LTS Termination Proof

by T2Cert

Input

• Initial Location: 6
• Transitions: (pre-variables and post-variables)

  0	0	1:		0 ≤ 0 ∧ 0 ≤ 0 ∧ y_0 ≤ 0 ∧ x_0 ≤ 0 ∧ −1 + p_post ≤ 0 ∧ 1 − p_post ≤ 0 ∧ p_0 − p_post ≤ 0 ∧ − p_0 + p_post ≤ 0 ∧ − y_post + y_post ≤ 0 ∧ y_post − y_post ≤ 0 ∧ − y_0 + y_0 ≤ 0 ∧ y_0 − y_0 ≤ 0 ∧ − x_0 + x_0 ≤ 0 ∧ x_0 − x_0 ≤ 0
  0	1	2:		0 ≤ 0 ∧ 0 ≤ 0 ∧ 1 − y_0 ≤ 0 ∧ 1 − y_0 + y_post ≤ 0 ∧ −1 + y_0 − y_post ≤ 0 ∧ y_0 − y_post ≤ 0 ∧ − y_0 + y_post ≤ 0 ∧ − x_0 + x_0 ≤ 0 ∧ x_0 − x_0 ≤ 0 ∧ − p_post + p_post ≤ 0 ∧ p_post − p_post ≤ 0 ∧ − p_0 + p_0 ≤ 0 ∧ p_0 − p_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_0 + x_0 ≤ 0 ∧ x_0 − x_0 ≤ 0 ∧ − p_post + p_post ≤ 0 ∧ p_post − p_post ≤ 0 ∧ − p_0 + p_0 ≤ 0 ∧ p_0 − p_0 ≤ 0
  3	3	2:		x_0 ≤ 0 ∧ − y_post + y_post ≤ 0 ∧ y_post − y_post ≤ 0 ∧ − y_0 + y_0 ≤ 0 ∧ y_0 − y_0 ≤ 0 ∧ − x_0 + x_0 ≤ 0 ∧ x_0 − x_0 ≤ 0 ∧ − p_post + p_post ≤ 0 ∧ p_post − p_post ≤ 0 ∧ − p_0 + p_0 ≤ 0 ∧ p_0 − p_0 ≤ 0
  3	4	4:		0 ≤ 0 ∧ 0 ≤ 0 ∧ 1 − x_0 ≤ 0 ∧ −1 − p_0 + p_post ≤ 0 ∧ 1 + p_0 − p_post ≤ 0 ∧ p_0 − p_post ≤ 0 ∧ − p_0 + p_post ≤ 0 ∧ − y_post + y_post ≤ 0 ∧ y_post − y_post ≤ 0 ∧ − y_0 + y_0 ≤ 0 ∧ y_0 − y_0 ≤ 0 ∧ − x_0 + x_0 ≤ 0 ∧ x_0 − x_0 ≤ 0
  5	5	3:		0 ≤ 0 ∧ 0 ≤ 0 ∧ p_post ≤ 0 ∧ − p_post ≤ 0 ∧ p_0 − p_post ≤ 0 ∧ − p_0 + p_post ≤ 0 ∧ − y_post + y_post ≤ 0 ∧ y_post − y_post ≤ 0 ∧ − y_0 + y_0 ≤ 0 ∧ y_0 − y_0 ≤ 0 ∧ − x_0 + x_0 ≤ 0 ∧ x_0 − x_0 ≤ 0
  6	6	5:		− y_post + y_post ≤ 0 ∧ y_post − y_post ≤ 0 ∧ − y_0 + y_0 ≤ 0 ∧ y_0 − y_0 ≤ 0 ∧ − x_0 + x_0 ≤ 0 ∧ x_0 − x_0 ≤ 0 ∧ − p_post + p_post ≤ 0 ∧ p_post − p_post ≤ 0 ∧ − p_0 + p_0 ≤ 0 ∧ p_0 − p_0 ≤ 0

Proof

1 Invariant Updates

The following invariants are asserted.

The invariants are proved as follows.

IMPACT Invariant Proof

• nodes (location) invariant:

  0		(0)		p_post ≤ 0 ∧ − p_post ≤ 0 ∧ p_0 ≤ 0 ∧ − p_0 ≤ 0 ∧ x_0 ≤ 0
  1		(1)		−1 + p_post ≤ 0 ∧ 1 − p_post ≤ 0 ∧ −1 + p_0 ≤ 0 ∧ 1 − p_0 ≤ 0 ∧ x_0 ≤ 0 ∧ y_0 ≤ 0
  2		(2)		p_post ≤ 0 ∧ − p_post ≤ 0 ∧ p_0 ≤ 0 ∧ − p_0 ≤ 0 ∧ x_0 ≤ 0
  3		(3)		p_post ≤ 0 ∧ − p_post ≤ 0 ∧ p_0 ≤ 0 ∧ − p_0 ≤ 0
  4		(4)		1 − x_0 ≤ 0
  5		(5)		TRUE
  6		(6)		TRUE

• initial node: 6
• cover edges:
• transition edges:

  0	0 1
  0	1 2
  2	2 0
  3	3 2
  3	4 4
  5	5 3
  6	6 5

2 Switch to Cooperation Termination Proof

3 Transition Removal

We remove transitions 0, 3, 4, 5, 6 using the following ranking functions, which are bounded by −17.

4 Location Addition

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

2* 10 2: − y_post + y_post ≤ 0 ∧ y_post − y_post ≤ 0 ∧ − y_0 + y_0 ≤ 0 ∧ y_0 − y_0 ≤ 0 ∧ − x_0 + x_0 ≤ 0 ∧ x_0 − x_0 ≤ 0 ∧ − p_post + p_post ≤ 0 ∧ p_post − p_post ≤ 0 ∧ − p_0 + p_0 ≤ 0 ∧ p_0 − p_0 ≤ 0

5 Location Addition

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

2 8 2_var_snapshot: − y_post + y_post ≤ 0 ∧ y_post − y_post ≤ 0 ∧ − y_0 + y_0 ≤ 0 ∧ y_0 − y_0 ≤ 0 ∧ − x_0 + x_0 ≤ 0 ∧ x_0 − x_0 ≤ 0 ∧ − p_post + p_post ≤ 0 ∧ p_post − p_post ≤ 0 ∧ − p_0 + p_0 ≤ 0 ∧ p_0 − p_0 ≤ 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 { 0, 2, 2_var_snapshot, 2* }.

6.1.1 Transition Removal

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

6.1.2 Transition Removal

We remove transitions 8, 10, 2 using the following ranking functions, which are bounded by −3.

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

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

• version: 1.0