LTS Termination Proof

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Input

Integer Transition System

Proof

1 Invariant Updates

The following invariants are asserted.

1: arg1P ≤ 0arg1 ≤ 0
2: TRUE
3: TRUE
4: 1 − arg1P ≤ 0arg3P ≤ 01 − arg1 ≤ 0arg3 ≤ 0
5: TRUE

The invariants are proved as follows.

IMPACT Invariant Proof

2 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
3 8 3: arg3P + arg3P ≤ 0arg3Parg3P ≤ 0arg3 + arg3 ≤ 0arg3arg3 ≤ 0arg2P + arg2P ≤ 0arg2Parg2P ≤ 0arg2 + arg2 ≤ 0arg2arg2 ≤ 0arg1P + arg1P ≤ 0arg1Parg1P ≤ 0arg1 + arg1 ≤ 0arg1arg1 ≤ 0
4 15 4: 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.

3 Transition Removal

We remove transitions 1, 2, 4, 7 using the following ranking functions, which are bounded by −17.

5: 0
2: 0
1: 0
4: 0
3: 0
5: −6
2: −7
1: −8
4: −9
4_var_snapshot: −9
4*: −9
3: −12
3_var_snapshot: −12
3*: −12
Hints:
9 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]
16 lexWeak[ [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] ]
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] ]
6 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] ]
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] ]
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] ]
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] ]
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] ]

4 Location Addition

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

3* 11 3: 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.

3 9 3_var_snapshot: 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.

4* 18 4: 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.

4 16 4_var_snapshot: arg3P + arg3P ≤ 0arg3Parg3P ≤ 0arg3 + arg3 ≤ 0arg3arg3 ≤ 0arg2P + arg2P ≤ 0arg2Parg2P ≤ 0arg2 + arg2 ≤ 0arg2arg2 ≤ 0arg1P + arg1P ≤ 0arg1Parg1P ≤ 0arg1 + arg1 ≤ 0arg1arg1 ≤ 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 { 4, 4_var_snapshot, 4* }.

8.1.1 Transition Removal

We remove transition 6 using the following ranking functions, which are bounded by 4.

4: 1 + 5⋅arg2 + 3⋅arg3
4_var_snapshot: 5⋅arg2 + 3⋅arg3
4*: 2 + 5⋅arg2 + 3⋅arg3
Hints:
16 lexWeak[ [0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 5, 0, 0, 0, 0, 0] ]
18 lexWeak[ [0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 5, 0, 0, 0, 0, 0] ]
5 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 5, 0] ]
6 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 5, 0, 0, 5, 0, 3, 0] , [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

8.1.2 Transition Removal

We remove transitions 16, 18 using the following ranking functions, which are bounded by 0.

4: arg1 + arg3
4_var_snapshot: −1 + arg1 + arg3
4*: 1 + arg1 + arg3
Hints:
16 lexStrict[ [0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0] , [0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]
18 lexStrict[ [0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0] , [0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]
5 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0] ]

8.1.3 Transition Removal

We remove transition 5 using the following ranking functions, which are bounded by 0.

4: 0
4_var_snapshot: arg1P
4*: 0
Hints:
5 lexStrict[ [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, 0, 0, 0] ]

8.1.4 Splitting Cut-Point Transitions

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

8.1.4.1 Cut-Point Subproblem 1/1

Here we consider cut-point transition 15.

8.1.4.1.1 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 transition 3 using the following ranking functions, which are bounded by 2.

3: 1 + 3⋅arg1
3_var_snapshot: 3⋅arg1
3*: 2 + 3⋅arg1
Hints:
9 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0] ]
11 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0] ]
3 lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 3, 3, 0, 0, 0, 0, 0] , [0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0] ]

8.2.2 Transition Removal

We remove transition 9 using the following ranking functions, which are bounded by −1.

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

8.2.3 Transition Removal

We remove transition 11 using the following ranking functions, which are bounded by −1.

3: −1
3_var_snapshot: 0
3*: 0
Hints:
11 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] ]

8.2.4 Splitting Cut-Point Transitions

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

8.2.4.1 Cut-Point Subproblem 1/1

Here we consider cut-point transition 8.

8.2.4.1.1 Splitting Cut-Point Transitions

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

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