# Tautologies and Quantification

Are the following two questions a) tautologically valid, b) logically valid but not tautologically valid, or c) invalid.

Here's what I have so far: For 10.3: It is not tautologically valid since the truth functional form is not. But it is logically valid since we know there is no such object such that it is small, therefore the antecedent of premise one must also be false and not everything is a cube. Therefore there is an object that is not a cube. Or is it not logically valid since if we had a world with one large cube and a large tetrahedron all the premises would be satisfied and still the conclusion would be false?

For 10.6: It is also not tautologically valid. But would it be logically valid since due to premises 2 and 3, we know the consequent of premise 1 is false and therefore the antecedent must be false which would always make the conclusion follow from the premises?

Thanks in advance for the help!

• 10.3 valid: by contraposition, from the first premise you get "not exists to not for all" and you can derive "not for all" with second premise by modus ponens. Finally, "not for all" is equivalent to "exists not". May 26 '17 at 16:28
• 10.6 valid: basically, you can use the same approach above, plus De Morgan. May 27 '17 at 15:33

I don't know what rules you're allowed to use, but here's an outline that should help.

10.3 is logically valid:

1. ∀xCube(x) → ∃ySmall(y)
2. ¬∃ySmall(y)
3. ¬∀xCube(x) (From 1,2: p → q and ¬q entail ¬p)
4. ∃x¬Cube(x) (From 3: ¬∀xFx = ∃x¬Fx)

10.6 is also logically valid:

1. ∃x(Cube(x) Λ Large(x)) → (Cube(c) Λ Large(c))
2. Tet(c) → ¬Cube(c)
3. Tet(c)
4. ¬Cube(c) (From 2,3: modus ponens).
5. ¬(Cube(c) Λ Large(c)) (From 4)
6. ¬∃x(Cube(x) Λ Large(x)) (From 1,5: p → q and ¬q entail ¬p)
7. ∀x¬(Cube(x) Λ Large(x)) (From 6: ∀x¬Fx = ¬∃xFx)