How Can Logic Laws Remove Implications in Expressions?

[werebilby]

Hello Everyone,

This is my first post, so please me kind :D

I have an assignment and have been stuck on this thing for 2 days. I have no problems working this out with a truth table but we are not allowed to use this. So this it the expression.

View attachment 5619

I understand that I have to remove the implication but this doesn't really make sense to me and I have been using Youtube etc to try and get more info but just doesn't make sense as to how to apply the Laws of Logic. Help please?!

Regards
Werebilby

 

[solakis1]

Hello Everyone,

This is my first post, so please me kind :D

I have an assignment and have been stuck on this thing for 2 days. I have no problems working this out with a truth table but we are not allowed to use this. So this it the expression.
I understand that I have to remove the implication but this doesn't really make sense to me and I have been using Youtube etc to try and get more info but just doesn't make sense as to how to apply the Laws of Logic. Help please?!

Regards
Werebilby

Which are the laws of propositional logic that you can use

 

[werebilby]

Which are the laws of propositional logic that you can use

Ok so the laws I can use are :-

View attachment 5620

 

[solakis1]

Ok so the laws I can use are :-

O.k

Note some of the laws you have mentioned can be proved using the others.

Coming now to your problem the 1st step is to substitute the "=>" with "v"

Can you do that??

 

[werebilby]

O.k

Note some of the laws you have mentioned can be proved using the others.

Coming now to your problem the 1st step is to substitute the "=>" with "v"

Can you do that??

Ok let me try that again. So this is what I have worked out so far.

View attachment 5623

 

[solakis1]

Ok let me try that again. So this is what I have worked out so far.

http://mathhelpboards.com/attachments/discrete-mathematics-set-theory-logic-15/5623-laws-logic-q1-iv-1-png

I can see you are completely in the dark.

Before we continue is that a University course ??

 

[werebilby]

I can see you are completely in the dark.

Before we continue is that a University course ??

Yep it is a uni course. It's Ok I have submitted the assignment. I have completed most of the q's so this one was just 2 points. Yes I am completely in the dark about this. I have a fraction of info from the textbook, study book and the lectures but doesn't actually explain how this concept actually works.

 

[solakis1]

Yep it is a uni course. It's Ok I have submitted the assignment. I have completed most of the q's so this one was just 2 points. Yes I am completely in the dark about this. I have a fraction of info from the textbook, study book and the lectures but doesn't actually explain how this concept actually works.

I am going to complete the problem for you and then you tell me what you do not understand

Working on the LHS of the relation we have:
[(p=>q)=>(p^r)]=

= [(~pvq)=>(p^r)]=......by implication on (p=>q)

= ~(~pvq)v (p^r)=......by implication on [(~pvq)=>(p^r)]

= (~~p^~q)v (p^r)=......by D.Morgan on ~(~pvq)

= (p^~q)v (p^r) =........ by D.Negation on ~~p

= p^(~qvr)=.........by distributive property

[Note: if you expand p^(~qvr) using the distributive property you will get (p^~q)v (p^r)]

=p^(q=>r).........by implication on (~qvr)

So starting from the LHS of the identity we ended on the RHS of the idendity.

This is a usual procedure when proving identities ,whether in real Algebra or propositional Algebra

 

[soroban]

Show that:

$$(p \to q) \to (p \to r) \;\equiv\;p \wedge (q \to r)$$

$$\begin{array}{ccccc} 1. & (p \to q) \to (p \to r) && 1 . & \text{Given, LHS} \\ 2. & (\sim p \vee q) \to (\sim p \vee r) && 2. & \text{Implication} \\ 3. & \sim(\sim p \vee q) \vee (\sim p \vee r) && 3. & \text{Implication} \\ 4. & (p\: \wedge \sim q) \vee \sim p \vee r && 4. & \text{DeMorgan} \\ 5. & (p\:\vee \sim p) \wedge (\sim q\: \vee \sim p) \vee r && 5.& \text{Distributive} \\ 6. & t \wedge (\sim q \:\vee \sim p) \vee r && 6. & \text{Inverse} \\ 7. & \sim q \:\vee \sim p \vee r && 7. & \text{Identity} \\ 8. & \sim p \vee (\sim q \vee r) && 8. &\text{Comm. Assoc.} \\ 9. & \sim p \vee (q \to r) && 9. & \text{Implication} \\ 10. & p \to (q \to r) && 10. & \text{implication, RHS} \end{array}$$