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How does the following proof of argument is valid and makes sense?

1. (R • C)  [It is raining and It is cloudy]
2. ~R [It is not raining]  
Therefore, S [It is snowing]

According to proof by contradiction,

3. Let ~S as our assumption
4. R {Using AND Simplification rule break 1)
5. C {Using AND Simplification rule break 1)
6. S {From 3, 2 contradicts 4}

It says if its not raining then it is snowing which doesn't make sense to me but how does mathematically the proof comes valid?

  • Be aware ! It says : "if it is raining and ... and it is not raining, then it is snowing". – Mauro ALLEGRANZA Oct 1 '15 at 15:05
  • So, when in any argument the premises contradict the conclusion have no effect? I think the argument should regarded as invalid in such cases. It's no different than just saying 1.) R 2.) ~R 3.) X. – user963241 Oct 1 '15 at 15:32
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    ??? From R you cannot derive ¬ R with a valid rule. But if the premises are contardictory, i.e. contain a contradiction, like R ∧ ¬ R, you can derive absolutely anything : S, ... and also R as well as ¬ R. – Mauro ALLEGRANZA Oct 1 '15 at 15:38
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It is an application of the Reductio Ad Absurdum rule.

RAA is a formulation of the principle of proof by contradiction: if one derives a contradiction from the hypothesis ¬ϕ, then one has a derivation of ϕ (without the hypothesis ¬ϕ) :

assume ¬ ϕ --- step 3)

derive a contradicition : --- in this case steps 2) and 4)

conclude with "rejecting" the assumption, i.e. with ϕ --- step 6).


If we agree on the truth-functional definition of connectives (as per classical logic), the rule is valid : if we derive a contradiction from the assumption ¬ ϕ, this implies that ¬ ϕ is false.

Then, by double Negation, ϕ must be true.

In fact, the rule is equivalent to the Double negation eleimination rule :

¬¬ ϕ ⊢ ϕ.


Intuitionism rejects as "fallacoius" this type of argument : in Intuitionistic Logic RAA and Double Negation are not valid rules of inference.


See also Ex falso quodlibet (or Principle of explosion) :

any statement can be proven from a contradiction.

The two premises R ∧ C and ¬ R are contradictory, while S is not present in the premises; thus, the proof shows that from contradictory premises we can derive absolutely anything as a conclusion.

With this logic law we can derive S in a simpler way :

1) and 2) : as above

3) R --- from 1)

4) R ∧ ¬ R --- from 3) and 2) by Conjunction introduction

5) ⊢ R ∧ ¬ R → S --- Ex falso quodlibet

6) S --- from 4) and 5) by Mmodus ponens.

Note : Ex falso quodlibet is intuitionistically valid. For logics that do not allow it, see Paraconsistent Logic.

  • Yes, but how does the conclusion "It is snowing" makes sense at all? I think it would be more appropriate if it were "It is not cloudy". – user963241 Oct 1 '15 at 13:42
  • @user963241 - because the two premises R ∧ C and ¬ R are (already) contradictory. – Mauro ALLEGRANZA Oct 1 '15 at 13:48
  • In that case I could put anything non-sense in conclusion and the argument would valid. – user963241 Oct 1 '15 at 13:52
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    ... the second approach is to recognize that a conclusion like "it is snowing" is only "useful" if the axioms are true. You can always explore hypothetical situations, like the one here where it is effectively raining and not raining, but they don't apply to reality until you actually find a place where the proposition and its negation may safely both be assumed to be true (which there is a philosophical law from Aristotle's era that presumes this can never possibly happen in the real world). – Cort Ammon Oct 1 '15 at 15:48
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    You can always rephrase this as an "if" statement, making the conclusion result from the axioms. You can say "If (it is raining and is cloudy) and (it is not raining) then (it is snowing)" and defend that statement with the proof you wrote above. However, you will find it difficult to find situations where you can apply that (true) statement because it is only applicable if it is raining and is not raining. – Cort Ammon Oct 1 '15 at 15:50
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Kant turned on this issue at various points, for example in the Critique of Pure Reason, A 789-91|B 817-19. The main problem of these apagogical arguments is, that there can be subjective and/or incomplete presuppositions, for example because we forgot one or simply cannot know about it at all. The outcome is that the (logical) truth of this proof can be seen within the laws of logic, but the (empirical) truth cannot be understood in a way it could produce knowledge in sensu strictu. There is no possible way to say anything more for the correctness of this kind of proof than "It is logically correct."

The empirical truth of these proofs, understood as knowledge about the world, is based on the empirical truth and completeness of the presuppositions. This is why it can be argued by these means, but only for the possibility of the presuppositions if the conclusion includes one of them (= not contradictionary), not the truth.

It is a basic problem of this way of proof. In respect of the logical correctness of this proof, see above. Especially the point "everything can be followed out of a contradiction".

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Let's try to make it make sense to you.

I'm sure you have had times when it was raining and cloudy.

I'm sure you have also had times when it was not raining.

Have you ever encountered the situation that it was (raining and cloudy) and at the same time it was (not raining)? I don't think so. It's not possible.

Have you ever encountered the situation that it was (raining and cloudy), and at the same time it was (not raining), AND at the same time it was not snowing? I don't think so. You never have, and you never will. It's just as impossible.

So every time when it is at the same time (raining and cloudy), and (not raining), it is snowing. It will never happen that this isn't true.

OK, normally when we correctly say "if A and B then C" you will find that there is a logical connection between A, B and C. C is often caused by A and B. For example A = "I am walking through the rain", B = "I have no umbrella", C = "I get wet".

There are cases when A or B could be irrelevant: A = "I walk through the rain without an umbrella", B = "Today is Thursday", C = "I get wet". This is still true, although B is totally irrelevant: But whenever "if A then C" is true then "if A and B then C" will also be true, no matter what B is.

Another unexpected situation happens when C is always true. A = "I walk through the rain", B = "I have no umbrella", C = "the name of the day ends in Y". It's true because C is true anyway, so "if A and B then C" is true whatever A and B are. (An interesting one if B = "Today is Wednesday". C is always true, but it is also true because of B).

And finally the situation that you have here: A and B cannot be true at the same time. In that case "if A and B then C" is true, no matter what C is.

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