propositional calculus

[chrislav]

prove in propositional calculus the De morgan rule;

[math]\neg(p\wedge q)\implies \neg q\vee\neg q[/math]
You may consider the above as challenging problem
Do not use the other De Morgan rule:
[math]\neg(p\wedge q)\implies \neg p\vee \neg q[/math]

 

[lex]

Can you check the question, or post the source.
(When p is False, and q True, it evaluates to False).

 

[blamocur]

prove in propositional calculus the De morgan rule;

[math]\neg(p\wedge q)\implies \neg q\vee\neg q[/math]
You may consider the above as challenging problem
Do not use the other De Morgan rule:
[math]\neg(p\wedge q)\implies \neg p\vee \neg q[/math]

Are you allowed to use truth tables?

 

[chrislav]

Can you check the question, or post the source.
(When p is False, and q True, it evaluates to False).

1st of all ,when i said the other De Morgan rule i meant the rule:

[math]\neg (p\vee q)\implies (\neg p\wedge \neg q)[/math]
Now coming to your question we have : p is false and q is true hence p and q is false and [math]\neg(p\wedge q)[/math] is true .This is for the LHS
Now for the RHS:
Since p is false [math]\neg p[/math] is true
And since q is true [math]\neg q[/math] is false
Hence :[math]\neg p\vee\neg q[/math] is true
Therefor we have : LHS Is true and RHS TRUE as well
But true impling true is true

Are you allowed to use truth tables?

No because true tables can only establish that the above is provable according to the central theorem of propositional calculus.
They do not provide the actual proof

 

[pka]

Yours's is a confusing question. I have been involved with symbolic logic since 1964, as a student and as university faculty teaching it. But i have never encountered "central theorem of propositional calculus". So here is from Copi's textbook.
"Any conditional with a true antecedent and a false consequent must be false. Hence any conditional, if p then q, is is known to be false in case the conjunction [imath] p\wedge\neg q[/imath] is known to be true. Copi also gives this summery:
A false statement implies any statement is true.
A true statement is implies by any statement.


[imath][/imath][imath][/imath]

 

[Dr.Peterson]

Now for the RHS:
Since p is false [math]\neg p[/math] is true
And since q is true [math]\neg q[/math] is false
Hence :[math]\neg p\vee\neg q[/math] is true
Therefor we have : LHS Is true and RHS TRUE as well
But true impling true is true

But the statement you said you had to prove was

[math]\neg(p\wedge q)\implies \neg q\vee\neg q[/math]

with q twice in the RHS. I presume you didn't mean that; did you not carefully proofread as you were asked??

Please state the entire problem again, correctly!

 

[Dr.Peterson]

No because true tables can only establish that the above is provable according to the central theorem of propositional calculus.
They do not provide the actual proof

In order to help, we'll need to see what rules you are operating under. A Google search for "central theorem of propositional calculus" yields no source except a question on a site like this. And truth tables can be a valid form of proof, though it's perfectly reasonable for your teacher to require something else.

There are different ways in which logic can be taught; see, for example, Wikipedia, which talks about various examples of propositional calculus; it also mentions using a truth table as a proof.

So, again, please show us what your book or instructor says about what is required for a proof. At the least, we need to see your axioms; also, it could be very helpful if you showed their proof of "the other De Morgan rule", which will probably serve as a useful model for your proof. (In fact, I would expect it to be so similar that this should not be called a challenge -- unless they don't give a proof at all.)

 

[chrislav]

this is as i said a challenging problem
I have no instractor
The problem again is:
prove using the laws of propositional calculus:

[math]\neg(p\wedge q)\implies \neg p\vee\neg\ q[/math]
Do not use the other De morgan rule :

[math]\neg(p\vee q)\implies \neg p\wedge\neg q[/math]
For this particular proof you can use contradiction,
Disjunction introduction [math]A\implies A\vee B[/math]Conjuction introduction :from A,B we can infer [math]A\wedge B[/math]Negation elimination [math]\neg\neg A\implies A[/math]
Page 69 of Angelo Margaris book:FIRST ORDER MATHEMATICAL LOGIC
States the follwing central theorem of the propositional calculus
A formula is a theorem of the statement calculus if and only if it is a tautology
TRUTH tables can show you if the formula you want to prove is a tautology and according to the above theorem is then provable ( a theorem) .They do not provide a way to a proof
Propositional calculus is decidable theory BECAUSE of the truth table
Simple high school algebra is not a decidable theory
WIKI i repeat again is not a reliable source of information sometimes