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I'm trying to prove that $p \lor \lnot p$ is true using natural deduction. I want to do this without using any premises.

As it's done in a second using a truth table and because it is so intuitive, I would think that this proof shouldn't be too difficult, but I not able to construct one.

Git Gud
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Kasper
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  • I would assume that $p$ is any wff, am I correct? – Mr Mathster Jul 09 '13 at 00:35
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    Is it OK to make a premise and then discharge it? If not, I don't think it can be done. – Fred Rickey Jul 09 '13 at 00:40
  • There does exist something of a way to do this, in a certain sense. – Doug Spoonwood Jul 09 '13 at 12:33
  • Truth tables and proofs don't necessarily correspond all that well. The truth table for EEEpqrEpEqr only takes up 8 rows, but proving that in a natural deduction system, from what I know of them takes a lot more work. – Doug Spoonwood Jul 10 '13 at 03:53
  • You are going to need something in your natural deduction system that implies classical rather than intuitionistic logic. (See http://en.wikipedia.org/wiki/Intuitionistic_logic.) E.g., double negation elimination: $\lnot \lnot p \implies p$ or the law of excluded middle: $p \lor \lnot p$ (i.e., the statement you want to prove). – Rob Arthan Nov 22 '13 at 01:15

5 Answers5

12

I would like not to use de morgan law, as you would need to include that as a premise. I was thinking of this proof.

    • $\lnot (p \lor \lnot p) \quad \quad \quad (H)$
      • $p \quad \quad \quad \quad \quad (H)$
      • $p \lor \lnot p \quad \quad \; \;(\lor \text{I} 2)$
      • $ \bot \quad \quad \quad \quad \;\;(\lnot \text{E}1,3)$
    • $\lnot p \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \; (\lnot \text{I}2 - 4)$
    • $p \lor \lnot p \;\;\;\;\;\;\;\;\; \;\;\;\;\;\,(\lor \text{I}5)$
    • $\bot \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,(\lnot \text{E}1,6)$
  8. $p \lor \lnot p \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,\;\;\;(\bot \text{E}1-7)$
Kasper
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  • Yes, I see now. I also wasn't aware of what "natural deduction" meant. I have deleted my answer to reflect this. – Ryan Sullivant Jul 10 '13 at 07:16
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    This is not correct. Line 8 should read $\lnot\lnot(p \lor \lnot p)$. – Rob Arthan Nov 22 '13 at 00:59
  • The question is to prove the law of excluded middle. But you used the law of excluded middle in step 5 and 8. Why is this acceptable? – Mygod Apr 18 '17 at 16:34
  • @RobArthan it is correct. If they had used the negation elimination rule to derive 8, you were right. But they used RAA. That is exactly what RAA states. If you can derive falsity from negation of p, then you can conclude p. – Pharaz Fadaei Jun 26 '23 at 23:39
  • The proof explicitly states that step 8 uses $\bot$-elimination which says that if you can deduce $\bot$ from $\phi$ then you may conclude $\lnot\phi$. – Rob Arthan Jun 27 '23 at 21:41
3

The "standard" Natural Deduction proof of it is :

1) $\lnot (p \lor \lnot p)$ --- assumed [a]

2) $p$ --- assumed [b]

3) $p \lor \lnot p$ --- 2) : $\lor$I

4) $\bot$ --- 1), 3) : $\rightarrow$E

5) $\lnot p$ --- 2), 4) : $\rightarrow$I, discharging [b]

6) $p \lor \lnot p$ --- 5) : $\lor$I

7) $\bot$ --- 1), 6) : $\rightarrow$E

8) $p \lor \lnot p$ --- 1), RAA, discharging [a]

As expected, we need the classical rule for indirect proof (RAA, which is equivalent to Double Negation) to prove it, because Excluded Middle : $p \lor \lnot p$ is not intuitionistically valid.

Mauro ALLEGRANZA
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0

It's a proof by contradiction. Some of these answers are wrong because they (wrongly) miss the double negation that one would get by negating the original premise, that is, ¬(p∨¬p). Others are more complicated than they need to be. The strategy one employs is to exploit the fact that given an assumption one is always valid in introducing a disjunction, because we know from our assumption that at least one disjunct is true. So assume the negation of P ∨ ¬P, show that a contradiction arises for either P or ¬P, thus contradicting the original premise.

The proof is:

  1. ¬(P ∨ ¬P) --- (Hyp)
  2. P ------------- (Hyp)
  3. P ∨ ¬P ------- (2, ∨ introduction)
  4. ⊥ ------------- (1, 3, contradiction)
  5. ¬P------------ (2, 4, negation introduction) (Perhaps 1 is true, but 2 is false.)
  6. P ∨ ¬P ------- (5, ∨ introduction)
  7. ⊥ ------------- (1, 6, contradiction) (Since we only assume 1 at this point, we now must reject it)
  8. ¬¬(P ∨ ¬P)-- (1, negation introduction)
  9. P ∨ ¬P ------- (8, negation elimination)

QED.

Daniel Soutar
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  • They are not wrong. What they do, is precisely allowed by the RAA (proof by contradiction) rule. You use the negation introduction rule in combination with the double negation elimination (DNE) rule instead. Although RAA and DNE can be shown equal, the original system of Gentzen, includes RAA but not DNE. – Pharaz Fadaei Jun 26 '23 at 23:34
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First assume any thesis $\alpha$ as an assumption. Then show that the thesis can have the same scope as ApNp. Then infer C$\alpha$ApNp. Then prove $\alpha$ and use conditional-out. I use Polish notation, and I can give you formation rules if you like:

    • Cpp assumption
      • NApNp assumption
        • p assumption
        • ApNp 3 A-in
        • ⊥ 3, 4. K-in
      • Np 3-5 N-in
      • ApNp 6 A-in
      • ⊥ 2, 7 K-in
    • ApNp 2-8 N-out
  10. CCppApNp 1-9 C-in
    • p assumption
  12. Cpp 11-11 C-in
  13. ApNp 10, 12 C-out

Note that we did not prove |-ApNp at any point (though we did prove Cpp|-ApNp).

Doug Spoonwood
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-1

If you have the definition Cpq := ANpq and you have that A-commutes, and Cpp, then you can do this really quickly. Since we have Cpp, by the definition we have ANpp. By A-commutation we then have ApNp.

More formally, I'll first point out that natural deduction allows for definitions as well as uniform substitution on theses. So, again, here's our definition

Cpq := ANpq. (or C := AN for short)

First one of the lemmas:

    • p hypothesis

  2. Cpp 1-1 C-in

Now for another of the lemmas:

    • Apq hypothesis
      • p hypothesis
      • Aqp 2 A-in
    • CpAqp 2-3 C-in
      • q hypothesis
      • Aqp 5 A-in
    • CqAqp 5-6 A-in
    • Aqp 1, 4, 7 A-out
  9. CApqAqp

Now we have the following sequence of theses:

1 Cpp by the above

2 CApqAqp by the above

3 ANpp definition applied to thesis 1

4 CANppApNp 2 p/Np, q/p (meaning in thesis 2 we substitue p with Np and q with p)

5 ApNp 3, 4 C-out

Doug Spoonwood
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