Post 2: MIT6_042JF10_assn01

1)
a)
$$\exists x \in X. S(x) \land A(x)$$
b)
$$\forall x \in X. T(x) \land S(x) \implies A(x)$$
c)
$$\forall x \in X.\neg T(x) \land \neg A(x)$$
d)
$$\exists x\exists y \exists z(\neg E(x,y) \land \neg E(x,z) \land \neg E(y,z) \land (T(x) \land \neg S(x)) \land (T(y) \land \neg S(y)) \land (T(z) \land \neg S(z)))$$ 2)
a)
First I'll evaluate the left hand side, then I'll evaluate the right hand site of the equation
Left hand side:

$$Q$$	$$R$$	$$Q \land R$$
true	true	true
true	false	false
false	true	false
false	false	false

$$P$$	$$Q \land R$$	$$P \lor (Q \land R)$$	$$\neg(P \lor (Q \land R))$$
true	true	true	false
true	false	true	false
false	true	true	false
false	false	false	true

right hand side:

$$\neg Q$$	$$\neg R$$	$$Q \lor R$$
false	false	false
false	true	true
true	false	true
true	true	true

$$\neg P$$	$$Q \lor R$$	$$(\neg P) \land (\neg Q \lor \neg R)$$
false	false	false
false	true	false
true	false	false
true	true	true

So this statement is true: $$\neg(P \lor (Q \land R)) = (\neg P) \land (\neg Q \lor \neg R)$$ This equation could also have been solved with De Morgan's law. De Morgan's law was explained in my previous post already.