- (f) Premise (\forall x)(\exists y) Gyx

Conclusion ¬(\exists y) (\forall z) Gyz - (h) Premise (\forall x)[ Ex \biconditional ¬Ox ]

Premise Et

Conclusion ¬Ot - (l) Premise (\forall x)[ Sxy -> Lxy ]

Conclusion (\forall x)(\forall y) [ ( Lxy & [Sxz & Syw] ) -> Lzw ]

- (f) (\forall x)(\forall y)[ ¬Ixa & ¬Iya ->( (\forall z) Pxyz -> ¬Iza) ]
- (h) (\forall x)(\forall y)(\forall z)[ ( Lxy & Laz ) - >

(\forall w) (\forall u)( [ Pxzw &Pyzu]-> Lwu) ]

- (j) (\exists x)[Pqx & Prx]
- (l) (\forall x) ¬ Pxx

(Fxay v (¬Gxa -> (\forall z)Hzwb)) / \ Fxay (¬Gxa -> (\forall z)Hzwb) / \ ¬Gxa (\forall z)Hzwb | | Gxa HzwbThe question gives instructions for defining the set

By 1.I have shown the steps of distinct lines. However, if I was 'solving' such a problem, I would just write out theFree(Gxa) = {x} andFree(Hzwb)={z,w} Then: by 2Free(¬Gxa) = {x} by 1. by 4Free((\forall z)Hzwb) = {z,w} - {z} = {w} Then by 1Free(Fxay) = {x,y} by 3Free((¬Gxa -> (\forall z)Hzwb)) = {x} U {w} = {x,w} Finally by 3:Free((Fxay v (¬Gxa -> (\forall z)Hzwb))) = {x,y} U {x,w} = {x,y,w}

Gabc -> ¬(Fa&Hm) / \ Gabc ¬(Fa&Hm) | (Fa&Hm) / \ Fa Hm We can now define SubWffs recursively up from the leaves of this tree: By 1 SubWffs(Fa) = {Fa} SubWffs(Hm) = {Hm} By 3 SubWffs(Fa&Hm) = {Fa&Hm} U {Fa} U {Hm} = {Fa&Hm,Fa,Hm} By 2 SubWffs(¬(Fa&Hm)) = {¬(Fa&Hm),Fa&Hm,Fa,Hm} By 1 SubWffs(Gabc) = {Gabc} By 3 SubWffs(Gabc -> ¬(Fa&Hm)) = {Gabc -> ¬(Fa&Hm), Gabc, ¬(Fa&Hm), Fa&Hm, Fa, Hm} Again, if I were actually solving this, I would just write out the parsing tree and then work back up writing the SubWffs beside each node. Therefore the Size is 6.

- (f) [Ga -> (\exists x)Fxc]
- (h) [ Kaa v Fbac]
- (j) [(\exists x) Sx \biconditional (\exists y) Kay]

- (b) (\exists y) Fabc, (\exists y) Faby.
- (d) (\exists y) (\exists x) Lxc, (\exists y) (\exists x) Lxy.
- (f) (\exists y) [ (\exists x)Kxc -> Fc], (\exists y) [ (\exists x)Kxy -> Fc],

(\exists y) [ (\exists x)Kxc -> Fy], (\exists y) [ (\exists x)Kxy -> Fy].

- (b) (\exists x)[ Fxa \biconditional Fax] is T:

Faa is T and Faa is T. Therefore the biconditional Faa \biconditional Faa is T.

This gives us (\exists x)[ Fxa \biconditional Fax] is T - (d) (\forall x)(\exists y)Fxy is F:

Consider b/x. By the definition of the truth value assignment, Fba, Fbb, Fbc, .... are all F. Therefore (\exists y)Fby is F.

One failure is enough to make (\forall x)(\exists y)Fxy to be F - (f) (\forall x)(\exists y)Fyx is T

For any instantiation d/x, Fad is T.

Therefore (\exists y)Fyd is T for all choices of d

Therefore (\forall x)(\exists y)Fyx is T

To demonstrate this we need to offer a 'model' in which the Premise of the argument is true and the Conclusion is false.

The whole point of logic is to preserve any 'Truth' of the Premises into 'Truth' for the Conclusion!

One way to communicate the 'model' is to list the truth value assignment for the critical ground atomic sentences (for a restricted finite set of constants that appear in the formulae).

We choose a model with the universal domain labeled U = { a, b }.

We assign the following tva:

Fa is T, Fb is F, Ga is F, Gb is T.

With this choice Fa v Ga is T and Fb v Gb is T.

We conclude that (\forall x)Fx v (\forall x)Gx is F.
**Note:** To figure out the counterexample, you actually first see what it takes to make

(\forall x)Fx v (\forall x)Gx False.

From our current work on PD+, we know that this is the same as making:
¬[(\forall x)Fx v (\forall x)Gx] T, or equivalently.

(\exists) x ¬ Fx & (\exists) x ¬ Gx True.

This suggests Fb is F and Ga is F for some a, b.

The next step is to make (\forall x)[ Fx v Gx] T.

This requires Fa v Ga is T, and since Ga is F, we must have Fa T.
This also requires Fb v Gb is T, and since Fb is F, we must have Gb T.

We choose a model with the universal domain labeled U = { a, b }.

We assign the following tva:

Fa is T, Fb is T, Ga is T, Gb is F.

With this assignment: (\exists x)Gx is T (because of the one instantiation Gb);

Therefore Fa -> (\exists x)Gx is also T.

However, (\forall x)[Fa ->Gx] is False because of the one instantiation:
[Fa ->Gb] which is F (with Fa T and Gb F).

**Note:** To figure out the counterexample, you actually first see what it takes to make

(\forall x)[Fa ->Gx] F: Fa T and Gb F for some b.

Then look for a way to make Fa -> (\exists x)Gx T - which now requires

(\exists x)Gx T or Gc T for some instantiation.

I chose to use *a* for this instantiation just to keep things small!

A

Many people thought this was false - but they were really answering (c)

The 'counterexamples' such as (\forall x)Fx v (\forall x)Gx
do not have overlapping scopes but * this is not a quantified formula*.

This is not a sentence since the x in Bxy is Free (not inside the scope of any quantifier with x as the variable).

The universal generalization would be (\forall x)[Fx -> Gc].

This is different in appearance from (\forall x)Fx -> Gc.

That difference in appearance is enough to demonstrate the sentence is false, since 'universal generalization' is a syntactic operation - pure appearance!

By the way, if you think about it, in terms of models or chapter 8, the universal generalization is also different in meaning from the formula they gave. That fact is nice to know, but not relevant to this question!