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Discrete Math: Truth Tables?
Show that each of these conditional statements is a tautology by using truth tables.
• (p∧q)→p
• p→(p∨q)
• p ̄ → ( p → q )
• (p∧q)→(p→q)
• (p→q)→p
• ( p → q ) → q ̄
I just need help figure out how i could prove this. Would you create one table or multiple tables and how would you go about creating it?
1 Answer
- ?Lv 79 years agoFavorite Answer
I'd create a separate table for each case.
Let's take the first example for demonstration. It expresses the logical Rule of Inference known as "Simplification." The truth table might be more convincing (that is, more obviously correct) if we include a column for the p∧q component, so let's do that:
p . q . p∧q . (p∧q)→p
T . T . . T . . . . . T
T . F . . F . . . . . T
F . T . . F . . . . . T
F . F . . F . . . . . T
What makes it a tautology (and therefore the basis for a Rule of Inference) is that the whole expression evaluates as True no matter what the truth-values of the primitive components, p and q, happen to be.
I'll leave the others to you.
