There are usually two types of nested proofs, indirect and conditional proofs. For all intensive purposes, I am going to give you the general layout of the proof. So the proof that I am giving you is in no way correct. I could, but that would take a bit longer to formulate a valid proof. But the main thing is to show you what the format is.

An indirect proof is done as follows.


1. We start off in step 1 with the proof all set up. The conclusion is ~B.
2.In second step, we begin the conditional proof. The first step in the conditional proof is to indent the proof and assume the opposite of ~B. Put down on line 3 the negation of the conclusion and cite it as an assumed premise. It is important to note that you can put down two forms of ~B in negated form, ~~B or B. It is more useful to put ~~B because you can use a double negation if you need to anyway. Why do you indent? The reason is because whatever you do within that indention cannot leave it until you discharge the conditional. More about this will come later.
3.Put down the negated conclusion ~~B
4.Once you have the negated conclusion, ~~B, use the inference and replacement rules (i.e. shown as lines 4,5,6) to derive a contradiction. The contradiction does not need to be the same as the conclusion or what you originally started out with, it just has to end in a conjunction of two identical but opposite expressions.
5.You can now discharge the indirect proof as whatever you want (the conclusion). But be sure to cite the proof will all the lines involved in the indention, so in our example IP 3-7.

A conditional proof is done as follows.

1. Like the indirect, we have the proof set up and the conclusion is B-->C
2.We set up the indention
3.We start with an assumed premise, which is the premise of the conclusion.
4.Through the use of inference and replacement rules, we eventually will get C, the consequent of the conclusion.
5.We can close out the indention and get the full conditional B-->C and cite the indention CP 3-7.

Now, when we have nested proofs, we are taking what we know as far as indirect and conditional proofs and taking it a step farther.



In a nested proof, we have an indention already. But suppose that even with an indention we need another variable that we just cannot get. In the example of a nested proof pic, we get to line 5 and realize that we need a nested proof. We CAN open another proof inside another indented proof AS LONG AS what is in that nested proof is discharged before moving on to solve the indented proof. Lines 6 -10 are the nested proof. The C is assumed and the conditional proof that we applied will help us solve the main conditional proof. Line 11 is the result of the nested proof which will help us reach line 14, which will help us discharge the proof line 15.

So to answer some of the questions taking all of this into account, I have given you an example of a nested conditional proof, as well as a recap of conditional and indirect proofs. The antecedent of the nested proof is coming from you arbitrarily. You can choose whatever you want because it is an assumed premise. The only thing required from you is to complete conjunction after you discharge the proof. So it basically comes out of nowhere, but appears after the proof has been discharged.

Can you do a nested proof for an indirect proof? Yes... but it is a bit harder to do. But suffice to say that it can be done.

I'm sure I missed a few points or was a bit vague in some areas. Please let me know if you need any more clarification because I am happy to help.

So, could you give me a proof that requires nested conditional proofs to solve? Or a proof that requires nest indirect proofs?

Couldn't that proof be done like this:

1. (A -> B) & C
2. ~(B & C) /B -> C
3. B CPA
4. C 1 Simp.
5. B -> C 3-4 CP

Is this correct?

Keep in mind that doing a conditional or an indirect proof is just like doing a regular nested proof. The only rules to keep in mind are that you cannot discharge the assumption if there are open nested proofs within it and every assumption must be discharged before you can finish the proof.

The reason why I didn't do a full proof is because I would have to construct the problem backwards in order to make sure it could work. It takes a fair bit of time to construct. A regular conditional or indirect proof is not a problem, like this BK proof ;

1.~[(A&~B) &~C]
2. ~(BvC) / ~A
3.~~A AP
4.~[A&(~B&~C] Association 1
5. ~[A&~(~~B&~~C] DeMorgan 4
6.~[A&~(BvC)] Double negation 5
7.~~ {~Av~~(BvC)] DeMorgan 6
8.~A v ~~ (BvC) Double Negation 7
9. ~~(BvC) Disjunctive Syllogism 3,8
10. BvC Double Negation 9
11. (BvC) & ~(BvC) Conjunction 10,2
12. ~A Indirect proof 3-11

For the conditional proof you want with a nested proof incorporated, you could do it this way. Note that the problem could be done a lot differently, but this is just to give you the example. Also, it may not be 100% correct because this is done off the bat. The bold part is the nested part of the proof.

1.(A-->B)-->~(E-->S)
2.(~EvS)vH
3.H-->J / B-->J
4.|B AP
5.|Bv~A Addition 4
6.|~AvB Communication 5
7.||A AP
8.||A&B Conjunction 4,7
9.||B Simplification 8
10.| A-->B Conditional proof 7-9
11.|~(EvS) Implication 1, 10
12.|H Disjunctive Syllogism 2,11
13.|J Modus Ponens 3,12
14.B-->J Conditional proof 4-13

As to you solution in post #4;

1.(A -> B) & C
2. ~(B & C) /B -> C
3. B CPA
4. C 1 Simp.
5. B -> C 3-4 CP

Yup! In line 3, you can assume the premise of B if you wish to open up a conditional proof. You can also simplify C from (A-->B) & C. And then you can also discharge the conditional assumption from lines 3-4. But to tell the truth, I just put down simple random sets. Conditional and indirect proofs make life a little easier when you take a problem like this with three variables into account. But with more of them, unlike the one I gave you that is for a simple conditional proof, it gets tougher.