LTS Termination Proof

by T2Cert

Input

Integer Transition System

Initial Location: 3

Transitions: (pre-variables and post-variables)

0	0	1:	0 ≤ 0 ∧ 0 ≤ 0 ∧ 1 − arg2 ≤ 0 ∧ − x2 ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ − x7 + x7 ≤ 0 ∧ x7 − x7 ≤ 0 ∧ − x6 + x6 ≤ 0 ∧ x6 − x6 ≤ 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
1	1	2:	0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 1 − arg2 ≤ 0 ∧ − x6 ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ −99 + x6 − 100⋅x7 ≤ 0 ∧ − x6 + 100⋅x7 ≤ 0 ∧ − arg1P + x6 − 100⋅x7 ≤ 0 ∧ arg1P − x6 + 100⋅x7 ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − x2 + x2 ≤ 0 ∧ x2 − x2 ≤ 0
2	2	2:	0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 1 − arg1 ≤ 0 ∧ −1 − arg1P + arg1 ≤ 0 ∧ 1 + arg1P − arg1 ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − x7 + x7 ≤ 0 ∧ x7 − x7 ≤ 0 ∧ − x6 + x6 ≤ 0 ∧ x6 − x6 ≤ 0 ∧ − x2 + x2 ≤ 0 ∧ x2 − x2 ≤ 0
3	3	0:	0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ 0 ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0 ∧ − arg2P + arg2 ≤ 0 ∧ arg2P − arg2 ≤ 0 ∧ − x7 + x7 ≤ 0 ∧ x7 − x7 ≤ 0 ∧ − x6 + x6 ≤ 0 ∧ x6 − x6 ≤ 0 ∧ − x2 + x2 ≤ 0 ∧ x2 − x2 ≤ 0

Proof

1 Invariant Updates

The following invariants are asserted.

0:	TRUE
1:	1 − arg1 ≤ 0 ∧ 1 − arg2 ≤ 0 ∧ − x2 ≤ 0
2:	− x2 ≤ 0 ∧ − x6 ≤ 0
3:	TRUE

The invariants are proved as follows.

IMPACT Invariant Proof

nodes (location) invariant:

0	(0)	TRUE
1	(1)	1 − arg1 ≤ 0 ∧ 1 − arg2 ≤ 0 ∧ − x2 ≤ 0
2	(2)	− x2 ≤ 0 ∧ − x6 ≤ 0
3	(3)	TRUE

initial node: 3
cover edges:
transition edges:

0 0 1

1 1 2

2 2 2

3 3 0

2 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:

− x7 + x7 ≤ 0 ∧ x7 − x7 ≤ 0 ∧ − x6 + x6 ≤ 0 ∧ x6 − x6 ≤ 0 ∧ − x2 + x2 ≤ 0 ∧ x2 − x2 ≤ 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, 1, 3 using the following ranking functions, which are bounded by −13.

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

4 Location Addition

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

2* 7 2: − x7 + x7 ≤ 0 ∧ x7 − x7 ≤ 0 ∧ − x6 + x6 ≤ 0 ∧ x6 − x6 ≤ 0 ∧ − x2 + x2 ≤ 0 ∧ x2 − x2 ≤ 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.

2 5 2_var_snapshot: − x7 + x7 ≤ 0 ∧ x7 − x7 ≤ 0 ∧ − x6 + x6 ≤ 0 ∧ x6 − x6 ≤ 0 ∧ − x2 + x2 ≤ 0 ∧ x2 − x2 ≤ 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 { 2, 2_var_snapshot, 2* }.

6.1.1 Transition Removal

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

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

6.1.2 Transition Removal

We remove transitions 5, 7 using the following ranking functions, which are bounded by −1.

2:	0
2_var_snapshot:	−1
2*:	1

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 4.

6.1.3.1.1 Splitting Cut-Point Transitions

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

Tool configuration

T2Cert

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