Which of the following logic statements are valid?
Revision as of 18:53, 23 February 2014 by Arjun Suresh (talk | contribs)

Predicate logic formulas without quantifiers can be verified using derivation. But when it comes to first order logic (predicate logic with quantifiers), the simplest way is to apply logical reasoning. Most of these questions asked will be for very small formulas and we can easily apply logical reasoning to check if they are valid. The method is to convert the logical formula into its corresponding English meaning. This is shown in the following examples:

Which of the following predicate logic statements is/are valid?

(1) <math>\forall (x) P(x) \vee \forall(x)Q(x) \implies \forall (x) (P(x) \vee Q(x))</math>

(2) <math>\exists (x) P(x) \wedge \forall(x)Q(x) \implies \exists (x) (P(x) \wedge Q(x))</math>

(3) <math>\exists (x) (P(x) \vee Q(x)) \implies \forall (x) P(x) \vee \forall (x) Q(x)</math>

(4) <math>\exists (x) (P(x) \vee Q(x)) \implies \sim \forall (x) </math>


Solution

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Predicate logic formulas without quantifiers can be verified using derivation. But when it comes to first order logic (predicate logic with quantifiers), the simplest way is to apply logical reasoning. Most of these questions asked will be for very small formulas and we can easily apply logical reasoning to check if they are valid. The method is to convert the logical formula into its corresponding English meaning. This is shown in the following examples:

Which of the following predicate logic statements is/are valid?[edit]

(1) <math>\forall (x) P(x) \vee \forall(x)Q(x) \implies \forall (x) (P(x) \vee Q(x))</math>

(2) <math>\exists (x) P(x) \wedge \forall(x)Q(x) \implies \exists (x) (P(x) \wedge Q(x))</math>

(3) <math>\exists (x) (P(x) \vee Q(x)) \implies \forall (x) P(x) \vee \forall (x) Q(x)</math>

(4) <math>\exists (x) (P(x) \vee Q(x)) \implies \sim \forall (x) </math>


Solution[edit]