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Natural Deduction. Homework #1. ~(~P & ~Q). ~(~P & ~Q). ~(~P & ~Q). ~(~P & ~Q). ~(~P & ~Q). ~(~P & ~Q). ~(~P & ~Q). (~(~P & ~Q) ↔ (P v Q)). So “~(~P & ~Q)” has the same truth-table as “(P v Q).” Why is that?
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(~(~P & ~Q) ↔ (P v Q)) So “~(~P & ~Q)” has the same truth-table as “(P v Q).” Why is that? Suppose I say: “you didn’t do your homework and you didn’t come to class on time.” When is this statement false? When either you did your homework or you came to class on time.
(~(P → ~Q) ↔ (P & Q)) So “~(P → ~Q)” has the same truth-table as “(P & Q).” Why is that? Suppose I say: “If you eat this spicy food, you will cry.” You might respond by saying “No, that’s not true: I will eat the spicy food and I will not cry.”
((~P v Q) ↔ (P → Q)) So (~P v Q) has the same truth-table as (P → Q). Why is that? Premise: Either I didn’t take my keys with me when I left home, or I lost them on the way to the office. Conclusion: Therefore, if I took them with me, I must have lost them on the way.
((~P v Q) ↔ (P → Q)) So (~P v Q) has the same truth-table as (P → Q). Why is that? Premise: If my mother remembers my birthday, I will get a letter in the mail. Conclusion: Therefore, either she won’t remember, or I’ll get a card.
So “(((P → Q) → P) → P)” is always true. Why is that? I’ll leave you to think about that. Later we’ll prove that it’s true.
The Truth-Table Test for Validity We know that an argument is deductively valid when we know that if it is true, then its conclusion must be true. We can use truth-tables to show that certain arguments are valid.
The Test Suppose we want to show that the following argument is valid: (P → Q) ~Q Therefore, ~P We begin by writing down all the possible truth-values for the sentence letters in the argument.