1. ## Contradictions imply everything (a proof)

First, we'll prove that either a proposition P holds, or else it implies everything.

To do so, we will admit that we can quantify over and form sets of propositions and that we have the axiom of choice. We'll let T stand for a true proposition, whichever you like.

Code:
```1) Consider the set of propositions Q such that P or Q. This is non-empty since it at least contains T.
2) Consider the set of propositions R such that either P or else (R implies everything). This is non-empty since it at least contains the proposition "everything holds".
3) Pair the above sets and use the axiom of choice to pick out a Q and R respectively.
4) Assume P
5) Then the sets considered above are the same set: the universal set of propositions.
6) Thus choice picked out only one proposition above. In other words, Q = R
7) Assume Q
8) Assume R implies everything
9) Hence Q implies everything (from 6 and 8)
10) Everything (from 7 and 9)
11) If Q and R implies everything, then everything (from 7-10)
12) If P, Q and R implies everything, then everything (from 4-11).
13) (if Q and R implies everything) then P implies everything (reordering the antecedents of 12)
14) (P or Q) and (P or R implies everything) (from 1, 2 and 3)
15) P or (Q and R implies everything) (from 14 by distributivity)
16) P or P implies everything (from 13 and 15)```
As a corollary:

Code:
```1) Assume P.
2) Assume not-P.
3) Have P or P implies everything (from the proof above)
4) P implies everything (disjunctive syllogism from 2 and 3).
5) Anything. (from 1 and 4).```

2. Are P and Q collectively exhaustive? Like P and not P are

3. Originally Posted by fast
Are P and Q collectively exhaustive? Like P and not P are
P and Q cannot both be false. They may both be true.

4. Originally Posted by A Toy Windmill
Originally Posted by fast
Are P and Q collectively exhaustive? Like P and not P are
P and Q cannot both be false. They may both be true.
If P and Q cannot both be false, then one must be true rendering Q equivalent to not P, but if they may both be true, Q isn’t equivalent to not P.

5. I am used to using trtuth tables and symbolic logic.

OR function. A || B = C

A B C
F F F
F T T
T F T
T T T

A || !A = TRUE From the truth table for OR any true in gives a true out. If A is false !A is true if A is false !A is true.

6. Originally Posted by fast
Originally Posted by A Toy Windmill
Originally Posted by fast
Are P and Q collectively exhaustive? Like P and not P are
P and Q cannot both be false. They may both be true.
If P and Q cannot both be false, then one must be true rendering Q equivalent to not P
That doesn't follow. Not sure if it's just a silly mistake, but you might like to check its invalidity using a truth-table as steve suggests.

7. Originally Posted by A Toy Windmill
Originally Posted by fast
If P and Q cannot both be false, then one must be true rendering Q equivalent to not P
That doesn't follow. Not sure if it's just a silly mistake, but you might like to check its invalidity using a truth-table as steve suggests.
Is there any part of what I said you agree with?

8. Originally Posted by fast
Originally Posted by A Toy Windmill
Originally Posted by fast
If P and Q cannot both be false, then one must be true rendering Q equivalent to not P
That doesn't follow. Not sure if it's just a silly mistake, but you might like to check its invalidity using a truth-table as steve suggests.
Is there any part of what I said you agree with?
No. Everything was wrong.

I'm happy to help with basic logic, but given the OP, this isn't the appropriate thread.

9. Thanks for trying to do that.

One would need to restrict the propositions, though, otherwise a universal set of propositions runs into trouble (proper classes won't help, either). Do you have some restrictions in mind?

In any case, (P&¬P)->Q is more basic than the axiom of choice. But I feel like every attempt at proving it will fail to persuade those who don't see it.

10. Originally Posted by Angra Mainyu
Thanks for trying to do that.

One would need to restrict the propositions, though, otherwise a universal set of propositions runs into trouble (proper classes won't help, either). Do you have some restrictions in mind?
Assume that P and Q are carved out of subsets of X = {T, all propositions hold}. Line 5 then changes to

5) Then the sets considered above are the same set: X.

In any case, (P&¬P)->Q is more basic than the axiom of choice. But I feel like every attempt at proving it will fail to persuade those who don't see it.
The features of the proof I was going for are:

1) The absence of clear irrelevancies, cases where a Q comes out of nowhere. Instead, I consider "everything holds" (∀p. p).
2) No reliance on the definition of material implication. I don't assume that everything implies true, or that false implies everything.

So even if mathematicians are using an intuitive logic without 1 and 2 (it wouldn't surprise me), they are still forced to conclude that inconsistencies imply everything by axioms of logic to which they explicitly hold, specifically choice.

Obviously, I don't think this way, and I take it as basic that falsehoods imply everything. My favourite axiom system for classical propositional logic is this one:

1) P → (Q → P)
2) (P → Q) → (Q → R) → (P → R)
3) ((P → Q) → P) → P
4) ⊥ → P

Here, negation isn't even a basic connective. Contradictions are captured directly by a special proposition ⊥ whose single defining characteristic is that it implies everything. Negation is then defined as P → ⊥.

I like to justify this in a few ways. First, you can read ⊥ as an arbitrary absurdity like "I'm the Queen of Sheba", and so negation is just everywhere the colloquialism "if that's true, then I'm the Queen of Sheba!" Secondly, it understands negation as a kind of bet. To declare a proposition false, I must put up a forfeit should I be mistaken and the proposition turn out to be true. And that forfeit is ⊥, a proposition designed to render the entire enterprise pointless and boring by being the ultimate cheat code for all other problems, the master key to the dungeon.

As Hardy declared of proof by contradiction, the mathematician is willing to sacrifice the game (⊥).

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