Answer to Quiz #2
Show that  ⎕φ → F⎕φ is invalid
Validity is truth in all models. A valid formula cannot evaluate to falsehood in any model. A formula that is not valid is said to be invalid. So, it suffices to show that ⎕φ → F⎕φ can evaluate to falsehood in some model.
Let M be <A,W,T,<,F> such that
	A = {a,b}
	W = {w1, w2, w3}
	T = {t1, t2, t3}
	<  is such that  t1< t2 < t3
	F is such that:	
	F(d)(<w,t>)=a  and F(j)(<w,t>)=b  for each  <w,t>  in W X T; that is, d and j 			are rigid designators
	F(B)(<w,t>)={a,b}  for each  <w,t>  in W X T.
Let g be an arbitrary value assignment. Let φ be B(d).
Evidently, [[⎕B(d)]]M,w1,t3,g=1 since [[B]]M,w,t,g([[d]]M,w,t,g)=1 for all <w,t> in W X T.
However, [[F⎕B(d)]]M,w1,t3,g=0 since there is no t such that t3 < t and [[⎕B(d)]]M,w1,t,g=1. (We need not even consider the latter half of the conjunct since the first half is false.)
Hence [[⎕φ → F⎕φ]]M,w1,t3,g=0, and since g is an arbitrary value assignment
[[⎕φ → F⎕φ]]M,w1,t3 = 0, that is, the formula is false with respect to model M world w1 and time t3. It is therefore, invalid.