Sentential logic exercise from 'How to Prove it - A Structured Approach'

In summary, the conversation discusses an exercise from the book "How to Prove it - A Structured Approach" and the laws that have been covered so far, including DeMorgan, Absorption, Idempotent, Double Negation, Commutative, Associative, Distributive, Tautology, Contradiction, Conditional, and Contrapositive. The person is struggling with the exercise and seeking help. There is also a discussion about using truth tables and trying to prove a formula using the laws.
  • #1
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Homework Statement



This is an exercise from 'How to Prove it - A Structured Approach' (Exercise 7a, page 54) . So far a a really great book.

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Homework Equations



The DeMorgan, Absorption, Idempotent, Double Negation, Commutative, Associative, Distributive, Tautology, Contradiction, Conditional and Contrapositive law is what the book has gone through so far.

The Attempt at a Solution



I have been banging my head at this for hours. Constructing a truth table is trivial. But i haven't been able to go from one formula to the other by use of the stated laws. All exercises up to this have just required a few applications of the laws. So all my attempts at solving this so far have been mostly pages of jumping between formulas using the applicable laws.

I really feel like I am missing something. So any help would be greatly appreciated!
 
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  • #2
[itex]\varphi[/itex][itex]\rightarrow[/itex][itex]\psi[/itex] Is introduced as an abbreviation for or shown to be equivalent to [itex]\neg[/itex][itex]\varphi[/itex] v [itex]\psi[/itex] earlier in that chapter, and [itex]\varphi[/itex][itex]\leftrightarrow[/itex][itex]\psi[/itex] for ([itex]\varphi[/itex][itex]\rightarrow[/itex][itex]\psi[/itex]) ^ ([itex]\psi[/itex][itex]\rightarrow[/itex][itex]\varphi[/itex]), so what you have to show is that

([itex]\neg[/itex]P v Q) ^ ([itex]\neg[/itex]Q v R) [itex]\Leftrightarrow[/itex] ([itex]\neg[/itex]P v R) ^ (((P ^ Q) v ([itex]\neg[/itex]P ^ [itex]\neg[/itex]Q)) v ((R ^ Q) v ([itex]\neg[/itex]R ^ [itex]\neg[/itex]Q)))

Does that help?
 
  • #3
Im sorry I've shouldve been more clear. That formula is what i have tried to get to but with no success.
 
  • #4
yeah I remember doing this one. Velleman isn't too clear on what he wants. I made a table but that did seem a little too simple. I tried hard to try to transform that first part into the second part but it is quite tricky. I wouldn't worry too much about it.
 
  • #5
I would start by assuming (P→Q) and (Q→R) and then try to prove the rest. I can get you started in this direction. For example,

Suppose P. Then Q. Then R. Thus, P→R.

Then show the disjunction with the bijections is true too and then the "backwards" direction of assuming RHS and proving LHS of the bijection that you're trying to establish.

(Didn't realize how old this thread is -_-)
 
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What is sentential logic?

Sentential logic, also known as propositional logic, is a branch of logic that deals with the logical relationships between simple statements or propositions, using logical operators such as "and," "or," and "not."

How is sentential logic used?

Sentential logic is used to analyze and evaluate arguments based on logical reasoning. It is also used in computer science and mathematics for its simplicity and precision.

What is the purpose of exercises in sentential logic?

The purpose of exercises in sentential logic is to help develop critical thinking skills and strengthen one's ability to construct and evaluate valid arguments.

What are the common logical operators used in sentential logic?

The most common logical operators used in sentential logic are "and" (∧), "or" (∨), and "not" (¬). Other important operators include "if-then" (→) and "if and only if" (↔).

What are the rules for constructing a valid argument in sentential logic?

In sentential logic, a valid argument must follow the rules of logical deduction, such as the law of detachment and the law of contrapositive. It must also have a clear and valid structure with correct use of logical operators.

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