User talk:EricBright

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1 Images
2 Redoing the table
3 → is closer to what we use in...
4 Material equivalence

[edit] Images

I have deleted Image:Propositional Logic.png as you requested but bear in mind that all that means is that I have removed a db tag. The image is on the Commons and you must go over there to request the image itself be removed.

I do not like to see different articles or images on titles which would be identical if threated case-insensitively. But actual deletion of Image:Propositional Logic.png will solve that problem.

I strongly disapprove of this image (either version) - it is totally anti-colaborative. It should be rendered as a normal Wiki table. As well as allowing others to contribute, it is more flexible allowing for different formatting depending on the browser and window size in use. I am sure the Wiki <math> function (which I believe is just LaTex) is fully capable of handling the formula. Please consider re-doing as a table. -- RHaworth 10:47, 7 October 2005 (UTC)

[edit] Redoing the table

Dear RHaworth,

Thank you for your warm welcome. I am considering redoing the 'propositional logic' table using Wiki table. I do it as soon as I figured its how-to out. It would look like the following table:

Common name	Secuent	Description
Basic argument forms of the calculus
Modus Ponens	$[(p \supset q) \wedge p] \supset q$	if p then q; p; therefore q
Modus Ponens	[(p ⊃ q) ∧ p] ⊃ q	if p then q; p; therefore q
Modus Ponens	[(p → q) ∧ p] ⊢ q	if p then q; p; therefore q
Modus Tollens	[(p → q) ∧ ¬q] ⊢ ¬q	if p then q; not q; therefore not p
Hypothetical Syllogism	[(p → q) ∧ (q → r)] ⊢ (p → r)	if p then q; if q then r; therefore, if p then r
Disjunctive Syllogism	(p ∨ q), ¬p ⊢ q	Either p or q; not p; therefore, q
Constructive Dilemma	(p → q) ∧ (r → s), (p ∨ r) ⊢ (q ∨ s)	If p then q; and if r then s; but either p or r; therefore either q or s
Destructive Dilemma	(p → q) ∧ (r → s), (¬q ∨ ¬s) ⊢ (¬p ∨ ¬r)	If p then q; and if r then s; but either not q or not s; therefore rather not p or not r
Simplification	(p ∧ q) ⊢ p	p and q are true; therefore p is true
Conjunction	p, q ⊢ (p ∧ q)	p and q are true separately; therefore they are true conjointly
Addition	p ⊢ (p ∨ q)	p is true; therefore the disjunction (p or q) is true
Composition	(p → q) ⊢	If p then q; and if p then r; therefore if p is true then q and r are true
De Morgan's Theorem (1)	¬(p ∧ q) ⊢ (¬p ∨ ¬ q)	The negation of (p and q) is equiv. to (not p or not q)
De Morgan's Theorem (2)	¬(p ∨ q) ⊢ (¬p ∧ ¬ q)	The negation of (p or q) is equiv. to (not p and not q)
Commutation (1)	(p ∨ q) ⊢ (q ∨ p)	(p or q) is equiv. to (q or p
Commutation (2)	(p ∧ q) ⊢ (q ∧ p)	(p and q) is equiv. to (q and p)
Association (1)	[p ∨ (q ∨ r)] ⊢ [(p ∨ q) ∨ r]	p or (q or r) is equiv. to (p or q) or r
Association (2)	[p ∧ (q ∧ r)] ⊢ [(p ∧ q) ∧ r]	p and (q and r) is equiv. to (p and q) and
Distribution (1)	[p ∧ (q ∨ r)] ⊢ [(p ∧ q) ∨ (p ∧ r]	p and (q or r) is equiv. to (p and q) or (p and (p or r)
Distribution (2)	[p ∨ (q ∧ r)] ⊢ [(p ∨ q) ∧ (p ∨ r]	p or (q and r) is equiv. to (p or q) and (p or r)
Double Negation	p ⊢ ¬¬p	p is equivalent to the negation of not p
Transposition	(p → q) ⊢ (¬q → ¬p)	If p then q is equiv. to if not q then not p
Material Implication	(p → q) ⊢ (¬p ∨ q)	If p then q is equiv. to either not p or q
Material Equivalence (1)	(p :⇔ q) ⊢ [(p → q) ∨ (q → p)	(p is equiv. to q) means, either (if p is true then q is true) or (if q is true then p is true)
Material Equivalence (2)	(p :⇔ q) ⊢ [(p ∧ q) ∨ (¬q ∧ ¬p)	(p is equiv. to q) means, either (p and q are true) or ( both p and q are false)
Exportation	[(p ∧ q) → r] ⊢ [p → (q → r)]	from (if p and q are true then r is true) we can prove (if q is true then r is true, if p is true)
Importation	[p → (q → r)] ⊢ [(p ∧ q) → r]
Tautology	p :⇔ (p ∨ p)	p is true is equiv. to p is true or p is true

Eric 02:20, 11 October 2005 (UTC)

Excellent. I have redone the first line using <math> as described in meta:Help:Formula - and it does not look as good. I have redone it again using ⊃ ⊃ - see the ISO 8859-1 list. &equiv; ≡ is another you need. -- RHaworth 15:49, 11 October 2005 (UTC)

[edit] → is closer to what we use in...

Dear RHaworth,

Thanks for your encouragement. I revised the table and I think this symbol ⊃ that usually means 'proper superset of...' is not exactly what we mean by → in 'propositional logic'; we rather use → for what we call 'if...then...'. However, some people use ⊃ in their writings for 'if...then...'. In that case, it becomes confusing when it comes to the Axiomatic propositional logic, where we use both ⊃ and → with the aforementioned meanings. Therefore, if you agree, I keep the arrow '→' untouched.

We also use turnstile, ⊢, to convey something like the following:

p ∨ q, p → r, q → s, ¬s ⊢ r

and we read this methalinguistic claim like:

There is a proof of the sentence on the right of the turnstile (r) from the ensemble of sentences on the left of it (p ∨ q, p → r, q → s, ¬s). Then, turnstile is not metalinguistically equal to 'if...then...'.

However we can prove:

⊢ Σ → A

from:

Σ ⊢ A

using the axiomatic logic. Such a thing is called The Deduction Theorem [DT] that says:

For any set Σ ⊆ Φ, and for any wffs α, β ∈ Φ, if Σ ∪ {α} ⊢ β, then Σ ⊢ α → β.

By the way, ≡ is exactly what I wanted. I substituted :⇔ with ≡. Now we have a more accurate table.

I also want to know if there is any way to refer to this table from inside of 'Deductive Logic' article (it uses the PNG format yet). Then we don't have to reproduce and edit the same table twice.

Eric 01:27, 12 October 2005 (UTC)

[edit] Material equivalence

Material Equivalence (1)
- (p :⇔ q) ⊢ [(p → q) ∨ (q → p)]
- (p is equiv. to q) means, either (if p is true then q is true) or (if q is true then p is true)

There is an error on material equivalence, both conditions are necessary to make p equivalent to q, so the simbol AND is mistaken by simbol OR.

image: (p is equiv. to q) means, either (if p is true then q is true) or (if q is true then p is true)
- $(p \equiv q) \equiv [(p \rightarrow q) \or (q \rightarrow p)]$
correct: (p is equiv. to q) means, (if p is true then q is true) and (if q is true then p is true)
- $(p \equiv q) \equiv [(p \rightarrow q) \and (q \rightarrow p)]$