Sunday, 17 June 2007

Proof of DeMorgan's Laws

I couldn't find a formal proof of deMorgan's laws anywhere on the internet. If I could have, it might have saved me a lot of time as I had to end up doing it myself. Here they are in case they prove useful to someone.

Just click on the images to see them fully. (They were constructed using Fitch, a program that comes with the text Language, Proof & Logic.)

I'm not saying that the proofs are the most elegant or shortest. But they are all complete and correct.

Proof of ¬(P∧Q) → ¬P∨¬Q:



Proof of ¬(P∨Q) → ¬P∧¬Q:



Proof of ¬P∨¬Q → ¬(P∧Q):



Proof of ¬P∧¬Q → ¬(P∨Q):



Notation:

¬ Negation 'not'
∧ Conjunction 'and'
∨ Disjunction 'or'
→ Conditional 'implies/only if'
⊥ Contradiction

Each vertical line (bar the outermost one) represents a subproof. Each subproof starts with an assumption.

7 comments:

Bao said...

you're awesome!! thank youuu

Unknown said...

Could you please explain any one of these more fully? I have just started Intermediate Logic (the Nance book) and am interested in this proof. The book doesn't require it (we use shorter truth tables to show equivalence). I am not sure what the rationale was for some of your steps. Thank you.

Unknown said...

Thank you so much!! These are so helpful, just even in understanding how deMorgan's Laws work.

Unknown said...

Hello

Can you explain your fourth proof? It's unclear to me what rules you used there.

Unknown said...
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John Doe said...
This comment has been removed by the author.
John Doe said...

Better is 4.
1 | ~P^~Q
. -------
2 | ~P ^E:1
3 | ~Q. ^E:1
4 | | PvQ
. | -----
5 | | | P
. | | ---
6 | | | ⊥ ⊥I:2,5
7 | | | Q
. | | ---
8 | | | ⊥ ⊥I:3,7
9 | | ⊥ vE:4,5-6,7-8
10| ~(PvQ) ~I:4-9