Question
Question: The following statement \((p \to q) \to \left[ {(\sim p \to q) \to q} \right]\) is A. Tautology ...
The following statement (p→q)→[(∼p→q)→q] is
A. Tautology
B. Equivalent to ∼p→q
C. Equivalent to p→∼q
D. A fallacy
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 means negation p or not p and (∼p→q) means if not p then q that is if not p is true then q 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 (p→q)→[(∼p→q)→q] =x
Let us draw the truth table
p | q | ∼p | p→q | ∼p→q | (∼p→q)→q | x |
---|---|---|---|---|---|---|
T | T | F | T | T | T | T |
T | F | F | F | T | F | T |
F | T | T | T | T | T | T |
F | F | T | T | F | T | T |
Hence, the given question which we took as x is (p→q)→[(∼p→q)→q] 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 is always of p and p→q means if p is true the q have to be true.