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how do you put p=0 and q=0

and why the other part !p&!q left without values?

in * TF "Emb. Sys. and Rob." by (370 points)

1 Answer

+1 vote

This is due to the following laws that you can find on page 69 on the chapter on Kripke structures that was given there as a tipp for computing the predecessors and successors symbolically: 

φ ∧ ⇔ φ 1∧ x   and  φ ∧ ¬ ⇔ φ 0∧ x

by (143k points)
This law is only valid when we are computing existential predecessor and successor? Because in the slides it is given for existential
OR
Is it valid everywhere?

Secondly in the above mentioned question we have already quantified over the primed variables, why would we assign values now?
The mentioned laws are valid and can be used wherever. If phi is expanded to Shannon decomposition, i.e., (x & phi1 | !x & phi0) then phi&x becomes (x & x & phi1 | x & !x & phi0) = x & x & phi1 | 0 = x & x & phi1 = x & phi1. About the second question, yes, the quantification is already eliminated, and therefore one may stop here unless we want to compute a nice DNF.
Instead can i do this?

= !((p↔q) ∧ (!p∨q)) ∧ !(p∨q)
=!((!p∨q) ∧ (p∨!q) ∧ (!p∨q)) & !(p∨q)
=!((!p∨q) ∧ (p∨!q)) ∧ !(p∨q)
=!((p∧q) ∨ (!p∧!q)) ∧ !(p∨q)
=!(p∧q) ∧ !(!p∧!q) ∧ !(p∨q)
=!(p∧q) ∧ (p∨q) ∧ !(p∨q)

Since (p∨q) ∧ !(p∨q) = 0 the whole answer becomes 0.

=0.
Thanks for the clarification.
Sure that is also fine.
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