Assignment 1 - A

1. Assignment 1 - A Let M = Monitoring T = Test state O = Operating Substituting the exclusive OR expression into the 2nd expression (as shown in the text book), we start with : (M Λ ~T V T Λ ~M) Λ ~T => O [ (M Λ ~T) V (T Λ ~M) ] Λ ~T => O ( order of operation with AND first) [ ( M Λ ~T) Λ ~T) ] V [ (T Λ ~M) Λ~T] => O (distrib. Λ V rule) (M Λ ~T Λ ~T ) V (T Λ ~M Λ ~T) => O (assoc. rule) ( M Λ ~T ) V ( ~M ΛT Λ~T) => O (AND same; assoc. rule) ( M Λ ~T) V (~M Λ False) => O (Contradiction rule) ( M Λ ~T) V (False) => O (AND False rule) ( M Λ ~T) => O (OR False rule) ( Monitoring Λ Not Testing ) => Operating

2. Assignment 1-A: “common mistake” ( M Λ ~T V T Λ ~M) Λ ~T => O ( M Λ True Λ ~M) Λ ~T => O ( M Λ ~M) Λ ~T => O ( False ) Λ ~T => O False => O False implies operating Using OR first and Ignoring operator Precedence !

3. Assignment 1-A : “Interesting mistake” [ ( M Λ ~T) V ( T Λ ~M) ] Λ ~T => O [ (M V (T Λ ~M)) Λ (~T V (T Λ ~M)) ] Λ ~T [ ( (M V T) Λ (M V ~M) ) Λ ( (~T V T) Λ (~T V ~M) ) ] Λ ~T [ ( (M V T) Λ (True) ) Λ ( (True) Λ (~TV~M) ) ] Λ ~T [ ( M V T ) Λ ( ~T V ~M) ] Λ ~T (M V T ) Λ (~T V ~M) Λ ~T (~T V ~M) Λ [ (T V M) Λ ~T ] (~T V ~M) Λ [ M ] (~T Λ M) V ( ~M Λ M) (~T Λ M ) V (False) (M Λ ~T ) => O ( Monitoring and Not Testing) implies Operating Using disjunctive syllogism (T v M) Λ~T=> M is true; BUT (T v M) Λ ~T<=> M is not true !

4. Answer borrowed from Joel (spring ’08 student) (M Λ ~T V T Λ ~M) Λ ~T => O [ (M Λ ~T) V (T Λ ~M) ] Λ ~T => O ( M Ve T ) Λ ~T => O (M Λ ~T) Ve (T Λ ~T) => O (M Λ ~T) Ve (false) => O (M Λ ~T) => O Look at this replacement with “Exclusive OR” !