LTS Termination Proof

by T2Cert

Input

Integer Transition System

Initial Location: 4

Transitions: (pre-variables and post-variables)

0	0	1:	1 + ___const_10_0 − x_0 ≤ 0 ∧ − y_post + y_post ≤ 0 ∧ y_post − y_post ≤ 0 ∧ − y_0 + y_0 ≤ 0 ∧ y_0 − y_0 ≤ 0 ∧ − x_post + x_post ≤ 0 ∧ x_post − x_post ≤ 0 ∧ − x_0 + x_0 ≤ 0 ∧ x_0 − x_0 ≤ 0 ∧ − ___const_10_0 + ___const_10_0 ≤ 0 ∧ ___const_10_0 − ___const_10_0 ≤ 0
0	1	2:	0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ − ___const_10_0 + x_0 ≤ 0 ∧ − ___const_10_0 + x_0 + y_post ≤ 0 ∧ ___const_10_0 − x_0 − y_post ≤ 0 ∧ −1 − x_0 + x_post ≤ 0 ∧ 1 + x_0 − x_post ≤ 0 ∧ x_0 − x_post ≤ 0 ∧ − x_0 + x_post ≤ 0 ∧ y_0 − y_post ≤ 0 ∧ − y_0 + y_post ≤ 0 ∧ − ___const_10_0 + ___const_10_0 ≤ 0 ∧ ___const_10_0 − ___const_10_0 ≤ 0
2	2	0:	− y_post + y_post ≤ 0 ∧ y_post − y_post ≤ 0 ∧ − y_0 + y_0 ≤ 0 ∧ y_0 − y_0 ≤ 0 ∧ − x_post + x_post ≤ 0 ∧ x_post − x_post ≤ 0 ∧ − x_0 + x_0 ≤ 0 ∧ x_0 − x_0 ≤ 0 ∧ − ___const_10_0 + ___const_10_0 ≤ 0 ∧ ___const_10_0 − ___const_10_0 ≤ 0
3	3	2:	0 ≤ 0 ∧ 0 ≤ 0 ∧ −1 + x_post ≤ 0 ∧ 1 − x_post ≤ 0 ∧ x_0 − x_post ≤ 0 ∧ − x_0 + x_post ≤ 0 ∧ − y_post + y_post ≤ 0 ∧ y_post − y_post ≤ 0 ∧ − y_0 + y_0 ≤ 0 ∧ y_0 − y_0 ≤ 0 ∧ − ___const_10_0 + ___const_10_0 ≤ 0 ∧ ___const_10_0 − ___const_10_0 ≤ 0
4	4	3:	− y_post + y_post ≤ 0 ∧ y_post − y_post ≤ 0 ∧ − y_0 + y_0 ≤ 0 ∧ y_0 − y_0 ≤ 0 ∧ − x_post + x_post ≤ 0 ∧ x_post − x_post ≤ 0 ∧ − x_0 + x_0 ≤ 0 ∧ x_0 − x_0 ≤ 0 ∧ − ___const_10_0 + ___const_10_0 ≤ 0 ∧ ___const_10_0 − ___const_10_0 ≤ 0

Proof

1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:

− y_post + y_post ≤ 0 ∧ y_post − y_post ≤ 0 ∧ − y_0 + y_0 ≤ 0 ∧ y_0 − y_0 ≤ 0 ∧ − x_post + x_post ≤ 0 ∧ x_post − x_post ≤ 0 ∧ − x_0 + x_0 ≤ 0 ∧ x_0 − x_0 ≤ 0 ∧ − ___const_10_0 + ___const_10_0 ≤ 0 ∧ ___const_10_0 − ___const_10_0 ≤ 0

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

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

3 Location Addition

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

2* 8 2: − y_post + y_post ≤ 0 ∧ y_post − y_post ≤ 0 ∧ − y_0 + y_0 ≤ 0 ∧ y_0 − y_0 ≤ 0 ∧ − x_post + x_post ≤ 0 ∧ x_post − x_post ≤ 0 ∧ − x_0 + x_0 ≤ 0 ∧ x_0 − x_0 ≤ 0 ∧ − ___const_10_0 + ___const_10_0 ≤ 0 ∧ ___const_10_0 − ___const_10_0 ≤ 0

4 Location Addition

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

2 6 2_var_snapshot: − y_post + y_post ≤ 0 ∧ y_post − y_post ≤ 0 ∧ − y_0 + y_0 ≤ 0 ∧ y_0 − y_0 ≤ 0 ∧ − x_post + x_post ≤ 0 ∧ x_post − x_post ≤ 0 ∧ − x_0 + x_0 ≤ 0 ∧ x_0 − x_0 ≤ 0 ∧ − ___const_10_0 + ___const_10_0 ≤ 0 ∧ ___const_10_0 − ___const_10_0 ≤ 0

5 SCC Decomposition

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

5.1 SCC Subproblem 1/1

Here we consider the SCC { 0, 2, 2_var_snapshot, 2* }.

5.1.1 Transition Removal

We remove transition 1 using the following ranking functions, which are bounded by −2.

0:	−1 + 4⋅___const_10_0 − 4⋅x_0
2:	1 + 4⋅___const_10_0 − 4⋅x_0
2_var_snapshot:	4⋅___const_10_0 − 4⋅x_0
2*:	2 + 4⋅___const_10_0 − 4⋅x_0

Hints:

6	lexWeak[ [0, 0, 0, 0, 0, 0, 0, 4, 4, 0] ]
8	lexWeak[ [0, 0, 0, 0, 0, 0, 0, 4, 4, 0] ]
1	lexStrict[ [0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 4, 0, 0, 4, 0] , [0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]
2	lexWeak[ [0, 0, 0, 0, 0, 0, 0, 4, 4, 0] ]

5.1.2 Transition Removal

We remove transitions 6, 8, 2 using the following ranking functions, which are bounded by −1.

0:	−1
2:	1
2_var_snapshot:	0
2*:	2

Hints:

6	lexStrict[ [0, 0, 0, 0, 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, 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] ]

5.1.3 Splitting Cut-Point Transitions

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

5.1.3.1 Cut-Point Subproblem 1/1

Here we consider cut-point transition 5.

5.1.3.1.1 Splitting Cut-Point Transitions

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

Tool configuration

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