Question

Establish these logical equivalences, where x does not occur as a free variable in A. Assume that the domain is nonempty.

a)\(\forall x\left( {A{\rm{ }} \to {\rm{ }}P{\rm{ }}\left( x \right)} \right){\rm{ }} \equiv {\rm{ }}A{\rm{ }} \to {\rm{ }}\forall xP{\rm{ }}\left( x \right)\)

b) \(\exists x\left( {A{\rm{ }} \to {\rm{ }}P{\rm{ }}\left( x \right)} \right){\rm{ }} \equiv {\rm{ }}A{\rm{ }} \to {\rm{ }}\exists xP{\rm{ }}\left( x \right)\)

Short answer

a) \(\forall x(A \to P(x)) \equiv A \to \forall xP(x)\) are both propositions are equivalent

b) \(\exists x(A \to P(x)) \equiv A \to \exists xP(x)\) are both propositions are equivalent

Step-by-step solution

Logical Equivalences

\(p \to q \equiv \neg p \vee q\)

\(\forall x\left( {A{\rm{ }} \to {\rm{ }}P{\rm{ }}\left( x \right)} \right){\rm{ }} \equiv {\rm{ }}A{\rm{ }} \to {\rm{ }}\forall xP{\rm{ }}\left( x \right)\)

a) \(A \to \forall xP(x)\)is logically equivalent with \(\neg A \vee (\forall xP(x))\)by logical equivalence \((1)\).

\(A \to \forall xP(x) \equiv \neg A \vee (\forall xP(x))\)

According to exercise \(46\) this is logically equivalent with \(\forall x(\neg A \vee P(x))\)which is also logically equivalent with \(\forall x(A \to P(x))\)by logical equivalence\((1)\).

\( \equiv \forall x(\neg A \vee P(x))\)

\( \equiv \forall x(A \to P(x))\)

And thus \(A \to \forall xP(x)\)is logically equivalent with \(\forall x(A \to P(x))\)

\(\exists x\left( {A{\rm{ }} \to {\rm{ }}P{\rm{ }}\left( x \right)} \right){\rm{ }} \equiv {\rm{ }}A{\rm{ }} \to {\rm{ }}\exists xP{\rm{ }}\left( x \right)\)

b) \(A \to \exists xP(x)\)is logically equivalent with \(\neg A \vee (\exists xP(x))\) by logical equivalence \((1)\).

\(A \to \exists xP(x) \equiv \neg A \vee (\exists xP(x))\)

According to exercise\(46\)this is logically equivalent with\(\exists x(\neg A \vee P(x))\)which is also logically equivalent with\(\exists x(A \to P(x))\)by logical equivalence\((1)\).

\( \equiv \exists x(\neg A \vee P(x))\)

\( \equiv \exists x(A \to P(x))n\)

And thus \(A \to \exists xP(x)\)is logically equivalent with \(\exists x(A \to P(x)).\)