Question 1090173: How do I solve ~Q → (L → F), Q → ~A, F → B, L, therefore, ~A v B with either reductio ad absurdum or conditional proof?
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
To do a reductio ad absurdum argument, I'm going to use a proof by contradiction.
Proof By Contradiction:
This is an informal paragraph style proof:
Assume the opposite of the conclusion is true
The conclusion is ~A v B
The opposite of the conclusion is ~(~A v B) = ~~A & ~B = A & ~B through the use of De Morgan's Law
If we assume A & ~B is true, then A is certainly true and so is ~B (Keep the fact that ~B is true in mind). Both parts of a conjunction must be true if the whole thing is true.
If A is the case, then so is ~~A
By modus tollens, we can arrive that ~Q is also the case.
Then through modus ponens, we can use ~Q and ~Q -> (L -> F) to find that L -> F is the case
Now use the premise L and L -> F to find that F is true (use modus ponens again)
Finally use F and F -> B to find B is true (another application of modus ponens)
But wait, earlier I said that ~B was true (In a previous note above). So how can B also be true at the same time? This is where the contradiction lies. Therefore, the expression ~(~A v B) cannot be true so the original ~A v B must be true.
So in short, we've assumed a condition -- assumed that ~(~A v B) was true -- but it led to an absurdity of B and ~B being true at the same time.
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Here's a more formal way to do the proof using a derivation table
| Number | Statement | Lines Used | Reason |
|---|
| 1 | | ~Q -> (L -> F) | | | | 2 | | Q -> ~A | | | | 3 | | F -> B | | | | 4 | | L | | | | :. | ~A v B | | | | 5 | ~(~A v B) | | Assumption for Indirect Proof | | 6 | ~~A & ~B | 5 | De Morgan's Law | | 7 | A & ~B | 6 | Double Negation | | 8 | A | 7 | Simplication | | 9 | ~B | 7 | Simplication | | 10 | ~~A | 8 | Double Negation | | 11 | ~Q | 2,10 | Modus Tollens | | 12 | L -> F | 1,11 | Modus Ponens | | 13 | F | 12,4 | Modus Ponens | | 14 | B | 3,13 | Modus Ponens | | 15 | B & ~B | 14,9 | Conjunction | | 16 | | ~A v B | 5-15 | Indirect Proof |
Note: "Indirect Proof" is another term for "Proof by Contradiction"
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If you want to use a conditional proof, then you first need to realize that ~A v B is logically equivalent to A -> B through the material implication rule.
An informal proof would go like this:
Start by assuming A. We can't directly jump to B with A since that would be too easy and we cannot use the conclusion as part of the premises. That would lead to cicular reasoning.
Instead turn A into ~~A (double negation). That would allow us to pull out ~Q by modus tollens. As you can probably guess by this point, the steps are very similar to those shown above.
We use ~Q to get L -> F (modus ponens)
We use L and L -> F to get F (modus ponens)
We use F and F -> B to get B (modus ponens)
This is where the proof differs than the section above. Instead of a contradiction, we have essentially arrived at the proper conclusion we want based on the assumption provided.
Basically we started with A and we did a bunch of logical steps to arrive at B. If we assume A is true, then somewhere down the line B is true. So naturally if A, then B follows. That is written as A -> B which is equivalent to ~A v B
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Here's a formal derivation table
| Number | Statement | Lines Used | Reason |
|---|
| 1 | | ~Q -> (L -> F) | | | | 2 | | Q -> ~A | | | | 3 | | F -> B | | | | 4 | | L | | | | :. | ~A v B | | | | 5 | A | | Assumption for Conditional Proof | | 6 | ~~A | 5 | Double Negation | | 7 | ~Q | 2,6 | Modus Tollens | | 8 | L -> F | 1,7 | Modus Ponens | | 9 | F | 8,4 | Modus Ponens | | 10 | B | 3,9 | Modus Ponens | | 11 | | A -> B | 5-10 | Conditional Proof | | 12 | | ~A v B | 11 | Material Implication |
Note how this derivation table has a lot in common with the previous table.
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