LTS Termination Proof

by T2Cert

Input

Integer Transition System

Initial Location: 2

Transitions: (pre-variables and post-variables)

0	0	1:	0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ − arg1P ≤ 0 ∧ − arg2 ≤ 0 ∧ − arg2P ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ 2 − arg3P ≤ 0 ∧ −2 + arg3P ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − arg3P + arg3 ≤ 0 ∧ arg3P − arg3 ≤ 0 ∧ − x7 + x7 ≤ 0 ∧ x7 − x7 ≤ 0 ∧ − x12 + x12 ≤ 0 ∧ x12 − x12 ≤ 0
1	1	1:	0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ − arg3 ≤ 0 ∧ 42 − x7 ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ 1 − arg2 ≤ 0 ∧ −1 − arg1P + arg1 ≤ 0 ∧ 1 + arg1P − arg1 ≤ 0 ∧ 1 − arg3P + arg3 ≤ 0 ∧ −1 + arg3P − arg3 ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg3P + arg3 ≤ 0 ∧ arg3P − arg3 ≤ 0 ∧ − x12 + x12 ≤ 0 ∧ x12 − x12 ≤ 0 ∧ − arg2P + arg2P ≤ 0 ∧ arg2P − arg2P ≤ 0 ∧ − arg2 + arg2 ≤ 0 ∧ arg2 − arg2 ≤ 0
1	2	1:	0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ − arg3 ≤ 0 ∧ 1 − arg2 ≤ 0 ∧ − x12 ≤ 0 ∧ − arg1P ≤ 0 ∧ −41 + x12 ≤ 0 ∧ −1 − arg2P + arg2 ≤ 0 ∧ 1 + arg2P − arg2 ≤ 0 ∧ 2 − arg3P + arg3 ≤ 0 ∧ −2 + arg3P − arg3 ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − arg3P + arg3 ≤ 0 ∧ arg3P − arg3 ≤ 0 ∧ − x7 + x7 ≤ 0 ∧ x7 − x7 ≤ 0
2	3	0:	0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − arg3P + arg3 ≤ 0 ∧ arg3P − arg3 ≤ 0 ∧ − x7 + x7 ≤ 0 ∧ x7 − x7 ≤ 0 ∧ − x12 + x12 ≤ 0 ∧ x12 − x12 ≤ 0

Proof

1 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:

− x7 + x7 ≤ 0 ∧ x7 − x7 ≤ 0 ∧ − x12 + x12 ≤ 0 ∧ x12 − x12 ≤ 0 ∧ − arg3P + arg3P ≤ 0 ∧ arg3P − arg3P ≤ 0 ∧ − arg3 + arg3 ≤ 0 ∧ arg3 − arg3 ≤ 0 ∧ − arg2P + arg2P ≤ 0 ∧ arg2P − arg2P ≤ 0 ∧ − arg2 + arg2 ≤ 0 ∧ arg2 − arg2 ≤ 0 ∧ − arg1P + arg1P ≤ 0 ∧ arg1P − arg1P ≤ 0 ∧ − arg1 + arg1 ≤ 0 ∧ arg1 − arg1 ≤ 0

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

2 Transition Removal

We remove transitions 0, 3 using the following ranking functions, which are bounded by −11.

2:	0
0:	0
1:	0
2:	−4
0:	−5
1:	−6
1_var_snapshot:	−6
1*:	−6

Hints:

5	lexWeak[ [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] ]
2	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	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] ]
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, 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* 7 1: − x7 + x7 ≤ 0 ∧ x7 − x7 ≤ 0 ∧ − x12 + x12 ≤ 0 ∧ x12 − x12 ≤ 0 ∧ − arg3P + arg3P ≤ 0 ∧ arg3P − arg3P ≤ 0 ∧ − arg3 + arg3 ≤ 0 ∧ arg3 − arg3 ≤ 0 ∧ − arg2P + arg2P ≤ 0 ∧ arg2P − arg2P ≤ 0 ∧ − arg2 + arg2 ≤ 0 ∧ arg2 − arg2 ≤ 0 ∧ − arg1P + arg1P ≤ 0 ∧ arg1P − arg1P ≤ 0 ∧ − arg1 + arg1 ≤ 0 ∧ arg1 − arg1 ≤ 0

4 Location Addition

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

1 5 1_var_snapshot: − x7 + x7 ≤ 0 ∧ x7 − x7 ≤ 0 ∧ − x12 + x12 ≤ 0 ∧ x12 − x12 ≤ 0 ∧ − arg3P + arg3P ≤ 0 ∧ arg3P − arg3P ≤ 0 ∧ − arg3 + arg3 ≤ 0 ∧ arg3 − arg3 ≤ 0 ∧ − arg2P + arg2P ≤ 0 ∧ arg2P − arg2P ≤ 0 ∧ − arg2 + arg2 ≤ 0 ∧ arg2 − arg2 ≤ 0 ∧ − arg1P + arg1P ≤ 0 ∧ arg1P − arg1P ≤ 0 ∧ − arg1 + arg1 ≤ 0 ∧ arg1 − arg1 ≤ 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 { 1, 1_var_snapshot, 1* }.

5.1.1 Transition Removal

We remove transition 2 using the following ranking functions, which are bounded by 0.

1:	arg2
1_var_snapshot:	arg2
1*:	arg2

Hints:

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

5.1.2 Transition Removal

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

1:	1 + 3⋅arg1
1_var_snapshot:	3⋅arg1
1*:	2 + 3⋅arg1

Hints:

5	lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0] ]
7	lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0] ]
1	lexStrict[ [0, 0, 0, 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, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

5.1.3 Transition Removal

We remove transition 5 using the following ranking functions, which are bounded by −1.

1:	0
1_var_snapshot:	−1
1*:	1

Hints:

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] ]
7	lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

5.1.4 Transition Removal

We remove transition 7 using the following ranking functions, which are bounded by −1.

1:	−1
1_var_snapshot:	0
1*:	0

Hints:

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] ]

5.1.5 Splitting Cut-Point Transitions

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

5.1.5.1 Cut-Point Subproblem 1/1

Here we consider cut-point transition 4.

5.1.5.1.1 Splitting Cut-Point Transitions

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

Tool configuration

T2Cert

version: 1.0