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Question: The following statement \((p \to q) \to \left[ {(\sim p \to q) \to q} \right]\) is A. Tautology ...

The following statement (pq)[(pq)q](p \to q) \to \left[ {(\sim p \to q) \to q} \right] is
A. Tautology
B. Equivalent to pq\sim p \to q
C. Equivalent to pqp \to \sim q
D. A fallacy

Explanation

Solution

A tautology is a formula or assertion that is true in every possible interpretation. Whereas a fallacy is a condition that is always false in every possible interpretation. Equivalent means similar and p\sim p means negation pp or not pp and (pq)(\sim p \to q) means if not pp then qq that is if not pp is true then qq is true. For example: The dog is either brown, or the dog is not brown. The sentence is always true because one or the other is so. This is different from a statement that says “the dog is either or the dog is white” because dogs can be black, gray or a mix of colors. Note that when we put both halves of the logical tautology together it feels a bit reluctant, just like a verbal tautology.

Step-by-step solution: The statement is given to us is (pq)[(pq)q](p \to q) \to \left[ {(\sim p \to q) \to q} \right] =x = x
Let us draw the truth table

ppqqp\sim ppqp \to qpq\sim p \to q(pq)q(\sim p \to q) \to qxx
TTTTFFTTTTTTTT
TTFFFFFFTTFFTT
FFTTTTTTTTTTTT
FFFFTTTTFFTTTT

Hence, the given question which we took as xx is (pq)[(pq)q](p \to q) \to \left[ {(\sim p \to q) \to q} \right] a tautology. Since all the results in the truth table come out to be true.

Note:
In this type of question students often make mistakes while writing the truth table and end up getting the wrong answer. Students should remember that p\sim p is always of pp and pqp \to q means if pp is true the qq have to be true.