LTS Termination Proof

by T2Cert

Input

Integer Transition System

Initial Location: 4

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 ∧ − k_208_0 ≤ 0 ∧ −1 + ___patmp1_post ≤ 0 ∧ 1 − ___patmp1_post ≤ 0 ∧ ___patmp2_post − k_208_0 ≤ 0 ∧ − ___patmp2_post + k_208_0 ≤ 0 ∧ − ___patmp1_post + len_263_post ≤ 0 ∧ ___patmp1_post − len_263_post ≤ 0 ∧ − ___patmp2_post + k_243_post ≤ 0 ∧ ___patmp2_post − k_243_post ≤ 0 ∧ 1 − k_243_post ≤ 0 ∧ ___patmp1_0 − ___patmp1_post ≤ 0 ∧ − ___patmp1_0 + ___patmp1_post ≤ 0 ∧ ___patmp2_0 − ___patmp2_post ≤ 0 ∧ − ___patmp2_0 + ___patmp2_post ≤ 0 ∧ k_243_0 − k_243_post ≤ 0 ∧ − k_243_0 + k_243_post ≤ 0 ∧ len_263_0 − len_263_post ≤ 0 ∧ − len_263_0 + len_263_post ≤ 0 ∧ − k_208_0 + k_208_0 ≤ 0 ∧ k_208_0 − k_208_0 ≤ 0
4	2	0:		− len_263_post + len_263_post ≤ 0 ∧ len_263_post − len_263_post ≤ 0 ∧ − len_263_0 + len_263_0 ≤ 0 ∧ len_263_0 − len_263_0 ≤ 0 ∧ − k_243_post + k_243_post ≤ 0 ∧ k_243_post − k_243_post ≤ 0 ∧ − k_243_0 + k_243_0 ≤ 0 ∧ k_243_0 − k_243_0 ≤ 0 ∧ − k_208_0 + k_208_0 ≤ 0 ∧ k_208_0 − k_208_0 ≤ 0 ∧ − ___patmp2_post + ___patmp2_post ≤ 0 ∧ ___patmp2_post − ___patmp2_post ≤ 0 ∧ − ___patmp2_0 + ___patmp2_0 ≤ 0 ∧ ___patmp2_0 − ___patmp2_0 ≤ 0 ∧ − ___patmp1_post + ___patmp1_post ≤ 0 ∧ ___patmp1_post − ___patmp1_post ≤ 0 ∧ − ___patmp1_0 + ___patmp1_0 ≤ 0 ∧ ___patmp1_0 − ___patmp1_0 ≤ 0

Proof

1 Invariant Updates

The following invariants are asserted.

0:	TRUE
1:	−1 + ___patmp1_post ≤ 0 ∧ 1 − ___patmp1_post ≤ 0 ∧ 1 − k_243_post ≤ 0 ∧ −1 + len_263_post ≤ 0 ∧ − k_208_0 ≤ 0 ∧ −1 + ___patmp1_0 ≤ 0 ∧ 1 − ___patmp1_0 ≤ 0 ∧ 1 − k_243_0 ≤ 0 ∧ −1 + len_263_0 ≤ 0
4:	TRUE

The invariants are proved as follows.

IMPACT Invariant Proof

nodes (location) invariant:

0	(0)	TRUE
1	(1)	−1 + ___patmp1_post ≤ 0 ∧ 1 − ___patmp1_post ≤ 0 ∧ 1 − k_243_post ≤ 0 ∧ −1 + len_263_post ≤ 0 ∧ − k_208_0 ≤ 0 ∧ −1 + ___patmp1_0 ≤ 0 ∧ 1 − ___patmp1_0 ≤ 0 ∧ 1 − k_243_0 ≤ 0 ∧ −1 + len_263_0 ≤ 0
4	(4)	TRUE

initial node: 4
cover edges:
transition edges:

0 0 1

4 2 0

2 Switch to Cooperation Termination Proof

We consider the following cutpoint-transitions:

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 −8.

4:	0
0:	0
1:	0
4:	−4
0:	−5
1:	−6

Hints:

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] ]
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] ]

4 SCC Decomposition

There exist no SCC in the program graph.

Tool configuration

T2Cert

version: 1.0