In the case of a contradicted conditional given:

[pandaBee]

Homework Statement

If I have a given in a proof in the form:
A or B or C ... etc. etc. and if I choose to approach this given in a case by case basis: (assuming one of the A,B,C... one at a time) and if one or more of the assumptions contradicts some other given in the proof does that mean that I can simply ignore that possibility?

Homework Equations

The Attempt at a Solution

For example: if A or B or C

case1: Assume A
Contradiction
Therefore B or C

I was under the impression that in order for a case-by-case proof to suceed you need to ensure that all the possible cases come out to be true, however what about this situation:

Assume not A

A or B or C
case1: A
However, not A, contradiction

case 2: B
...
Goal is true
case 3: C
...
Goal is true

 

[vela]

Are you talking about a statement like: If A or B or C, then D? To show that statement is true, you need to show A→D, B→D, and C→D.

You can't infer that if A is false, then B or C is true. It's possible for A, B, and C to all be false, right?

 

[LCKurtz]

I think he might be thinking Given (A or B or C) and not(A) then B or C.

 

[pandaBee]

Kurtz is correct, the given would be of a form
(A or B or C)
Though it doesn't have to have three possibilities, it could have as arbitrarily many, it doesn't really matter for the sake of this discussion, just as long as there are 2 or more.

 

[RUber]

I was under the impression that in order for a case-by-case proof to suceed you need to ensure that all the possible cases come out to be true, however what about this situation:

Assume not A

A or B or C
case1: A
However, not A, contradiction

You do need to ensure that each case comes out true. That means that ( A or B or C) can be made true in each case. If you are given A is false, you will not have a case where you assume A is true.

Normally if you have n statements, you will have $2^n$ possible combinations of T/F in a truth table. Constraints such as A is false simply reduce the number of potential cases to look at. Now there are only $2^2$ combinations to consider. Only one of which makes the first statement (A or B or C) false.