0 votes
This step is unclear to me:
AGa & AG(Fa & b)
= AG(Fa & b)

Shouldn't it be (?):
AGa & AG(Fa & b)
= AG(a & b)

As AGa should hold, AGFa is trivial, right?
in * TF "Emb. Sys. and Rob." by (130 points)

1 Answer

+1 vote

You are right. AGa & AG(Fa & b) = AG(a & Fa & b) = AG(a & b); an update of the example solution has been made.

by (91.8k points)

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