9.3.16. UD:={1,2} Rxx:= {(1,1), (2,2), (1,2), (2,1)} It is obvious that the first sentence is satisfied in this model. Also, the second sentence is satisfied: whenever you have x and y such that Rxy is true (for example x=1, y=2), then you can find z different from x (in this example z=2) such that Ryz is true. You should verify this for all possible values of x and y.

9.4.1.(h) False. For example, let \alpha= A, \beta = ¬A.

9.4.1.(j) True. Prove first that {(Ax)Fx v (Ax)Gx} |- (Ax)( Fx v Gx ). Then, by the soundness theorem: {(Ax)Fx v (Ax)Gx} |= (Ax)( Fx v Gx ). To prove the first statement construct the derivation.

Premise 1 (Ax)Fx v (Ax)Gx Assumption 2 | (Ax)Fx 3 | Fa 4 | Fa v Ga 5 | (Ax)( Fx v Gx ) Assumption 6 | (Ax)Gx 7 | Ga 8 | Fa v Ga 9 | (Ax)( Fx v Gx ) vE 1,2-5,6-9 10 (Ax)( Fx v Gx )

9.4.4.(d)

UD1 := {1,2} Lxy := {(1,1),(1,2)} This model satisfies the sentence. UD2 := {1,2} Lxy := {(1,2)} This model does not satisfy the sentence.

9.5.1.(b)

UD := {1,2} Fx := {1} Gx := {1} a := 1 b := 1

Note that to satisfy the last sentence you have to define Gy to be true only when y is b. The second sentence says: if Fx is true then Gx must be true. So, by the previous remark, Fx is true only when x is b. By the first sentence, Fa is true. Hence a=b.9.5.6 (d)

UD: = {0,1} Lxy: = {(0,0)} Gx : = {0} a:=0 b:=1

It is easy to see that in the above structure, Lab and Ga are true. So it is the model for both Lab and (Lab -> Ga). However, since Gx:={0}, it is not a model of (Ax )Gx. So, the above structure satisfies all the premises but not the conclusion. Therefore, the argument is NOT valid.

Consider the three sentences:

- (Ax)[ Fx -> Gx ] ;

- (Ax) Fx -> (Ax) Gx ;

- (Ex)[ Fx -> Gx ] .

*six*possible arguments: {\alpha } |- \beta; where \alpha and \beta are distinct sentences from the list above.

For each such argument, if**valid**, give a derivation in PD+; if**invalid**, give a model that demonstrates this.Solutions: In the solutions below, "(1)|-(2)" will temparily mean an argument with the sentence in (1) as the only premise and the setence in (2) as the conclusion. The same convention will apply to other arguments.

- (1)|-(2): Valid. Here is a derivation.

Premise 1 (Ax)(Fx->Gx) Assume 2 +----(Ax)Fx AEa/x2 3 + Fa AEax1 4 + Fa->Ga ->E3,4 5 + Ga AIx/a 5 6 +----(Ax)Gx ->I2-3 7 (Ax)Fx -> (Ax)Gx

- (2)|-(1): invalid. Model:
UD:={a,b} Fx:={a} Gx:={b}

The above model obviously does not satisfy (Ax)Fx. Hence, it satisfies the sentence (Ax) Fx -> (Ax) Gx. However, when x is a, the model satisfies Fx but not Gx. (1)|-(3): Valid. This is just too obvious.

(3)|-(1): Invalid. Model

UD:= {a,b} Fx:={a} Gx:={b}

In the above model, b shows that (Ex)(Fx -> Gx) is satisfied. while a shows that (Ax)(Fx->Gx) is not satisfied(2)|-(3): Valid.

Premise 1 (Ax)Fx->(Ax)Gx Assume 2 +----¬(Ex)(Fx->Gx) IM2 3 + ¬(Ex)(¬Fx v Gx) QN3 4 + (Ax)¬(¬Fx v Gx) DM4 5 + (Ax)(¬¬Fx & ¬Gx) DN5 6 + (Ax)(Fx & ¬Gx) AEa/x6 7 + Fa & ¬Ga &EL7 8 + Fa AIa/x8 9 + (Ax)Fx ->E1,9 10+ (Ax)Gx AEa/x10 11+ Ga &ER7 12+---- ¬Ga ¬E2-11,2-12 13 (Ex)(Fx->Gx)

(3)|- (2): Invalid. Model

UD:={a,b} Fx:={a,b} Gx:={a}

Obviously, a shows that (Ex)(Fx->Gx) is satisfied. The model satisfies (Ax)Fx because Fx:={a,b} is the whole universe. The model does not satisfy (Ax)Gx because b is not in Gx. Hence the model does not satisfy

(Ax)Fx->(Ax)Gx.

- (Ax)[ Fx -> Gx ] ;