SEMANTICS
Assignment 13
Problem 6-1
1.	walk'(j) 			<t>
2.	change'(^j)			<t>
3.	λx[walk'(x)]			<e, t>
4.	λr[rise'(r)]			<<s, e>, t>
5.	λQ[Q = rise']			<<<s, e>, t>, t>
6.	λP[P =  ^walk']		<<s,<e,t>>,t>
7.	ˇP(ˇ^j)				<t>
8.	λP[ˇP(j)]			<<s , <e, t>>, t>
9.	λx[Sq(x)]			<e, e>
10.	^[λx[Gr(3)(x)]]		<s, <e, t>>
11.	λp[ˇp = rise' (^j)]		<<s,t>,t>
12.	λx[λp[ˇp = change' (^x)]]	<e,<<s,t>,t>>
13.	^ˇr				<s, e>
14.	ˇ^rise'				<<s, e>, t>
15.	λx[R(x)(Sq(x))]		<e, t>

Problem 6-2
1. [[λxB(x)]]M, w1, t1, g is that function h with domain De, such that for any object e in that domain, h(e) = [[Bx]] M, w1, t1, g' where g' is that  value assignment exactly like g with the possible difference that g'(x) is the object e.			…from B3
h(a) = [[Bx]]M,w1,ti,g’= B(g’(x)) = B(a) = 1
h(b) = [[Bx]]M,w1,ti,g’= B(g’(x)) = B(b) = 1
h(c) = [[Bx]]M,w1,ti,g’= B(g’(x)) = B(c) = 0
In other words, h is the characteristic function of the set of individuals {a,b}.

2. [[λxB(x)]]¢M, g is that function h with domain W x T such that for all <w, t> in W x T, h(<w, t>) is [[λxB(x)]]M,w,t,g			…C2

		<w1, t1>	→	{a, b}
		<w1, t2>	→	{a,c}
		<w1, t3>	→	{b,c}	
h =		<w2, t1>	→	{b,c}
		<w2, t2>	→	{a}
		<w2, t3>	→	{a,b,c}		

3. The function that maps each <w,t> in W X T to the function h defined in (2).
4. [[ B(m) AND ¬B(j) ]] ¢M, g is that function h with domain W x T such that for all <w, t> in W x T, h(<w, t>) is [[B(m) ^ ¬B(j)]]M,w,t,g		…C2

		<w1, t1>	→	0
		<w1, t2>	→	0
		<w1, t3>	→	1	
h =		<w2, t1>	→	1
		<w2, t2>	→	0
		<w2, t3>	→	0

5. [[^[B(m) AND ¬B(j)] ]]M, w2, t2, g = [[ B(m) ^ ¬B(j) ]] ¢M, g 	…C2
that is, the function defined in (4).
6. [[m = n]] M, w2, t2, g = 1 iff [[m]] M, w2, t2, g =  [[n]] M, w2, t2, g 		B.5.
    [[m]] M, w2, t2, g = [[n]] M, w2, t2, g = c
    Therefore, [[m = n]] M, w2, t2, g = 1
7. [[^m = ^n]] M, w2, t2, g = 1 iff [[^m]] M, w2, t2, g = [[^n]] M, w2, t2, g 
    [[^m]] M, w2, t2, g ≠ [[^n]] M, w2, t2, g
    because, [[m]] M, w1, t1, g ≠ [[n]] M, w1, t1, g 
    Hence, [[^m = ^n]] M, w2, t2, g = 0
Problem 6-3:	The answers would not change. Make sure you understand why.

Problem 6-5
(i)	[[ For(^B) ]]₡M,g = {<<w1,t1>, {}>, <<w2,t1>, {}>, <<w1,t2>, {b}>, <<w2,t2>, 	{b,c}>, <<w1,t3>,{a}>, <<w2,t3>, {}>}
(ii)	[[ ∀x[For(^B)(x) ↔ PB(x)] ]]M,w2,t2,g
∀x[For(^B)(x) → PB(x)] is a valid sentence, so only the right-to-left implication needs to be considered.
[[ PB(x) ]]M,w2,t2,g(x=a) = 0 since a is not in the extension of B at any time before t2. Hence, [[ [PB(x)→For(^B)(x) ]]M,w2,t2,g(x=a) = 1.				… 1
Both b and c are in the extension of For(^B) at <w2,t2>. 
Therefore, [[ For(^B)(x) ]]M,w2,t2,g(x=b) = 1 and  
[[ [For(^B)(x) ]]M,w2,t2,g(x=c) = 1.
	Consequently, [[ [PB(x)→For(^B)(x) ]]M,w2,t2,g(x=b) = 1, and			… 2
[[ [PB(x)→For(^B)(x) ]]M,w2,t2,g(x=c) = 1.				… 3
[[ ∀x [For(^B)(x) ↔ PB(x)] ]]M,w2,t2,g = 1			  … from 1,2 and 3
Hence, [[ ∀x[For(^B)(x) ↔ PB(x)] ]]M,w2,t2,g = 1
(iii)	[[ ∀x[For(^B)(x) ↔ PB(x)] ]] ¢M,g is a proposition, namely, that function h which maps every index into the denotation of ∀x[For(^B)(x) ↔ PB(x)], that is its truth-value, at that index.
Once again, since the left-to-right implication is valid, we consider only its right-	to-left counterpart, namely ∀x [PB(x)→For(^B)(x)]. PB(x) denotes 0 at both 	<w1,t1> and <w2,t1> because there is no time before t1. Consequently, 	PB(x)→For(^B)(x) denotes 1 at both of these indeces.
	[[ [PB(x)→For(^B)(x) ]]M,w1,t2,g(x=a) = 0 because [[ [PB(x) ]]M,w1,t2,g(x=a) = 1 but 
	[[ (^B)(x) ]]M,w1,t2,g(x=a) = 0. 
	Consequently, [[ ∀x[For(^B)(x) ↔ PB(x)] ]]M,w1,t2,g = 0. 
We have already seen 	that [[ ∀x[For(^B)(x) ↔ PB(x)] ]]M,w2,t2,g = 1.
	[[ ∀x[For(^B)(x) ↔ PB(x)] ]]M,w1,t3,g = 0 because [[ PB(x)] ]]M,w1,t3,g(x=b) = 1 and 
	[[ For(^B)(x) ]]M,w1,t3,g(x=b) = 0.
	[[ ∀x[For(^B)(x) ↔ PB(x)] ]]M,w2,t3,g = 0 because [[ PB(x)] ]]M,w1,t3,g(x=b) = 1 and 
	[[ For(^B)(x) ]]M,w1,t3,g(x=b) = 0.
So, [[ ∀x[For(^B)(x) ↔ PB(x)] ]] ¢M,g = {<<w1,t1>, 1>, <<w2,t1>, 1>, <<w1,t2>, 	0>, <<w2,t2>, 1>, <<w1,t3>, 0>, <<w2,t3>, 0>}
(iv)	[[ ^∀x[For(^B)(x) ↔ PB(x)] ]]M,w2,t2,g = [[ ∀x[For(^B)(x) ↔ PB(x)] ]] ¢M,g
[You will have noticed that my answers to 6-5 are different from those in the book. Can you guess why this is so?]