2

For any argument such as the following that we solve by "Proof by contradiction" method:

  1. (A v B)
  2. (A ⊃ C)
  3. (B ⊃ D)
    [ Therefore, (C v D)
  4. Assumption: ~(C v D)
  5. Therefore, ~C {from 4}
  6. Therefore, ~D {from 4}
  7. Therefore, ~A {from 2 and 5; Modus-tollens}
  8. Therefore, B {from 1 and 7; by Disjunctive syllogism}
  9. Therefore, ~B {from 3 and 6; Modus-tollens}
  10. Therefore, (C v D) {from 4; 8 contradicts 9 our argument is valid}

If we get one contradiction such as I got here with 8 and 9, how are we certain that this makes the argument valid? Don't we need to check if A, C, and D also contradict?

3

There is a point worth to be stressed in support of your concern; in logic we say that an argument is valid when we have a relation between premises and conclusion called : logical consequence.

We say that a sentence A is logical consequence of a set Γ of sentences, and we write : Γ ⊨ A, precisely when it is not possible that all the sentences in Γ are true and A is false.

In a Proof by Contradiction, we assume the negation of A : ¬A, and derive a contradiction; this shows that it is not possible that all the sentences in Γ and ¬A are simultaneously true.

But this implies that in every "situation" where all of Γ's are true (we say : all of Γ's are satisfied), ¬A is false; by properties of truth, if ¬A is false, then A is true; thus, we have established the relation of logical consequnce.


Bute there is a particular circumstance: when all of Γ's are already contradictory (i.e.unsatisfiable).

In this case it is not possible to find a "situation" where all of Γ's are simultaneously true (i.e.satisfied).

You can think at your example with an additional premise "on top" of (1)-(3) : (E ∧ ¬E).

What happen in this case ?

Well, applying the above definition of logical consequence, we have that :

if there is no "situation" that satisfies every member of Γ, then for any sentence A, it is vacuously true that Γ ⊨ A.

Thus, in presence of a contradcitory set of premises, like (1)-(3) with the additional (E ∧ ¬E), it is still true that they implies (C v D), simply because they, being contradictory, impliers everything.

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2

The propositions ~C, ~D, and ~A were not contradictions of any propositions that you derived. For example, up to those points, you had not derived C, D, or A. If you would have, you would have arrived at a contradiction sooner than when you arrived at B and ~B.

For your first question, here's a much a simpler problem, and with words, so it's easier to get the idea.

1. A          (It rains.)
2. A ⊃ B      (If it rains, then it will be wet.)
 [ ∴ B        (It will be wet.)
3. - Asm: ~B  (What happens if we assume that it won't be wet?)
4. - ∴ ~A     (It doesn't rain.)
5. B          (From 3. "It doesn't rain" contradicts "It rains.")

By assuming the negation of the conclusion, you arrive at a proposition that contradicts a true premise. (We assume our premises are true when the validity, and not the soundness, of the argument is in question. If the soundness were in question then we would also want to know if the premises really are true, but the soundness is not in question right now.) If ~B leads to a contradiction, then it must be false. In classical logic, if ~B is false then B is true.

What if we have all true premises and don't arrive at a contradiction by the same method?

1. C          (It snows.)
2. A ⊃ B      (If it rains, then it will be wet.)
 [ ∴ B        (It will be wet.)
3. - Asm: ~B  (What happens if we assume that it won't be wet?)
4. - ∴ ~A     (It doesn't rain.)

It means the premises didn't entail the conclusion, like it did above, and the argument is invalid.

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  • Yes, it could have been C, D or A instead of B in my example. But when we know ~B and B contradict, then is it necessary that if we derive C, D, or A they will also contradict? But we just stop at ~B and B without checking any further contradictions and say our proof is done. – cpx Oct 19 '14 at 9:36
  • That's correct, all you need is one contradiction and you know your "assumption" is false, and that the blocked-off conclusion now true. You might accidentally continue to derive more contradictions because you didn't see your first derived contradiction, but they make no difference. One contradiction is one reduction to absurdity, and all you need is one reduction to absurdity. – Dise Oct 19 '14 at 9:49
  • Can we somehow show that it really takes one contradiction and all the others aren't necessary? – cpx Oct 19 '14 at 10:21
  • You can show yourself by replacing the symbols with natural statements and realizing precisely everything that you're saying when you're going through the proof. When you know what you're saying, you'd realize that any need to go beyond one contradiction to check if there are others is dumbass. – Dise Oct 19 '14 at 10:53
  • Suppose if I just sneak up and add another non-sense line into the argument and if, luckily, we don't use that line and still get one contradiction and stop there, then I think we cannot still assume our argument is valid. That's why I was wondering why not go through everything we have to arrive at more contradictions. – cpx Oct 19 '14 at 13:40
1

In a basic logic system such as this one, all contradictory statements are logically equivalent. Any given contradiction can be substituted for any other. Thus, once you have B & ~B you know that you could have produced any other contradiction as well (such as A & ~A).

It's also worth noting that formal logic is a rule-based system, and in this case, the rule is that any contradiction stemming from an assumption entitles you to establish the negation of the assumption. We may believe that this reflects a basic aspect of how the universe operates, but in the context of the system, it works that way because that's how the system is defined.

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