SEMANTICS
Answers to assignment #12
Problem 5-8
i.	[[◇M(d)]]M, t, g = 1 iff for some i in I, [[M(d)]]M, i, g = 1.
	t1 ∈ I and [[M(d)]]M, t1, g = 1.
	Therefore, [[◇M(d)]]M, t3, g = 1
ii.	[[∀x[M(x) → □M(x)] ]]M, t1, g 
Hint: a cursory glance at the model shows that the sentence under consideration evaluates to 0, since a is in the extension of M at t1 but not so at t3.
	[[M(x)]]M, t1, g(x/a) = [[M]]M, t1, g(x/a)([[x]]M, t1, g(x/a)) = 1.
	But, [[□M(x)]]M, t1, g(x/a) = 0 since [[M(x)]]M, t3, g(x/a) = 0.
	Hence, [[∀x[M(x) → □M(x)] ]]M, t1, g(x/a) = 0.
	And consequently, [[∀x[M(x) → □M(x)] ]]M, t1, g = 0.
iii.	[[ □[M(d)→M(j)] ]]M, t1, g  = 1, iff for all i in I, [[M(d)→M(j)]]M, i, g = 1
	[[M(d)→M(j)]]M, t1, g = 1; [[M(d)→M(j)]]M, t2, g = 1; [[M(d)→M(j)]]M, t2, g = 1.
	Therefore, [[M(d)→M(j)]]M, t1, g = 1.
iv.	[[∃x□M(x)]]M, t3, g  = 1, iff there is some e in A such that [[□M(x)]]M, t3, g(x/e)  = 1.
	And this is true, iff for all i in I, [[M(x)]]M, i, g(x/e)  = 1.
	Considering each e in A one by one, we find that
	[[M(x)]]M, t3, g(x/a) = 0; [[M(x)]]M, t3, g(x/b) = 0; [[M(x)]]M, t2, g(x/c) = 0
	Therefore, [[∃x□M(x)]] M, t3, g  = 0
v.	[[□∃xM(x)]]M, t3, g  = 1 iff, for all i in I, [[∃xM(x)]]M, i, g  = 1, and the latter is true just in case for some e in A [[M(x)]]M, i, g(x/e) = 1.
	Considering each i in I, one by one, we find that
	[[∃xM(x)]]M, t1, g(x/a) = 1, because [[M(x)]]M, t1, g(x/a) = 1; [[∃xM(x)]]M, t2, g(x/a) = 1, 	because [[M(x)]]M, t2, g(x/a) = 1; and [[∃xM(x)]]M, t3, g(x/c) = 1, because 
	[[M(x)]]M, t3,g(x/c) = 1. 
	Therefore, [[□∃xM(x)]]M, t3, g  = 1.

Problem 5-9

i.	[[B(m)]] M, w1, t2, g = 0, as [[m]] M, w1, t2, g  is not in [[B]] M, w1, t2, g
	Therefore, [[ [B(m) ^ FB(j)] ]]M, w1, t2, g = 0.
ii.	[[ [◇P¬B(d)] ]]M, w1, t2, g  = 1 iff, for some w in W and some t in T,
	[[P¬B(d)]]M, w, t, g  = 1.
	[[¬B(d)]]M, w1, t2, g  = 1 since [[(d)]]M, w1, t2, g is not in [[B]]M, w1, t2, g
	[[P¬B(d)]]M, w1, t3, g  = 1, because t2 < t3 and [[¬B(d)]]M, w1, t2, g  = 1
	Therefore, [[ [◇P¬B(d)] ]]M, w1, t2, g  = 1
iii.	[[ □[FB(m)→PB(m)] ]]M, w1, t2, g = 1, iff for all w in W and t in T 
	[[FB(m)→PB(m)]]M, w, t, g = 1.
	[[FB(m)→PB(m)]]M, w1, t1, g = 0 because, [[FB(m)]]M, w1, t1, g = 1 but 
	[[PB(m)]]M, w1, t1, g = 0.
	Therefore, [[□[FB(m)→PB(m)] ]]M, w1, t2, g = 0.
iv.	[[□[ [¬B(m) ^ FB(m) ]→PB(m) ] ]]M, w1, t2, g = 1, iff for all w in W and t in T, 
	[[ [¬B(m) ^ FB(m)]→PB(m) ]]M, w, t, g  = 1.
	Evidently, [[ ¬B(m) ^ FB(m) ]]M, w, t3, g  = 0  for all w in W because 
	[[ FB(m)]]M, w, t3, g  = 0  for all w in W. Therefore [ [¬B(m) ^ FB(m)]→PB(m) ] 	denotes 1 in all t3-indeces (that is, indeces containing t3).
	[ [¬B(m) ^ FB(m)]→PB(m) ] denotes 1 in all t1-indeces since 
	[[ [¬B(m) ] ]]M, w, t1, g  = 0 for all w in W. 
	[[ [¬B(m) ^ FB(m)]→PB(m) ]]M, w1, t2, g  = 1 because [[ [¬B(m) ] ]]M, w1, t2, g  = 0.
	Finally, [[ [¬B(m) ^ FB(m)]→PB(m) ]]M, w2, t2, g  = 1 because 
	[[ PB(m) ]]M, w2, t2, g  = 1. That exhausts all the indeces in the model.
	Therefore, [[□[ [¬B(m) ^ FB(m) ]→PB(m) ] ]]M, w1, t2, g = 1.
v.	[[∃x□[B(x) v B(d)] ]]M, w1, t2, g = 1 iff there is some e in A such that 
	[[□[B(x) v B(d)] ]]M, w1, t2, g(x/e) = 1.
	This in turn is true iff for all w in W and t in T, [[B(x) v B(d)]]M, w, t, g(x/e) = 1 
	
