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- Thread starter kof9595995
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I think the reason no one has responded to your post is that no one can follow your reasoning or the meaning of your question. I certainly can't. What you say in the first sentence is correct. If P(F) and Q(T) then P-->Q (T) and if P(F) and Q(F) then P-->Q (T). Are you trying to argue that this is incorrect?

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Sorry, my words seemed to be messed up, let me state it this way:

1.P(F)=>Q(T) is true

[tex]\neg[/tex]Q is false

and

2.P(F)=>[tex]\neg[/tex]Q(F) is also true.

But I think 2 is the negation of 1, how can they both be true?

1.P(F)=>Q(T) is true

[tex]\neg[/tex]Q is false

and

2.P(F)=>[tex]\neg[/tex]Q(F) is also true.

But I think 2 is the negation of 1, how can they both be true?

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- #4

CRGreathouse

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The negation of P → Q isSorry, my words seemed to be messed up, let me state it this way:

1.P(F)=>Q(T) is true

[tex]\neg[/tex]Q is false

and

2.P(F)=>[tex]\neg[/tex]Q(F) is also true.

But I think 2 is the negative of 1, how can they both be true?

¬(P → Q), which can also be written

¬(¬P ∨ Q) or

P ∧ ¬Q.

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Perhaps I'm misunderstanding something here but isn't P^not Q the same as P(T) and Q(F)? In that case P--> Q (F). This is straightforward material implication. If your talking about modal logic (strict implication) you need the possibility operator.

EDIT: A strict implication would be P==>Q iff not$(P^not Q) where $ is the possibility operator.

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- #7

honestrosewater

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2 is not the negation of 1.Sorry, my words seemed to be messed up, let me state it this way:

1.P(F)=>Q(T) is true

[tex]\neg[/tex]Q is false

and

2.P(F)=>[tex]\neg[/tex]Q(F) is also true.

But I think 2 is the negation of 1, how can they both be true?

That two formulas share the same truth table is another way of saying that they are two names for the same object. The formulas "P -> Q" and "~P v Q" both refer to the same case x:

[tex]\begin{array}{|c|c|c|}

\hline \mathbf P&\mathbf Q&\mathbf x\\

\hline T&T&T\\

\hline T&F&F\\

\hline F&T&T\\

\hline F&F&T\\

\hline\end{array}

[/tex]

The rules for using the negation, disjunction, and conjunction operators (~, v, and &) are probably easier to remember and implement for a lot of people because of the familiarity and nice symmetry in the way that negation distributes over the latter two:

~(P & Q) <=> ~P v ~Q

~(P v Q) <=> ~P & ~Q

You can try to figure out how to distribute negation over implication by considering the definition of the implication operator, -> (which operates on truth values):

->(v

F iff v_{1} = T and v_{2} = F;

T otherwise}

T otherwise}

and looking at the possibilities in a truth table:

[tex]\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|}

\hline \mathbf P&\mathbf \neg P&\mathbf Q&\mathbf \neg Q

&{\mathbf P \rightarrow \neg Q }&{\mathbf Q \rightarrow \neg P}

&{\mathbf P \rightarrow Q }&{\mathbf \neg Q \rightarrow \neg P}

&{\mathbf \neg P \rightarrow \neg Q }&{\mathbf Q \rightarrow P}

&{\mathbf \neg P \rightarrow Q }&{\mathbf \neg Q \rightarrow P}\\

\hline T&F&T&F&F&F&T&T&T&T&T&T\\

\hline T&F&F&T&T&T&F&F&T&T&T&T\\

\hline F&T&T&F&T&T&T&T&F&F&T&T\\

\hline F&T&F&T&T&T&T&T&T&T&F&F\\

\hline\end{array}

[/tex]

Here are all formulas of the form t

If you really want one, you are free to define an operator, say <-, where t

I wonder why you thought 2 was the negation of 1. Did you read any rules for manipulating these symbols, or were you trying to figure out how others were using them?

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The OP uses the term "logical implication" as distinct from material implication. Could he or she be thinking of strict implication (modal logic)? (see post 6).I wonder why you thought 2 was the negation of 1. Did you read any rules for manipulating these symbols, or were you trying to figure out how others were using them?

