My question: Can I deduce 5. - 7. from the premis — Pippen

Yes.

But this can't be true since it leads to contradictions. Just an example:

1. premise: 1
2. premise: 2
3. premise: 3
...
5. 1 <-> 1
6. 1 <-> 2 .
7. 1 <-> 3 .

Here basically 1 equals 2 and 3 which is false. Somehow it can't be possible to introduce a biconditional with only the premises and this must be true even for implication. But I just don't know what the right rule says to deduce an imlication A -> B. What are the precise conditions of such an introduction "p -> q"?

In natural deduction, if that's what we're doing here, the introduction rule for → is roughly:
________ (a)
P
.
.
.
Q
________
P→Q (a)

That is, if assuming P and then doing stuff gets you Q, then you can say P→Q. (You are then said to "discharge" your assumption of P, marked here as "a.")

I still have no idea what's going on here though, because your premises are obviously inconsistent, so you can deduce whatever you want.

@srap: Yes, we talking about natural deduction, take this example:

1. premise: 1
2. premise: 2
3. premise: 3
4. 1 <-> 1
5. 1 <-> 2
6. 1 <-> 3

Would the deduction of 4.-6. correct and if not (what I think) then why?

Well, truth is truth. All true statements are (truth-functionally) equivalent.
Btw, I hope "1", "2", and "3" are statements, not the numbers 1, 2, and 3.

But this can't be true since it leads to contradictions. — Pippen

Actually, you can derive contradictions in a natural deduction system if you have inconsistent premises. That just shows that you have an inconsistent set of premises, not that the system is unsound. Here's a basic example:

1. premise: A
2. premise: ~A.
3. A & ~A (1, 2, &-intro)

@Nagase: What you do in 3. is using the AND-introduction. My question is if I could instead introduce an implication "A -> ~A". I doubt that. I doubt that you can just with two premises P1 and P2 follow P1 -> P2 and vice versa.

And I am confident that I am right, because look at my argument above. It is consistent, I just assume the natural numbers 1,2 and 3, but with implication-introduction I can inter a contradiction which seems absurd, because 1, 2 and 3 are obviously not inconsistent.

I just assume the natural numbers 1,2 and 3 — Pippen

You really need to get the basics squared away, man. 1, 2, and 3 do not have truth-values. They cannot be premises.

What you do in 3. is using the AND-introduction. My question is if I could instead introduce an implication "A -> ~A". I doubt that. I doubt that you can just with two premises P1 and P2 follow P1 -> P2 and vice versa. — Pippen

Actually, that's pretty straightforward:

1. premise: A
2. premise: ~A
| 3. Assumption: A (assumption for ->-intro)
| 4. ~A (p2)
5. A -> ~A (->-intro).

I used | to indicate a sub-derivation.

Here basically 1 equals 2 and 3 which is false. — Pippen

One needs to maintain the distinction between terms, which represent objects in the domain of discourse, and formulas, which (speaking roughly) have truth values.

In your OP, the implication is that A, B and C are formulas, in which case the OP can make sense.

In your second post, you appear to want '1', '2' and '3' to represent numbers. If so then they are terms, not formulas, hence statements like

Premise: 1

or

1<--> 1

are meaningless because '1' is not a proposition, and hence cannot be a premise.

If you are not familiar with the notions of 'terms' and 'formulas' just think of them as 'nouns' and 'sentences'. A term is like a noun, and means nothing unless uttered in a sentence.

Ok, let me give a more complete example:

1. premise: Pippen has a left hand (LH).
2. premise: Pippen has a right hand (RH).
3. premise: It is not the case that iff LH then RH, ~(LH <-> RH).
4. LH <-> RH, contradiction to 4.
5. Since 1.-2. seem true, 3. must be false, and so it follows: LH <-> RH, but that's absurd because it basically says that my left hand can only exist with my right hand and vice versa which is obviously wrong.

So this is a valid proof in natural deduction? (Remember that I also shortcut several things here, so don't be too picky and formal) Do you see why I'm puzzled?

Once again, you have inconsistent premises. P1 says LH is true; P2 says RH is true; P3 says LH and RH don't have the same truth-value. But you just said they do.

The problem is P3. You're trying to say that having a left hand or a right hand doesn't necessitate having the other. That's not what P3 says.

Take another shot at it.

I can give hints if you get stuck.

5. Since 1.-2. seem true, 3. must be false, and so it follows: LH <-> RH, but that's absurd because it basically says that my left hand can only exist with my right hand and vice versa which is obviously wrong. — Pippen

Bold emphasis mine.

