LTS Termination Proof

by T2Cert

Input

Integer Transition System

Proof

1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:
3 12 3: x2_post + x2_post ≤ 0x2_postx2_post ≤ 0x2_0 + x2_0 ≤ 0x2_0x2_0 ≤ 0x1_post + x1_post ≤ 0x1_postx1_post ≤ 0x1_0 + x1_0 ≤ 0x1_0x1_0 ≤ 0x0_post + x0_post ≤ 0x0_postx0_post ≤ 0x0_0 + x0_0 ≤ 0x0_0x0_0 ≤ 0oldX5_post + oldX5_post ≤ 0oldX5_postoldX5_post ≤ 0oldX5_0 + oldX5_0 ≤ 0oldX5_0oldX5_0 ≤ 0oldX4_post + oldX4_post ≤ 0oldX4_postoldX4_post ≤ 0oldX4_0 + oldX4_0 ≤ 0oldX4_0oldX4_0 ≤ 0oldX3_post + oldX3_post ≤ 0oldX3_postoldX3_post ≤ 0oldX3_0 + oldX3_0 ≤ 0oldX3_0oldX3_0 ≤ 0oldX2_post + oldX2_post ≤ 0oldX2_postoldX2_post ≤ 0oldX2_0 + oldX2_0 ≤ 0oldX2_0oldX2_0 ≤ 0oldX1_post + oldX1_post ≤ 0oldX1_postoldX1_post ≤ 0oldX1_0 + oldX1_0 ≤ 0oldX1_0oldX1_0 ≤ 0oldX0_post + oldX0_post ≤ 0oldX0_postoldX0_post ≤ 0oldX0_0 + oldX0_0 ≤ 0oldX0_0oldX0_0 ≤ 0
and for every transition t, a duplicate t is considered.

2 Transition Removal

We remove transitions 0, 2, 4, 5, 6, 7, 8, 9, 10, 11 using the following ranking functions, which are bounded by −17.

6: 0
5: 0
4: 0
2: 0
3: 0
0: 0
1: 0
6: −7
5: −8
4: −9
2: −10
3: −10
3_var_snapshot: −10
3*: −10
0: −14
1: −15
Hints:
13 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, 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, 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] ]
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, 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 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, 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, 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, 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, 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, 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, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]
5 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, 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, 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, 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, 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, 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, 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, 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, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]
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, 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, 0, 0, 0, 0, 0, 0] ]
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, 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, 0, 0, 0, 0, 0, 0] ]
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, 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, 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.

3* 15 3: x2_post + x2_post ≤ 0x2_postx2_post ≤ 0x2_0 + x2_0 ≤ 0x2_0x2_0 ≤ 0x1_post + x1_post ≤ 0x1_postx1_post ≤ 0x1_0 + x1_0 ≤ 0x1_0x1_0 ≤ 0x0_post + x0_post ≤ 0x0_postx0_post ≤ 0x0_0 + x0_0 ≤ 0x0_0x0_0 ≤ 0oldX5_post + oldX5_post ≤ 0oldX5_postoldX5_post ≤ 0oldX5_0 + oldX5_0 ≤ 0oldX5_0oldX5_0 ≤ 0oldX4_post + oldX4_post ≤ 0oldX4_postoldX4_post ≤ 0oldX4_0 + oldX4_0 ≤ 0oldX4_0oldX4_0 ≤ 0oldX3_post + oldX3_post ≤ 0oldX3_postoldX3_post ≤ 0oldX3_0 + oldX3_0 ≤ 0oldX3_0oldX3_0 ≤ 0oldX2_post + oldX2_post ≤ 0oldX2_postoldX2_post ≤ 0oldX2_0 + oldX2_0 ≤ 0oldX2_0oldX2_0 ≤ 0oldX1_post + oldX1_post ≤ 0oldX1_postoldX1_post ≤ 0oldX1_0 + oldX1_0 ≤ 0oldX1_0oldX1_0 ≤ 0oldX0_post + oldX0_post ≤ 0oldX0_postoldX0_post ≤ 0oldX0_0 + oldX0_0 ≤ 0oldX0_0oldX0_0 ≤ 0

4 Location Addition

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

3 13 3_var_snapshot: x2_post + x2_post ≤ 0x2_postx2_post ≤ 0x2_0 + x2_0 ≤ 0x2_0x2_0 ≤ 0x1_post + x1_post ≤ 0x1_postx1_post ≤ 0x1_0 + x1_0 ≤ 0x1_0x1_0 ≤ 0x0_post + x0_post ≤ 0x0_postx0_post ≤ 0x0_0 + x0_0 ≤ 0x0_0x0_0 ≤ 0oldX5_post + oldX5_post ≤ 0oldX5_postoldX5_post ≤ 0oldX5_0 + oldX5_0 ≤ 0oldX5_0oldX5_0 ≤ 0oldX4_post + oldX4_post ≤ 0oldX4_postoldX4_post ≤ 0oldX4_0 + oldX4_0 ≤ 0oldX4_0oldX4_0 ≤ 0oldX3_post + oldX3_post ≤ 0oldX3_postoldX3_post ≤ 0oldX3_0 + oldX3_0 ≤ 0oldX3_0oldX3_0 ≤ 0oldX2_post + oldX2_post ≤ 0oldX2_postoldX2_post ≤ 0oldX2_0 + oldX2_0 ≤ 0oldX2_0oldX2_0 ≤ 0oldX1_post + oldX1_post ≤ 0oldX1_postoldX1_post ≤ 0oldX1_0 + oldX1_0 ≤ 0oldX1_0oldX1_0 ≤ 0oldX0_post + oldX0_post ≤ 0oldX0_postoldX0_post ≤ 0oldX0_0 + oldX0_0 ≤ 0oldX0_0oldX0_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 { 2, 3, 3_var_snapshot, 3* }.

5.1.1 Transition Removal

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

2: −1 + 3⋅x0_0 − 3⋅x1_0
3: 1 + 3⋅x0_0 − 3⋅x1_0
3_var_snapshot: 3⋅x0_0 − 3⋅x1_0
3*: 1 + 3⋅x0_0 − 3⋅x1_0
Hints:
13 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 3, 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] ]
15 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 3, 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 lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 3, 0, 0, 3, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 3, 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, 3, 0, 0, 3, 0, 0, 0, 3, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 3, 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, 3, 0, 0, 3, 0, 0, 3, 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] ]

5.1.2 Transition Removal

We remove transitions 13, 15, 1 using the following ranking functions, which are bounded by −3.

2: 0
3: −2
3_var_snapshot: −3
3*: −1
Hints:
13 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, 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] ]
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, 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, 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, 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, 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] ]

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 12.

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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