Answers to Assignment 6

1.	Every man is-boring.
	(This is just a sentence. No need to enclose it within [[ ]] )
Syntax
Every man is-boring			…B5 

man		v1 is-boring		…B2

	is-boring		v1
Semantics
Let g be an arbitrary assignment of values to variables. (Note that proving something about an arbitrary member of a set amounts to proving that thing about every member of that set.)
[[Every man is-boring]]M,g = 1 iff, for all e in A such that [[man]]M, g (e) = 1, 
[[v1 is-boring]] M, gv1e = 1					…B5
There are exactly two e’s in A such that [[man]]M, g (e) = 1. These are AS and HK. 	[[man]]M,g (AS) = {F(man)}(AS) = 1, and			…A1
	[[man]]M,g (HK) = {F(man)}(HK) = 1			…A1
The only other e in A, namely QEII, does not qualify because
	[[man]]M,g (QEII) = {F(man)}(QEII) = 0.			…A1
Furthermore, both AS and HK are such that
	[[v1 is-boring]]M,gv1AS = [[is-boring]]M,gv1AS([[v1]]M,gv1AS)	…B2
	= {F(is-boring)}(gv1AS(v1)) = 1, and				…A2, A1
	[[v1 is-boring]]M,gv1HK = [[is-boring]]M,gv1HK([[v1]]M,gv1HK)	…B2
	= {F(is-boring)}(gv1HK(v1)) = 1				…A2, A1
Hence, [[Every man is-boring]]M,g = 1 for any g, and
[[Every man is-boring]]M = 1					…C1

2.	Some man snores.
Syntax
Some man snores			…B6 

man		v2 snores		…B2

	snores		v2
Semantics
For an arbitrary g, [[Some man snores]]M, g = 1 iff, for at least one e in A such that [[man]]M, g (e) = 1, [[v2 snores]] M, gv2e = 1			…B6
Anwar Sadat is such an e, since
[[man]]M,g (Anwar Sadat) = {F(man)}(Anwar Sadat)=1	…A1
and [[v2 snores]]M,gv2AS = [[snores]]M,gv2AS([[v2]]M,gv2AS)	…B2
	= {F(snores)}( gv2AS(v2)) = 1				…A1, A2
Therefore, for all g [[Some man snores]]M,g = 1, and
[[Some man snores]]M = 1						…C1

3.	The woman sleeps.
Syntax
The woman sleeps			…B7 

woman	v3 sleeps		…B2

	sleeps			v3
Semantics
For any arbitrary g, [[The woman sleeps]]M,g = 1 iff there is exactly one e in A such that [[woman]]M,g(e)=1, and furthermore, [[v3 sleeps]]M,gv3e = 1	…B7
There is indeed exactly one e in A, namely QEII, such that [[woman]]M, g(e) = 1, because
[[woman]]M, g(QEII) = {F(woman)}(QEII) = 1		…A2
[[woman]]M, g(AS) = {F(woman)}(AS) = 0			…A2
[[woman]]M, g(HK) = {F(woman)}(HK) = 0			…A3
and that exhausts all e’s in A.
Furthermore, [[v3 sleeps]]M, gv3QEII = [[sleeps]]M,gv3QEII ([[v3]]M, gv3QEII)	…B2
	= {F(sleeps)}(gv3QEII (v3)) = 1					…A2, A1
Therefore, [[The woman sleeps]]M,g = 1, from which it further follows that [[The woman sleeps]]M = 1. (Make sure you know why this is so.)

Problem 3-5
1.	Liz loves every man
Syntax
Liz loves every man					…B5 

man		Liz loves v1				…B2
			
	Liz			loves v1 		…B1

			loves			v1
Semantics
Let g be an arbitrary value assignment. [[Liz loves every man]]M,g = 1 iff, for all e in A such that [[man]]M,g (e) = 1, [[Liz loves v1]] M, gv1e = 1	…B5
There are exactly two entities e in the model such that [[man]]M, g (e)=1, namely AS and HK:
	[[man]]M, g (AS) = {F(man)}(AS) = 1			…A2
	[[man]]M, g (HK) = {F(man)}(HK) = 1			…A2
The only other entity in the model is QEII, but
	[[man]]M, g (QEII) = {F(man)}(QEII) = 0			…A2
But one of them, namely Henry Kissinger is such that:
[[Liz loves v1]]M, gv1HK = {[[loves]]M, gv1HK ([[v1]]M, gv1HK)} ([[Liz]]M, gv1HK)	…B2
	= [{F(loves)}(gv1HK(v1))] (F(Liz)) = 0			…A2, A1
Hence, [[Liz loves every man]]M,g = 0				…B5
And, [[Liz loves every man]]M = 0.					…C2
(How does the last line follow from C2?)

2.	The woman hates Hank.
Syntax
The woman hates Hank				…B7 

woman	v3 hates Hank			…B2
			
	v3		hates Hank			…B1

		hates			Hank
Semantics
For an arbitrary g, [[The woman hates Hank]]M, g = 1 iff, for exactly one e in A [[woman]]M, g (e) = 1, and furthermore, [[v3 hates Hank]] M, gv3e = 1	  …B7
There is indeed exactly one e in E, namely QEII, such that [[woman]]M, g (e) = 1. How do we justify this claim? See above.
Furthermore, 
[[v3 hates Hank]] M, gv3QEII = … follow earlier pattern
Hence, [[The woman hates Hank]]M, g = 1, and 
[[The woman hates Hank]]M = 1							C1