Math Problem Statement

Show that (PVQ)A] (1PA (QVR)) v ([PA]Q) v ([PA]R) is a tautology

Solution

To show that the expression:

(PQ)(¬P(QR))(¬PQ)(¬PR)(P \lor Q) \land (\neg P \land (Q \lor R)) \lor (\neg P \land Q) \lor (\neg P \land R)

is a tautology, we need to demonstrate that it is always true, regardless of the truth values of PP, QQ, and RR. A logical statement is a tautology if its truth table evaluates to true for all possible combinations of truth values of the variables involved.

We will construct the truth table step by step.

1. Breakdown of Components

  • PP
  • QQ
  • RR
  • PQP \lor Q
  • QRQ \lor R
  • ¬P\neg P
  • ¬P(QR)\neg P \land (Q \lor R)
  • ¬PQ\neg P \land Q
  • ¬PR\neg P \land R

2. Construction of the Truth Table

PPQQRRPQP \lor QQRQ \lor R¬P\neg P¬P(QR)\neg P \land (Q \lor R)¬PQ\neg P \land Q¬PR\neg P \land R(PQ)(¬P(QR))(¬PQ)(¬PR)(P \lor Q) \land (\neg P \land (Q \lor R)) \lor (\neg P \land Q) \lor (\neg P \land R)
TTTTTFFFFF
TTFTTFFFFF
TFTTTFFFFF
TFFTFFFFFF
FTTTTTTTFT
FTFTTTTTFT
FFTFTTTFTT
FFFFFTFFFF

3. Evaluation of the Final Expression

As we can see from the final column of the truth table, the expression evaluates to True in all cases where PP is false. This confirms that the logical statement is not a tautology because there are combinations of truth values where the expression evaluates to false.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Propositional Logic
Logical Operators
Tautologies

Formulas

Truth table for evaluating logical expressions

Theorems

Tautology definition

Suitable Grade Level

Grades 11-12 (Advanced Logic)