Proving ‘Law of Excluded Middle’ in Fitch System

logicnatural-deductionpropositional-calculus

I'm taking a course from Stanford in Logic. I'm stuck with an exercise where I'm doing some proof. The Fitch system I'm given only allows

  • $ \land $ introduction and elimination
  • $ \lor $ introduction and elimination
  • $ \to $ introduction and elimination
  • $ \equiv $ introduction and elimination
  • $ \lnot$ introduction and
  • $ \lnot \lnot $ elimination

but not

  • $ \bot $ introduction and elimination

I've been struggling to prove the law of excluded middle (“p ∨ ¬p`) within this system. All of the proofs I've seen online make use of $\bot$ elimination to prove it by contradiction:

| ~(p | ~p)  (assumption)
| | p        (assumption)
| | p | ~p   (or introduction)
| | $\bot$
|
| | ~p       (assumption)
| | p | ~p   (or introduction)
| | $\bot$
|
| (p | ~p)   (proof by contradiction)

Since I can't use $\bot$ elimination, I was trying to do this with $ \lnot$ introduction:

| ~(p | ~p)  (assumption)
| | ???
| ~(p | ~p) => x
|
| | ???
| ~(p | ~p) => ~x
| p | ~p     (not introduction)

but I just can't get it. I can't figure out how to get x and ~x.

Is this provable without $\bot$ elimination? If so, what are the steps?

Background:

What I'm ultimately trying to prove is this: Given p ⇒ q, use the Fitch System to prove ¬p ∨ q. I'm able to prove that p ⇒ ¬p ∨ q and that ¬p ⇒ ¬p ∨ q, but I need to be able to plug p ∨ ¬p into the proof to be able to do 'or elimination'.

Best Answer

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

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

3) $A \lor \lnot A$ --- from 1) by $\lor$-intro

4) $\lnot A$ --- from the contradiction : 2) and 3) by $\lnot$-intro, discharging [b]

5) $A \lor \lnot A$ --- $\lor$-intro

6) $\lnot \lnot (A \lor \lnot A)$ --- from the contradiction : 1) and 5) by $\lnot$-intro, discharging [a]

7) $A \lor \lnot A$ --- from 6) by $\lnot \lnot$-elim.

For the different "flavours" of the negation rules in Natural Deduction you can see :

Your idea regarding the proof : $P \to Q \vdash \lnot P \lor Q$ is correct: you can use the "derived" rule :

$$\dfrac { } {\lnot P \lor P}$$

Alternatively :

0) $A \to B$ --- premise

1) $\lnot A$ --- assumed [a]

2) $\lnot A \lor B$ --- by $\lor$-intro

3) $\lnot (\lnot A \lor B)$ --- assumed [b]

4) $\lnot \lnot A$ --- from 1) with contradiction : 2) and 3) by $\lnot$-intro, discharging [a]

5) $A$ --- from 4) by $\lnot \lnot$-elim

6) $B$ --- from 0) and 5) by $\to$-elim

7) $\lnot A \lor B$ --- $\lor$-intro

8) $\lnot \lnot (\lnot A \lor B)$ --- from 3) with contradiction : 7) and 3) by $\lnot$-intro, discharging [b]

9) $\lnot A \lor B$ --- from 8) by $\lnot \lnot$-elim

Thus, from 0) and 9) :

$A \to B \vdash \lnot A \lor B$.

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