Description
1. Consider the following expression
∀xP(x) ∧ Q(x) ↔ (∃xR(x) →∀x((S(x) ∧ Y (y)) ∨∃y(U(y) ∨∼ T(x))))
(a) For each occurrence of each variable, indicate whether the variable is free or bound. If the variable is bound, indicate whether it is bound to a ∀ or to a ∃.
P(x) bounded by ∀
Q(x) is a free variable
R(x) bounded by ∃
S(x) bounded by ∀
Y(y) is a free variable
U(y) is bounded by ∃
~T(x) is bounded by ∀
(b) Rename the variables so that distinct names are used for each distinct variable.
∀xP(x) = P
∀xS(x) = S
∀xT(x) = T
∃xR(x) = R
∃xU(y) = U
So, we get that ∀xP ^ Q(x) (∃xR(∀xS ^Y(y) V (∃yU V ~∀xT).
2. Show formally that ∼∃y(∀x∃zP (x, y, z) ∨∃x∀zQ(x, y, z)) is logically equivalent to ∀y(∃x∀z∼P(x,y,z) ∧∀x∃z∼Q(x,y,z))
∼∃y(∀x∃zP (x, y, z) ∨∃x∀zQ(x, y, z)) Premise
∀y ~(∀x∃zP (x, y, z) ∨∃x∀zQ(x, y, z)) Negation
∀y(~∀x∃zP(x,y,z) ∧~∃x∀z Q(x,y,z)) D.M law
∀y(∃x∀z∼P(x,y,z) ∧∀x∃z∼Q(x,y,z)) Negation
3. Find an interpretation to show that the following argument form is not valid. (∀x(P (x) → Q(x)) ∧ (∀x(P (x) → R(x)))) → ∀x((Q(x) → R(x))).
From the variables that questions gave, the argument has a domain ∀x that is unique individuals for each variable, each predicate of the expression defined, and all variables of the expression are bounded by ∀x.
If it is valid, then the expression should be true for all interpretations. So we get a chart.
x P(x) Q(x) R(x) P(x)Q(x) P(x)R(x) (P(x)Q(x)) ^(P(x)R(x)) Q(x)R(x)
a T T T T T T T
d T T F F F F T
The last column shows that ∀x((Q(x)R(x) is true, but from the 5th column ∀x(P(x)Q(x)) is false, the 6th column ∀x(P(x)R(x)) is false, so the 7th column ∀x ((P(x)Q(x))^(P(x)R(x))) is false. Thus, the ∀x((Q(x) → R(x))) expression is not valid.
