Ex falso quodlibet | Barış Özmen

Anything can be proven from contradiction.

Latin to English:

ex (out of) falso (false) quodlibet (whatever you want) -> from falsehood, anything follows

Formal Statement

If we accept that the statement P and its opposite are true at the same time, we can arrive any Q statement that we want following rules of logic. \[ \forall P, \forall Q, (P \land \neg P) \rightarrow Q \]

Formal Proof

Step	Statement	Justification
1	\(p \land \neg p\)	Hypothesis
2	\(p\)	Simplification, 1
3	\(p \lor q\)	Addition, 2
4	\(\neg p\)	Simplification, 1
5	\(q\)	Disjunctive Syllogism, 3,4

Logic rules used

Simplification (Conjunction Elimination)

If you have a conjunction (AND statement), you can derive either of its components individually.

Formally: \[(p \land q) \Rightarrow p\] and \[(p \land q) \Rightarrow q\] The intuition is straightforward: if you know two things are both true, you can conclude that each one is true individually.

For example:

"It's raining AND it's cold" (\(p \land q\))
Therefore "It's raining" (\(p\))
And also "It's cold" (\(q\))

Addition (Disjunction Introduction)

If you have any statement \(p\), you can always add another statement \(q\) to form a disjunction with OR.

Formally:

\[p \Rightarrow (p \lor q)\] Example:

"It's raining" (\(p\))
Therefore "It's raining OR I'm a giraffe" (\(p \lor q\))

Disjunctive Syllogism

If you have a disjunction (OR statement) \(p \lor q\) and you know that one of the disjuncts is false (\(\neg p\)), then the other disjunct (\(q\)) must be true.

Formally:

\[(p \lor q) \land \neg p \Rightarrow q\]

Example:

"Either it's raining OR I'm dreaming" (\(p \lor q\))
"It's not raining" (\(\neg p\))
Therefore, "I must be dreaming" (\(q\))

Implications

If there is a contradiction in a belief set (ideology), then adherents can be convinced of anything through logical inference.
- Marxism, Religions

References

https://en.wikipedia.org/wiki/Principle_of_explosion
video explanation
Popper's critique of dialectics