	For all w in W and t in T, [[B(x) v B(d)]]M, w, t, g(x/a) = 1, for since all w in W and t 	in T either [[d]]M, w, t, g(x/a) is in [[B]]M, w, t, g(x/a) or [[x]]M, w, t, g(x/a) is in 
	[[B]]M, w, t, g(x/a). Therefore, [[□[B(x) v B(d)] ]]M, w1, t2, g(x/a) = 1, and 
	[[∃x□[B(x) v B(d)] ]]M, w1, t2, g = 1
Problem 5-10
□φ→F□φ
The formula is invalid in any model in which time has an end because its consequent is false at all tf-indeces (that is indeces containing tf) where f is the last moment in time.
The following is a falsifying model for the formula:
Let M = <A, W, T, < F> where
	A = {a,b}
	W = {w1, w2}
	T = {t1, t2}
	< = {<t1, t2>}
	F(d)(<w,t>) = a  for all w in W and for all t in T
	F(j)(<w,t>) = b  for all w in W and for all t in T
	F(M)(<w,t>) = {a,b}  for all w in W and for all t in T
Let φ be M(d). Evidently, [[□M(d)]] M, w1, t2 = 1 but [[F□M(d)]] M, w1, t2  = 0.
Problem 5-11
Fφ→□φ
Invalid. 
Falsifying model:
Let M = <A, W, T, < F> where
	A = {a,b}
	W = {w1, w2}
	T = {t1, t2}
	< = {<t1, t2>}
	F(d)(<w,t>) = a  for all w in W and for all t in T
	F(j)(<w,t>) = b  for all w in W and for all t in T
	F(M)(<w,t2>) = {a,b}  for all w in W.
Let Let φ be M(d). Evidently [[FM(d)]]M, w1, t1 = 1 but [[□M(d)]] M, w1, t1  = 0. This is so because there is a t’, namely t2, such that t1<t2 and [[M(d)]]M, w1, t2 = 1, but it is not the case that for all w in W and for all t in T [[M(d)]]M, w, t = 1. 
Specifically, [[M(d)]]M, w1, t1 = 0 and [[M(d)]]M, w2, t1 = 0.
Problem 5-12
Gφ→◇φ
Hφ→◇φ
Both invalid.
Falsifying model:
Let M = <A, W, T, < F> where
	A = {a,b}
	W = {w1, w2}
	T = {t1, t2}
	< = {<t1, t2>}
	F(d)(<w,t>) = a  for all w in W and for all t in T
	F(M)(<w,t>) = {}  for all w in W and for all t in T.
Let Let φ be M(d). Evidently [[GM(d)]]M, w2, t2 = 1 but [[◇M(d)]] M, w2, t2  = 0. This is so because for all t’, such that t2<t’, [[M(d)]]M, w2, t’ = 1, but it is not the case that for some <w,t> in W X T [[M(d)]]M, w, t = 1. 
Similarly, it is clear that [[HM(d)]]M, w2, t1 = 1 but [[◇M(d)]] M, w2, t1  = 0. This is so because for all t’, such that t’<t1, [[M(d)]]M, w2, t’ = 1, but it is not the case that for some <w,t> in W X T [[M(d)]]M, w, t = 1.

This document involved a lot of cut-n-paste. Please look for typos and errors and bring them to my attention.