ 0arg3Parg3P ≤ 0arg3 + arg3 ≤ 0arg3arg3 ≤ 0arg2P + arg2P ≤ 0arg2Parg2P ≤ 0arg2 + arg2 ≤ 0arg2arg2 ≤ 0arg1P + arg1P ≤ 0arg1Parg1P ≤ 0arg1 + arg1 ≤ 0arg1arg1 ≤ 0
and for every transition t, a duplicate t is considered.

2 Transition Removal

We remove transitions 0, 2, 4, 6 using the following ranking functions, which are bounded by −19.

4: 0
0: 0
1: 0
2: 0
3: 0
4: −6
0: −7
1: −8
1_var_snapshot: −8
1*: −8
2: −11
2_var_snapshot: −11
2*: −11
3: −14
3_var_snapshot: −14
3*: −14
Hints:
8 lexWeak[ [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] ]
22 lexWeak[ [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, 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] ]
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 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] ]
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] ]
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] ]
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] ]

3 Location Addition

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

1* 10 1: arg3P + arg3P ≤ 0arg3Parg3P ≤ 0arg3 + arg3 ≤ 0arg3arg3 ≤ 0arg2P + arg2P ≤ 0arg2Parg2P ≤ 0arg2 + arg2 ≤ 0arg2arg2 ≤ 0arg1P + arg1P ≤ 0arg1Parg1P ≤ 0arg1 + arg1 ≤ 0arg1arg1 ≤ 0

4 Location Addition

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

1 8 1_var_snapshot: arg3P + arg3P ≤ 0arg3Parg3P ≤ 0arg3 + arg3 ≤ 0arg3arg3 ≤ 0arg2P + arg2P ≤ 0arg2Parg2P ≤ 0arg2 + arg2 ≤ 0arg2arg2 ≤ 0arg1P + arg1P ≤ 0arg1Parg1P ≤ 0arg1 + arg1 ≤ 0arg1arg1 ≤ 0

5 Location Addition

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

2* 17 2: arg3P + arg3P ≤ 0arg3Parg3P ≤ 0arg3 + arg3 ≤ 0arg3arg3 ≤ 0arg2P + arg2P ≤ 0arg2Parg2P ≤ 0arg2 + arg2 ≤ 0arg2arg2 ≤ 0arg1P + arg1P ≤ 0arg1Parg1P ≤ 0arg1 + arg1 ≤ 0arg1arg1 ≤ 0

6 Location Addition

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

2 15 2_var_snapshot: arg3P + arg3P ≤ 0arg3Parg3P ≤ 0arg3 + arg3 ≤ 0arg3arg3 ≤ 0arg2P + arg2P ≤ 0arg2Parg2P ≤ 0arg2 + arg2 ≤ 0arg2arg2 ≤ 0arg1P + arg1P ≤ 0arg1Parg1P ≤ 0arg1 + arg1 ≤ 0arg1arg1 ≤ 0

7 Location Addition

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

3* 24 3: arg3P + arg3P ≤ 0arg3Parg3P ≤ 0arg3 + arg3 ≤ 0arg3arg3 ≤ 0arg2P + arg2P ≤ 0arg2Parg2P ≤ 0arg2 + arg2 ≤ 0arg2arg2 ≤ 0arg1P + arg1P ≤ 0arg1Parg1P ≤ 0arg1 + arg1 ≤ 0arg1arg1 ≤ 0

8 Location Addition

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

3 22 3_var_snapshot: arg3P + arg3P ≤ 0arg3Parg3P ≤ 0arg3 + arg3 ≤ 0arg3arg3 ≤ 0arg2P + arg2P ≤ 0arg2Parg2P ≤ 0arg2 + arg2 ≤ 0arg2arg2 ≤ 0arg1P + arg1P ≤ 0arg1Parg1P ≤ 0arg1 + arg1 ≤ 0arg1arg1 ≤ 0

9 SCC Decomposition

We consider subproblems for each of the 3 SCC(s) of the program graph.

9.1 SCC Subproblem 1/3

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

9.1.1 Transition Removal

We remove transition 5 using the following ranking functions, which are bounded by −1000003.

3: −1 − 1000⋅arg2
3_var_snapshot: −2 − 1000⋅arg2
3*: −1000⋅arg2
Hints:
22 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 1000, 0, 0, 0, 0] ]
24 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 1000, 0, 0, 0, 0] ]
5 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1000, 0, 0, 0, 0, 0, 0, 1000, 0, 0] , [0, 0, 0, 0, 0, 0, 0, 0, 1000, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

9.1.2 Transition Removal

We remove transition 22 using the following ranking functions, which are bounded by −1.

3: 0
3_var_snapshot: −1
3*: 1
Hints:
22 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] ]
24 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

9.1.3 Transition Removal

We remove transition 24 using the following ranking functions, which are bounded by −1.

3: −1
3_var_snapshot: 0
3*: 0
Hints:
24 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] ]

9.1.4 Splitting Cut-Point Transitions

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

9.1.4.1 Cut-Point Subproblem 1/1

Here we consider cut-point transition 21.

9.1.4.1.1 Splitting Cut-Point Transitions

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

9.2 SCC Subproblem 2/3

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

9.2.1 Transition Removal

We remove transition 3 using the following ranking functions, which are bounded by −1000003.

2: −1 − 500⋅arg2 − 500⋅arg3
2_var_snapshot: −2 − 500⋅arg2 − 500⋅arg3
2*: −500⋅arg2 − 500⋅arg3
Hints:
15 lexWeak[ [0, 0, 0, 500, 0, 0, 0, 500, 0, 0, 0, 0] ]
17 lexWeak[ [0, 0, 0, 500, 0, 0, 0, 500, 0, 0, 0, 0] ]
3 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 500, 0, 0, 500, 0, 500, 0, 0, 0, 0, 500, 0, 500] , [0, 0, 0, 0, 0, 0, 0, 0, 1000, 0, 500, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

9.2.2 Transition Removal

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

2: 0
2_var_snapshot: −1
2*: 1
Hints:
15 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] ]
17 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

9.2.3 Transition Removal

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

2: −1
2_var_snapshot: 0
2*: 0
Hints:
17 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] ]

9.2.4 Splitting Cut-Point Transitions

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

9.2.4.1 Cut-Point Subproblem 1/1

Here we consider cut-point transition 14.

9.2.4.1.1 Splitting Cut-Point Transitions

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

9.3 SCC Subproblem 3/3

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

9.3.1 Transition Removal

We remove transition 1 using the following ranking functions, which are bounded by −2002.

1: −1 − 2⋅arg3
1_var_snapshot: −1 − 2⋅arg3
1*: −2⋅arg3
Hints:
8 lexWeak[ [0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0] ]
10 lexWeak[ [0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0] ]
1 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 2] , [0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

9.3.2 Transition Removal

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

1: 0
1_var_snapshot: −1
1*: 1
Hints:
8 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] ]
10 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

9.3.3 Transition Removal

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

1: −1
1_var_snapshot: 0
1*: 0
Hints:
10 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] ]

9.3.4 Splitting Cut-Point Transitions

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

9.3.4.1 Cut-Point Subproblem 1/1

Here we consider cut-point transition 7.

9.3.4.1.1 Splitting Cut-Point Transitions

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

Tool configuration

T2Cert