SEMANTICS
Answers to Assignment #11

Problem 5-1
[[ ∀x[M(x)→M(j)] ]]M, t1, g = 1 iff [[ M(x)→M(j) ]]M, t1, g’ = 1 for all g’ exactly like g except possibly for the value it assigns to x.
This condition is fulfilled since [[ M(j) ]]M, t1, g’ = 1 for all g’ exactly like g except possibly for the value it assigns to x. Therefore, [[ ∀x[M(x)→M(j)] ]]M, t1, g = 1. 
[[ ∀x[M(x)→M(j)] ]]M, t2, g = 1 for the same reasons. However, [[ ∀x[M(x)→
M(j)] ]]M, t3, g = 0 because [[ M(x) ]]M, t1, g(x=c) = 1 but [[ M(j) ]]M, t1, g(x=c) = 0.

Problem 5-2
[[∀x[FM(x)→PM(x)] ]]M, t1, g = 1 iff [[FM(x)→PM(x) ]]M, t1, g’ = 1 for all g’ exactly like g except possibly for the value it assigns to x.
But, [[FM(x)→PM(x) ]]M, t1, g(x=a) = 0 because [[FM(x)]]M, t1, g(x=a) = 1 but
[[PM(x) ]]M, t1, g(x=a) = 0 (since there is no t < t1).
Therefore, [[∀x[FM(x)→PM(x)] ]]M, t1, g = 0.
[[FM(x) ]]M, t2, g(x=a) = 0; [[FM(x) ]]M, t2, g(x=b) = 0; and [[PM(x) ]]M, t2, g(x=c) = 1. Therefore, [[∀x[FM(x)→PM(x)] ]]M, t2, g = 1.
[[FM(x) ]]M, t3, g(x=a) = 0, [[FM(x) ]]M, t3, g(x=b) = 0 and [[FM(x) ]]M, t3, g(x=c) = 0.
Therefore, [[∀x[FM(x)→PM(x)] ]]M, t3, g = 1.

Problem 5-3
i.	FPφ→[Fφ v φ v Pφ]
Suppose [[FPφ]]M,i,g =1 for some M, i and g. Then there is some i’ such that i<i’ and [[Pφ]]M,i',g =1 and there is some i”<i’ such that [[φ]]M,i”,g =1. Given the nature of <, either i”<i or i<i” or i=i”. Therefore, [[Fφ v φ v Pφ]]M,i,g =1

ii.	PFφ→[Fφ v φ v Pφ]
Suppose [[PFφ]]M, i, g =1 for some M, i and g. Then [[Fφ]]M, i', g =1 for some i’<i and [[φ]]M, i”, g =1 for some i” after i'. Given the formal properties of <, either i”<I, or i<i” or i=i”. Therefore, [[Fφ v φ v Pφ]] M, i, g =1.
[5-3(i-ii) are extremely elegant proofs that you should try to emulate. Study them carefully]

Problem 5-4
i. Fφ→FFφ
Let M be a model with exactly two times, t and t’. Suppose t < t’ and φ is true at t’. Then Fφ is true at t but not FFφ, since there is no time after t at which Fφ is true. The formula is valid in models with dense time. If we assume dense time then there will be a t” such that t < t” < t’ and Fφ will be true at t”, and consequently, FFφ will be true at t.

ii. Pφ→PPφ 
Let M be a model with exactly two times, t and t’. Suppose t < t’ and φ is true at t. Then Pφ is true at t’ but not PPφ, since there is no time before t’ at which Pφ is true. The formula is valid in models with dense time. If we assume dense time then there will be a t” such that t < t” < t’ and Pφ will be true at t”, and consequently, PPφ will be true at t’.

Problem 5-5
i. ∃uHφ→H∃uφ
Suppose [[∃uHφ]]M, i, g =1 for some M,i and g. Then for some e in A, 
[[Hφ]]M, i, g(u=e) =1, and for all i’ such that i’<i, [[φ]]M, i’,g(u=e) =1. 
Now let us assume that [[H∃uφ]]M, i, g = 0. This can be the case if and only if for some i’ < i, [[∃uφ]]M, i’, g = 0, that is, if and only if for no e in A, [[φ]]M, i’, g(u=e) = 1. But we have already seen that there is at least one e in A such that for all i’<i, 
[[φ]]M, i’,g(u=e) =1. Hence, the assumption that [[H∃uφ]]M, i, g = 0 is untenable, that is, [[H∃uφ]]M, i, g = 1. Therefore, ∃uHφ→H∃uφ is valid.

iii. ∀uHφ→H∀uφ
Suppose [[∀uHφ]]M, i, g =1 for some M,i and g. Then for all e in A, and for all i’<i, [[φ]]M,i’,g(x=e) =1.
Suppose now, [[H∀uφ]]M, i, g = 0. Then for some i’ such that i’<i, and some e in A [[φ]]M, i’,g(x=e) = 0. But this contradicts the conclusion arrived at earlier. Hence, the assumption is untenable and ∀uHφ→H∀uφ is valid.

iv. ∃uPφ→P∃uφ
Suppose [[∃uPφ]]M, i, g =1 for some M,i and g. Then for some e in A, and some i’<i, [[φ]]M,i’,g(x=e) =1.
Suppose now, [[P∃uφ]]M, i, g = 0. Then for all i’<i, and all e in A, [[φ]]M, i’,g(x=e) = 0. This, however, contradicts the earlier conclusion.
Therefore, [[P∃uφ]]M, i, g = 1 and ∃uPφ→P∃uφ is valid.

Please look for typos and errors and bring them to my attention.