Here you can ask questions and find or give answers to organizational, academic and other questions about studying computer science.

1.1k questions

1.2k answers

1.6k comments

546 users

0 votes

I tried solving it as follows:

A([d U 0] ∧ F [d U b]) 

= A(Gd ∧ F [d U b]) 

= A(Gd ∧ (FGd ∨ Fb)) 

= A ( (Gd ∧ FGd) ∨ (Gd ∧ Fb))

= A (Gd ∨ (Gd ∧ Fb))    {s.t FGd = Gd}

Why are we avoiding (Gd ∧ Fb)?

The solution provided is as follows:

1) A([d U 0] ∧ F [d U b]) 

2) A(Gd ∧ F [d U b]) 

3) A(Gd ∧ (FGd ∨ Fb)) 

4) AGd

From Step 3  to Step 4, how is
Gd ∧ (FGd ∨ Fb) = Gd  ?

in * TF "Emb. Sys. and Rob." by (640 points)

1 Answer

+1 vote
 
Best answer

Well for:

Gd ∧ (FGd ∨ Fb)

to hold, it is clear that Gd needs to hold. However Gd directly implies that FGd also holds due to:

Gd => FGd

so if Gd holds, FGd holds too and thus we don't need to care about the Fb part. That's the reason why the whole term reduces to Gd.

by (3.4k points)
selected by
Imprint | Privacy Policy
...