LTS Termination Proof

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Input

Integer Transition System

Initial Location: 4

Transitions: (pre-variables and post-variables)

0	0	1:	20 − j_0 ≤ 0 ∧ − j_post + j_post ≤ 0 ∧ j_post − j_post ≤ 0 ∧ − j_0 + j_0 ≤ 0 ∧ j_0 − j_0 ≤ 0 ∧ − i_post + i_post ≤ 0 ∧ i_post − i_post ≤ 0 ∧ − i_0 + i_0 ≤ 0 ∧ i_0 − i_0 ≤ 0
0	1	2:	−19 + j_0 ≤ 0 ∧ − j_post + j_post ≤ 0 ∧ j_post − j_post ≤ 0 ∧ − j_0 + j_0 ≤ 0 ∧ j_0 − j_0 ≤ 0 ∧ − i_post + i_post ≤ 0 ∧ i_post − i_post ≤ 0 ∧ − i_0 + i_0 ≤ 0 ∧ i_0 − i_0 ≤ 0
2	2	0:	0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ −2 − j_0 + j_post ≤ 0 ∧ 2 + j_0 − j_post ≤ 0 ∧ −2 + i_post − j_post ≤ 0 ∧ 2 − i_post + j_post ≤ 0 ∧ i_0 − i_post ≤ 0 ∧ − i_0 + i_post ≤ 0 ∧ j_0 − j_post ≤ 0 ∧ − j_0 + j_post ≤ 0
3	3	2:	0 ≤ 0 ∧ 0 ≤ 0 ∧ j_post ≤ 0 ∧ − j_post ≤ 0 ∧ j_0 − j_post ≤ 0 ∧ − j_0 + j_post ≤ 0 ∧ − i_post + i_post ≤ 0 ∧ i_post − i_post ≤ 0 ∧ − i_0 + i_0 ≤ 0 ∧ i_0 − i_0 ≤ 0
4	4	3:	− j_post + j_post ≤ 0 ∧ j_post − j_post ≤ 0 ∧ − j_0 + j_0 ≤ 0 ∧ j_0 − j_0 ≤ 0 ∧ − i_post + i_post ≤ 0 ∧ i_post − i_post ≤ 0 ∧ − i_0 + i_0 ≤ 0 ∧ i_0 − i_0 ≤ 0

Proof

1 Invariant Updates

The following invariants are asserted.

0:	TRUE
1:	20 − j_0 ≤ 0
2:	TRUE
3:	TRUE
4:	TRUE

The invariants are proved as follows.

IMPACT Invariant Proof

nodes (location) invariant:

0	(0)	TRUE
1	(1)	20 − j_0 ≤ 0
2	(2)	TRUE
3	(3)	TRUE
4	(4)	TRUE

initial node: 4
cover edges:
transition edges:

0 0 1

0 1 2

2 2 0

3 3 2

4 4 3

2 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:

− j_post + j_post ≤ 0 ∧ j_post − j_post ≤ 0 ∧ − j_0 + j_0 ≤ 0 ∧ j_0 − j_0 ≤ 0 ∧ − i_post + i_post ≤ 0 ∧ i_post − i_post ≤ 0 ∧ − i_0 + i_0 ≤ 0 ∧ i_0 − i_0 ≤ 0

and for every transition t, a duplicate t is considered.

3 Transition Removal

We remove transitions 0, 3, 4 using the following ranking functions, which are bounded by −13.

4:	0
3:	0
0:	0
2:	0
1:	0
4:	−5
3:	−6
0:	−7
2:	−7
2_var_snapshot:	−7
2*:	−7
1:	−8

Hints:

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

4 Location Addition

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

2* 8 2: − j_post + j_post ≤ 0 ∧ j_post − j_post ≤ 0 ∧ − j_0 + j_0 ≤ 0 ∧ j_0 − j_0 ≤ 0 ∧ − i_post + i_post ≤ 0 ∧ i_post − i_post ≤ 0 ∧ − i_0 + i_0 ≤ 0 ∧ i_0 − i_0 ≤ 0

5 Location Addition

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

2 6 2_var_snapshot: − j_post + j_post ≤ 0 ∧ j_post − j_post ≤ 0 ∧ − j_0 + j_0 ≤ 0 ∧ j_0 − j_0 ≤ 0 ∧ − i_post + i_post ≤ 0 ∧ i_post − i_post ≤ 0 ∧ − i_0 + i_0 ≤ 0 ∧ i_0 − i_0 ≤ 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 { 0, 2, 2_var_snapshot, 2* }.

6.1.1 Transition Removal

We remove transition 1 using the following ranking functions, which are bounded by −54.

0:	4 − 3⋅j_0
2:	2 − 3⋅j_0
2_var_snapshot:	−3⋅j_0
2*:	2 − 3⋅j_0

Hints:

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

6.1.2 Transition Removal

We remove transitions 6, 8, 2 using the following ranking functions, which are bounded by −3.

0:	−3
2:	−1
2_var_snapshot:	−2
2*:	0

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] ]
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] ]

6.1.3 Splitting Cut-Point Transitions

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

6.1.3.1 Cut-Point Subproblem 1/1

Here we consider cut-point transition 5.

6.1.3.1.1 Splitting Cut-Point Transitions

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

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