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 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ − arg3P ≤ 0 ∧ − arg2 ≤ 0 ∧ − arg4P ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ 1 − arg2P ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − arg3P + arg3 ≤ 0 ∧ arg3P − arg3 ≤ 0 ∧ − arg4P + arg4 ≤ 0 ∧ arg4P − arg4 ≤ 0
1	1	1:	0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ − arg4 ≤ 0 ∧ 1 − arg3 + arg4 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ −1 + arg2P − arg2 ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ 1 − arg2 ≤ 0 ∧ 1 − arg1P ≤ 0 ∧ 1 − arg2P ≤ 0 ∧ 2 − arg1 + arg3 ≤ 0 ∧ 2 − arg2 + arg4 ≤ 0 ∧ 1 − arg4P + arg4 ≤ 0 ∧ −1 + arg4P − arg4 ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − arg4P + arg4 ≤ 0 ∧ arg4P − arg4 ≤ 0 ∧ − arg3P + arg3P ≤ 0 ∧ arg3P − arg3P ≤ 0 ∧ − arg3 + arg3 ≤ 0 ∧ arg3 − arg3 ≤ 0
2	2	0:	0 ≤ 0 ∧ 0 ≤ 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 ∧ − arg4P + arg4 ≤ 0 ∧ arg4P − arg4 ≤ 0

Proof

1 Invariant Updates

The following invariants are asserted.

0:	TRUE
1:	1 − arg1P ≤ 0 ∧ 1 − arg2P ≤ 0 ∧ − arg3P ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ 1 − arg2 ≤ 0 ∧ − arg3 ≤ 0
2:	TRUE

The invariants are proved as follows.

IMPACT Invariant Proof

nodes (location) invariant:

0	(0)	TRUE
1	(1)	1 − arg1P ≤ 0 ∧ 1 − arg2P ≤ 0 ∧ − arg3P ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ 1 − arg2 ≤ 0 ∧ − arg3 ≤ 0
2	(2)	TRUE

initial node: 2
cover edges:
transition edges:

0 0 1

1 1 1

2 2 0

2 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:

− arg4P + arg4P ≤ 0 ∧ arg4P − arg4P ≤ 0 ∧ − arg4 + arg4 ≤ 0 ∧ arg4 − arg4 ≤ 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.

3 Transition Removal

We remove transitions 0, 2 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:

4	lexWeak[ [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	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] ]
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] ]

4 Location Addition

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

1* 6 1: − arg4P + arg4P ≤ 0 ∧ arg4P − arg4P ≤ 0 ∧ − arg4 + arg4 ≤ 0 ∧ arg4 − arg4 ≤ 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 Location Addition

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

1 4 1_var_snapshot: − arg4P + arg4P ≤ 0 ∧ arg4P − arg4P ≤ 0 ∧ − arg4 + arg4 ≤ 0 ∧ arg4 − arg4 ≤ 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

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 { 1, 1_var_snapshot, 1* }.

6.1.1 Transition Removal

We remove transition 1 using the following ranking functions, which are bounded by 3.

1:	1 + 4⋅arg3 − 4⋅arg4
1_var_snapshot:	4⋅arg3 − 4⋅arg4
1*:	2 + 4⋅arg3 − 4⋅arg4

Hints:

4	lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]
6	lexWeak[ [0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 4, 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, 4, 0, 0, 0, 0, 0, 0, 4, 0, 0, 4, 0] , [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ]

6.1.2 Transition Removal

We remove transition 4 using the following ranking functions, which are bounded by −1.

1:	0
1_var_snapshot:	− arg2P
1*:	0

Hints:

4	lexStrict[ [0, 1, 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, 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] ]

6.1.3 Transition Removal

We remove transition 6 using the following ranking functions, which are bounded by 0.

1:	0
1_var_snapshot:	0
1*:	arg2P

Hints:

6	lexStrict[ [0, 1, 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] ]

6.1.4 Splitting Cut-Point Transitions

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

6.1.4.1 Cut-Point Subproblem 1/1

Here we consider cut-point transition 3.

6.1.4.1.1 Splitting Cut-Point Transitions

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

Tool configuration

T2Cert

version: 1.0