SOLUTION: How do I solve ~Q → (L → F), Q → ~A, F → B, L, therefore, ~A v B with either reductio ad absurdum or conditional proof?

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.

-------------------------------------------------

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"
==============================================================================================================
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

-------------------------------------------------

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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SOLUTION: How do I solve ~Q &#8594; (L &#8594; F), Q &#8594; ~A, F &#8594; B, L, therefore, ~A v B with either reductio ad absurdum or conditional proof?

SOLUTION: How do I solve ~Q → (L → F), Q → ~A, F → B, L, therefore, ~A v B with either reductio ad absurdum or conditional proof?