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- #9

honestrosewater

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Perhaps. It sounded like they were talking about the classical interpretation of formulas containing an implication symbol. But some clarification would be helpful. Implication is central to logic and shows up in many ways. To me, it could mean several things:The OP uses the term "logical implication" as distinct from material implication. Could he or she be thinking of strict implication (modal logic)?

1) You have an implication symbol in your formal language, usually denoted by a right-arrow

2) The truth-functional operation that I mentioned, usually also denoted by a right-arrow. It is associated with the implication symbol in that this operation is what the symbol is taken to mean when your formal language is given an interpretation. This interpretation is signified by truth tables.

3) A syntactic/proof-theoretic implication relation on formulas and sets of formulas, which tells you what formulas are provable from what other formulas, often denoted by a turnstile |-.

4) A semantic/model-theoretic implication relation on formulas and sets of formulas, which tells you how interpreted formulas are related by truth, often denoted by a double-turnstile |= or double-right-arrow =>.

All of these "implications" are also variously called conditionals, entailments, and inferences. This general concept that two things

invite confusion.kof9595995 said:1.P(F)=>Q(T) is true

[tex]\neg[/tex] Q is false

and

2.P(F)=>[tex]\neg[/tex] Q(F) is also true.

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As I understand it, strict implication uses a "possibility operator" usually represented by a vertically elongated diamond (since I don't have that option I'm using $). If you use modal logic, you get a different result: P=>Q iff -$(P^notQ). In plain language P implies Q if and only if P^notQ is not possible (indicated by the minus sign). Material implication says P->Q if any of three conditions exist: P,Q; notP,Q; notP,notQ. Strict implication has application in some electronic logic networks.All of these "implications" are also variously called conditionals, entailments, and inferences. This general concept that two thingsxandyare connected in a way such that if you havex, in the most general sense of 'having', you also haveyalso underlies the idea of inference rules. So it helps to be clear about what type of objects you're dealing with (symbol, formula, operation, syntactic relation, semantic relation). So far, "material conditional", "logical implication", "p implies q", "~P v Q", "P → Q", "P => Q", and interpretations of implication have all been taken to be similar or the same, without saying which is the case. That and statements like

invite confusion.

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- #11

honestrosewater

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I get what you're saying, but again, a little care could avoid confusion. What is material logic? The definitions above are part of classical logic. In classical logic,Material logic says P->Q if any of three conditions exist: P,Q; notP,Q; notP,notQ.

I(

F iff I(*P*) = T and I(*Q*) = F;

T otherwise}

T otherwise}

and

The original problem seemed to only be a question of how to negate

Obviously, the negation of "p implies q" is "p does not imply q", but the negation was first stated as "p implies not q", and "does not imply" is not a common symbol or operator and is not defined anywhere or used again.I think the negative of "p implies q" is "p implies not q"

...

is the negative of "p implies q" "p does not imply q"?

The negation of a formula

1) The formula formed by applying a negation rule to

2) The interpretation of

3) Any of the formulas that are equivalent in some sense to (1) or (2).

(1) is straightforwardly

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Thanks for your comprehensive reply, but the quote you responded to contained an error. It should have been 'material implication' instead of 'material logic'. This has been corrected. The reason I brought up strict implication is because I couldn't follow the OP's argument and thought(1) is straightforwardly~(P -> Q). (2) is obtained just as easily by swapping the truth values ofI(P -> Q). Examples of (3) were given in a reply, but there seems to still be confusion about the interaction of negation and implication as symbols and operators. Other than this negation, the OP's reasoning is correct.

he was possibly confusing material implication with strict implication.

EDIT: Modes like possibility/ impossibility are are not synonymous with affirmation/negation. This can be shown by trying to express strict implication without the modal operator: P=>Q iff not(P^notQ). This reduces to (P=>Q) iff (notP^Q) which violates the definition of strict implication. Under strict implication P implies Q iff (P^not Q) is not possible. Modal logic is not reducible to "classical" logic. (P^Q), (notP^Q), and (notP^notQ)are not strong enough to support strict implication.

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- #13

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Thanks for the detailed answers, I think I understand it now.

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