/sci/ - Science & Math » Thread #11611334

9KiB, 330x153, images.png

View Same Google iqdb SauceNAO

LOGIC

Anonymous Tue 28 Apr 03:09:25 2020 No.11611334 View Reply Original Report

Quoted By: >>11611430 >>11611455 >>11611940 >>11611995 >>11612524 >>11612615

Can anyone tell me how to prove this using propositional logic?

-(P & -Q) : P -> Q

I've been banging my head against a wall trying to figure it out. I know its basic I just cant figure out how I'm supposed to approach it.

Anonymous

Anonymous Tue 28 Apr 2020 03:12:06 No.11611339 Report

Quoted By:

Replace P and Q by propositions that makes sense to us.

Anonymous

Anonymous Tue 28 Apr 2020 03:18:30 No.11611354 Report

Quoted By: >>11611491

at least it's true

~(P & ~Q) / P -> Q
https://programming.dojo.net.nz/study/truth-table-generator/index

Anonymous

Anonymous Tue 28 Apr 2020 03:32:28 No.11611403 Report

Quoted By: >>11611416

I don't understand what's the proposition you're trying to prove.

Anonymous

Anonymous Tue 28 Apr 2020 03:34:18 No.11611416 Report

Quoted By: >>11611437

>>11611403

-(P & -Q) is the premise. - means NOT and & is a conjunction

P -> Q is the conclusion and is a conditional. It means P implies Q or If P then Q

Anonymous

Anonymous Tue 28 Apr 2020 03:38:22 No.11611430 Report

Quoted By: >>11611435

>>11611334
I don't know logic but it looks like its saying P is incompatible with negation of Q, then you can deduce Q from existence of P because Q is vital for P.

Anonymous

Anonymous Tue 28 Apr 2020 03:39:47 No.11611435 Report

Quoted By:

>>11611430
(so being alive is incompatible with negation of having blood in your body, so anyone living surely has blood)

Anonymous

Anonymous Tue 28 Apr 2020 03:42:05 No.11611437 Report

Quoted By: >>11611451 >>11611491

>>11611416
Then your proposition is impossible to prove because it's not true. You can see this by noting that $(p \implies q) \iff \neg p \lor q$

Anonymous

Anonymous Tue 28 Apr 2020 03:46:38 No.11611451 Report

Quoted By: >>11611454 >>11611458 >>11611461

>>11611437
I think you just solved it unintentionally because -(P & -Q) <=> -P v Q

Anonymous

Anonymous Tue 28 Apr 2020 03:47:40 No.11611454 Report

Quoted By: >>11611458 >>11611461

>>11611451
Via DeMorgans

Anonymous

Anonymous Tue 28 Apr 2020 03:47:43 No.11611455 Report

Quoted By:

>>11611334
write out truth tables for values of p and q, this is basically a definition

Anonymous

Anonymous Tue 28 Apr 2020 03:47:52 No.11611456 Report

Quoted By:

Define P->Q because as far as I know, ¬(P&¬Q) is exactly how it's defined.

Anonymous

Anonymous Tue 28 Apr 2020 03:48:54 No.11611458 Report

Quoted By:

>>11611451
>>11611454
yeah yeah, sorry, I'm a retard, there you have it

Anonymous

Anonymous Tue 28 Apr 2020 03:50:55 No.11611461 Report

Quoted By: >>11611913

>>11611451
>>11611454
Nope. The tautology is $\neg(P \land Q) \iff \neg P \lor \neg Q$ , and not $\neg(P \land Q) \iff \neg P \lor Q$ which is what you would need.

Anonymous

Anonymous Tue 28 Apr 2020 03:53:10 No.11611462 Report

Quoted By:

Its -(P & -Q) not -(P & Q) so the tautology I said was correct per https://math.stackexchange.com/questions/924358/negation-outside-of-a-parenthesis

Anonymous

Anonymous Tue 28 Apr 2020 04:07:27 No.11611491 Report

Quoted By:

>>11611437
>it's not true
nope, it's a tautology >>11611354

Anonymous

View Same Google iqdb SauceNAO proof.png, 17KiB, 953x207

Anonymous Tue 28 Apr 2020 06:33:27 No.11611744 Report

Quoted By: >>11612010

There u go.
Link to proof: https://boxprover.utr.dk/index.html?proof=%25abbrev%0Aret%20%3A%20%7Bp%7D%7Bq%7D%0A%20%20proof%20(~%20(p%20%2F%5C%20~%20q%20)%20%2C%20%7C-%20p%20%3D%3E%20q)%20%3D%0A%5Bp%5D%5Bq%5D%0A~%20(p%20%2F%5C%20~%20q)%20premise%3B%20%5Bl1%5D%0A(%0A%20%20%20%20p%20assumption%3B%20%5Bl2%5D%0A%20%20%20%20(%0A%20%20%20%20%20%20%20%20~%20q%20assumption%3B%20%5Bl3%5D%0A%20%20%20%20%20%20%20%20p%20%2F%5C%20~%20q%20by%20con_i%20l2%20l3%20%3B%20%5Bl4%5D%0A%20%20%20%20%20%20%20%20bot%20by%20neg_e%20l4%20l1%0A%20%20%20%20)%3B%20%5Bb2%5D%0A%20%20%20%20~%20~%20q%20by%20neg_i%20b2%20%3B%20%5Bl5%5D%0A%20%20%20%20q%20by%20nne%20l5%0A)%3B%20%5Bb1%5D%0Ap%20%3D%3E%20q%20by%20imp_i%20b1%0A.

Anonymous

Anonymous Tue 28 Apr 2020 08:07:05 No.11611904 Report

Quoted By:

You'll get better help on /lit/

Anonymous

Anonymous Tue 28 Apr 2020 08:13:38 No.11611913 Report

Quoted By:

>>11611461
Can this brainlet please die

Anonymous

Anonymous Tue 28 Apr 2020 08:36:53 No.11611940 Report

Quoted By:

>>11611334
P->Q = !P or Q
!(P&!Q) = !P or Q

Anonymous

Anonymous Tue 28 Apr 2020 09:18:49 No.11611995 Report

Quoted By:

>>11611334
that is the definition of implies
you don't prove definition
unless you want a philosophical understanding of why a causal link between two events is defined that way in propositional logic, but that would take you beyond propositional logic

Anonymous

Anonymous Tue 28 Apr 2020 09:26:04 No.11612010 Report

Quoted By:

>>11611744
Which language is boxprover?
I'm surprised they don't tell on that page.

Anonymous

Anonymous Tue 28 Apr 2020 14:01:39 No.11612524 Report

Quoted By:

>>11611334
Not formal, but
the first statement means that either P is false or Q is true. If P is false then P -> Q is always true. If Q is true then P -> Q is also true. Done.

Anonymous

Anonymous Tue 28 Apr 2020 14:52:11 No.11612615 Report

Quoted By:

>>11611334
~(P & ~Q)
DeMorgan
~P V Q
^ is the definition of P -> Q

Capcode	All Only User Posts Only Moderator Posts Only Admin Posts Only Developer Posts
Show Posts	All Only With Images Only Without Images
Deleted Posts	All Only Deleted Posts Only Non-Deleted Posts
Ghost Posts	All Only Ghost Posts Only Non-Ghost Posts
Post Type	All Only Sticky Threads Only Opening Posts Only Reply Posts
Results	All Grouped By Threads
Order	Latest Posts First Oldest Posts First

Your latest searches