LTS Termination Proof

by T2Cert

Input

Integer Transition System

Transitions: (pre-variables and post-variables)

0	0	1:	0 ≤ 0 ∧ 0 ≤ 0 ∧ x_0 − y_0 ≤ 0 ∧ − x_0 + y_0 ≤ 0 ∧ −1 − x_0 + x_post ≤ 0 ∧ 1 + x_0 − x_post ≤ 0 ∧ x_0 − x_post ≤ 0 ∧ − x_0 + x_post ≤ 0 ∧ − y_0 + y_0 ≤ 0 ∧ y_0 − y_0 ≤ 0
1	1	0:	− y_0 + y_0 ≤ 0 ∧ y_0 − y_0 ≤ 0 ∧ − x_post + x_post ≤ 0 ∧ x_post − x_post ≤ 0 ∧ − x_0 + x_0 ≤ 0 ∧ x_0 − x_0 ≤ 0
2	2	0:	− y_0 + y_0 ≤ 0 ∧ y_0 − y_0 ≤ 0 ∧ − x_post + x_post ≤ 0 ∧ x_post − x_post ≤ 0 ∧ − x_0 + x_0 ≤ 0 ∧ x_0 − x_0 ≤ 0
3	3	2:	− y_0 + y_0 ≤ 0 ∧ y_0 − y_0 ≤ 0 ∧ − x_post + x_post ≤ 0 ∧ x_post − x_post ≤ 0 ∧ − x_0 + x_0 ≤ 0 ∧ x_0 − x_0 ≤ 0

We consider the following cutpoint-transitions:

− y_0 + y_0 ≤ 0 ∧ y_0 − y_0 ≤ 0 ∧ − x_post + x_post ≤ 0 ∧ x_post − x_post ≤ 0 ∧ − x_0 + x_0 ≤ 0 ∧ x_0 − x_0 ≤ 0

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

We remove transitions 2, 3 using the following ranking functions, which are bounded by −11.

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

0* 7 0: − y_0 + y_0 ≤ 0 ∧ y_0 − y_0 ≤ 0 ∧ − x_post + x_post ≤ 0 ∧ x_post − x_post ≤ 0 ∧ − x_0 + x_0 ≤ 0 ∧ x_0 − x_0 ≤ 0

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

0 5 0_var_snapshot: − y_0 + y_0 ≤ 0 ∧ y_0 − y_0 ≤ 0 ∧ − x_post + x_post ≤ 0 ∧ x_post − x_post ≤ 0 ∧ − x_0 + x_0 ≤ 0 ∧ x_0 − x_0 ≤ 0

We consider subproblems for each of the 1 SCC(s) of the program graph.

Here we consider the SCC { 0, 1, 0_var_snapshot, 0* }.

We remove transition 0 using the following ranking functions, which are bounded by −1.

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

We remove transition 7 using the following ranking functions, which are bounded by 0.

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

Here we consider cut-point transition 4.

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

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