Yes, that is obviously absurd, but it's not what is going on here. The bold part is employing modal reasoning (can only exist), which is not expressible in propositional logic. In the case at hand (sorry!), you are only entitled to conclude that, given that LH and RH are both true, then they have the same truth-value, which is obvious (they both have the value true). Of course, from this you are not entitled to [](LH <-> RH), which would be an adequate formalization of the bold part (with "[]" denoting "It is necessary that...").

@nagase: Ah, I see. So when I have a premise P1 and a premise P2 I can always introduce P1 <-> P2, just as I could introduce P1 & P2, in the calculus of natural deduction, yes?

You might have to introduce the two conditionals first, just depends on what rules you have, but yes.

Here's another question that kind of belong to this topic: Is equality ("=") and biconditionality ("<->") the same and you could interchange both symbols?

No, = is what you use between objects, ⟷ is what you use between statements. Another symbol sometimes used for equivalence of statements is ≡.

Here's another look at the previous argument:

1. premise: Pippen has a left hand (LH).
2. premise: Pippen has a right hand (RH).
3. premise: It is not the case that iff LH then RH, ~(LH <-> RH).
4. LH <-> RH, contradiction to 3.
5. Since 1.-2. seem true, 3. must be false, and so it follows: LH <-> RH, but that's absurd because it says that my left hand exists only if my right hand exists and vice versa which is obviously wrong. Here nagase can't come with modality. How would u answer this one?

(3) is just inconsistent with (1) and (2).

I understand that you're frustrated, because you're experiencing some of the weirdness of propositional logic for the first time.

I'll make it worse for you.
Let LH="Pippen has a left hand" and RH="Pippen has a right hand."
See if you can prove the following: (LH→RH) ∨ (RH→LH), with no other premises.

You want to say that having a left hand doesn't entail having a right hand and vice versa. While that's undoubtedly true, it's not really something material implication (→) all by itself is suited to express, as you should have just discovered. Happily ☐(P→Q)∨☐(Q→P) is not a tautology.

Material implication is odd, and takes some getting used to, because it is always true when the antecedent is false. That's useful for doing lots of things, but it's confusing when you're trying to express things like causal connections, for instance.

Here's one way (I learned from Michael Dummett) to think about it: material implication is more like a conditional command than a conditional bet.

Suppose you make the following wager: "If Brazil is in the finals, they will win the World Cup." If Brazil is beaten in the semi-finals, your bet never takes effect and you are not rewarded. (Analogous to False.)

On the other hand, if your mother tells you, "Do not leave the house without wearing a coat," you could rephrase that as "If you leave the house, wear a coat." In this case, not leaving the house counts as obeying the command. (Analogous to True.)

Here is what I originally wanted to proof:

If I have my perception p1 of the world then I cannot have a second perception p2 of the world that is different of p1.

My simplified proof was this:

1. premise: Pippen
2. premise: p1
3. premise: p2
4. premise: ~(p1 <-> p2)
5. p1 <-> p2 | biconditional introdcution which leads to a contradiction
6. Since 1.-3. seems alright, 4. must be false, so it follows p1 <-> p2 and that means that p1 and p2 are equal, so I cannot have two different perceptions.

It seems to me now that I'd need a stronger logic to prove my point, right? Modal logic with identity? How would one prove the above that since it seems so trivial. This is important to me since I am a solipsist that believes that nothing exists without me perceiving it, but of course I have to make sure I just have my one and only perception and not multiple ones.

If I understand you correctly, I think what you want is a definite description.

You've done some math? You know how in many proofs there is the existence component and the uniqueness component? That's how definite descriptions work, at least according to Russell.

You could formalize the idea something like this:

∃x [ Perception(x) & Pippen's(x) & ∀y ( Perception(y) & Pippen's(y) → y = x ) ]

That says that there is something that is a perception and it's Pippen's (existence) and anything else that is a perception and Pippen's is the same as that thing (uniqueness).

Don't think of this as a proof though. It's just one way of formalizing what ordinary language expresses by means of definite articles like "the".

"My nose" is probably a definite description because usually I only have one of those; "my shoulder" probably isn't because usually I have more than one and you need context to determine whether I meant either or meant one specifically. You would still have to argue that a perception of the world is the sort of thing you can only have one of (at a time?). Formalizing that, as above, at best clarifies what you need to provide arguments for. (Philosophy rarely deals in incontrovertible proof.)