LTS Termination Proof

by T2Cert

Input

Integer Transition System

Transitions: (pre-variables and post-variables)

0	0	1:	0 ≤ 0 ∧ 0 ≤ 0 ∧ − arg1P ≤ 0 ∧ arg1P ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0
1	1	1:	0 ≤ 0 ∧ 0 ≤ 0 ∧ −11 + arg1 ≤ 0 ∧ −19 + arg1 ≤ 0 ∧ 1 − arg1P + arg1 ≤ 0 ∧ −1 + arg1P − arg1 ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0
1	2	1:	0 ≤ 0 ∧ 0 ≤ 0 ∧ 12 − arg1 ≤ 0 ∧ −19 + arg1 ≤ 0 ∧ 1 − arg1P + arg1 ≤ 0 ∧ −1 + arg1P − arg1 ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0
2	3	0:	0 ≤ 0 ∧ 0 ≤ 0 ∧ − arg1P + arg1 ≤ 0 ∧ arg1P − arg1 ≤ 0

We consider the following cutpoint-transitions:

− arg1P + arg1P ≤ 0 ∧ arg1P − arg1P ≤ 0 ∧ − arg1 + arg1 ≤ 0 ∧ arg1 − arg1 ≤ 0

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

We remove transitions 0, 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.

1* 7 1: − arg1P + arg1P ≤ 0 ∧ arg1P − arg1P ≤ 0 ∧ − arg1 + arg1 ≤ 0 ∧ arg1 − arg1 ≤ 0

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

1 5 1_var_snapshot: − arg1P + arg1P ≤ 0 ∧ arg1P − arg1P ≤ 0 ∧ − arg1 + arg1 ≤ 0 ∧ arg1 − arg1 ≤ 0

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

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

We remove transitions 1, 2 using the following ranking functions, which are bounded by −58.

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

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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