1. ## Rule of Replacement

I am in Intro to Logic and I cannot for the life of me figure out these 4 problems. Can someone please help me out??? I am using symbles where I can, bit I need to use words for others.

1) Without using Disjunctive Syllogism
M v W
~W
There-for M

2) Without using absorption
A > (A and D)
there-for A > D

3) T if and only if (L > K)
~N and K
there-for T

4) Without the Distjunctive form of Equiv
M and W
There-for M if and only if W

2. You need to show some work. We'll help you get unstuck, but we don't do your homework for you. Also, generally you want to limit a thread to at most two problems. Just for future reference. So, what do you have so far?

3. Originally Posted by Ackbeet
You need to show some work. We'll help you get unstuck, but we don't do your homework for you. Also, generally you want to limit a thread to at most two problems. Just for future reference. So, what do you have so far?
All for the above is policy for this forum. But I need to add a comment.
I find this question very odd. You do not give us any help with definitions.
I been around this material a long time but do not know what the law of absorption is.
What is possibly the point of not using Disjunctive Syllogism?

I suggest you do all of these with truth tables.

I think the absorption laws are equivalent to the idempotency of OR and AND.

I agree that truth tables would work fine - I also think #3 would get a bit unwieldy, since it has 4 variables. There might be constraints on how ABP is supposed to do the proofs.

5. Plato,
I apologize for the definitions, this is my first time using this site for help and unlike math, the symbols I use are not necessarily recognized by everyone.

Key
v = Or
> = If...Then
~ = Nor or Negate
And is simply and
There For is the conclusion

The law of absorption is a part of the 9 rules of inference and defined as: p > q there for P (p and q) according to the 13th edition of Intro to Logic By Copi and Cohen.
The reason why we were instructed not to use Disjunctive Syllogism is because it makes the problem to proof to easy to solve and our professor wanted to make us work for the answer.

Ackbeet,
I apologize for presenting the question the way I did. I did not mean to imply that you do it for me. I will keep it to the first two questions.

This is what I have so far.

1) M v W
2) ~W There for M
3) ~W 2, Trans
4) W v M 1, Conj
5) [(W > M) and (M>W)] 4, Assoc
6) W > M 5, Simp
7) M 6, M.P.

Without using absorption
1) A > (A and D) /there for A > D
2) (A > A) and D 1, EXP
3) (A > A) and (D > A) 1,2 Add
4) A > D 3, H.S

I do not know how to do this one without using absorption, and I am pretty sure that if either one of you were here I would get a big old slap in the back of the head, but if you could help me out I would appreciate it.

6. Originally Posted by ABP
The law of absorption is a part of the 9 rules of inference and defined as: p > q there for P (p and q) according to the 13th edition of Intro to Logic By Copi and Cohen.
That is not in my copy of Copi’s Symbolic logic, 3ed.
But using some of his other notations here is what I would do.
$\displaystyle A\, \Rightarrow \,\left( {A \wedge D} \right) \equiv \neg A \vee \left( {A \wedge D} \right)$ Material Equivalance.
$\displaystyle \neg A \vee \left( {A \wedge D} \right) \equiv \left( {\neg A \vee A} \right) \wedge \left( {\neg A \vee D} \right)$, distribution.
$\displaystyle \left( T \right) \wedge \left( {\neg A \vee D} \right)$, tautology.
Therefore, $\displaystyle \left( {A\, \Rightarrow D} \right)$ Material Equivalance