On the meaning of logical implication within specific context/models

  • #1
cianfa72
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TL;DR Summary
logical implication makes sense only within specific context/models.
I'm still confused about the use of material implication (material conditional) vs logical implication .

From MSE the writing PQ makes the meta-logical assertion/statement that the logical statement Q is logically implied by the logical statement P.

First of all: saying for instance "logical statement Q...." actually implicitly asserts that it is True, I guess.

Second: the meta-logical assertion as above should only make sense/apply within a specific context/model/interpretation. Consider for instance: x>6x>2 This logical implication actually makes sense within the context/model/interpretation of mathematical theory of natural numbers, however it makes no sense in other contexts in which there is no notion of ordering (how should one interpreter the symbol > in such a case ?).

Therefore from the above logical implication follows (from a meta-logical theorem) that the material implication/conditional PQ: "If x>6 then x>2"

is true (i.e. it is a tautology) only within the specific context in which the logical implication applies.

What do you think about? Thanks.

Ps. note that I tend to use the english term "implication" for the logical implication and the form "If...then" for the material implication/conditional .
 
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  • #2
P→Q is a Boolean expression that is false when P is true and Q is false.
P⟹Q is a statement that P→Q is true - You are either presuming it to be true or you have proved it to be true.
 
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  • #3
.Scott said:
P⟹Q is a statement that P→Q is true - You are either presuming it to be true or you have proved it to be true.
Here your "it" in bold refers to the statement P⟹Q, right?
 
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  • #4
.Scott said:
P⟹Q is a statement that P→Q is true - You are either presuming it to be true or you have proved it to be true.
Proving P⟹Q (is true) means use a logic argument within a given model/interpreation showing that the combination P true and Q false is not the case (i.e. the row from the P→Q truth table where it is false is actually ruled out/doesn't apply).

The point I'd stress is that such a logic argument applies only within the specific model/interpretation employed.
 
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  • #5
cianfa72 said:
Here your "it" in bold refers to the statement P⟹Q, right?
No! It refers to PQ.

You are still misunderstanding modus ponens and modus tollens, the intent of which are to create new knowledge. Suppose you know PQ is true, and nothing else. That alone says nothing about P or Q. Now you are given that P is false. Since PQ is true regardless of the value of Q in the case that P is false, this new knowledge that P is false doesn't tell you anything else. If on the other hand you are given that P is true, Q must also be true to be consistent with both PQ and P being true. Finally, if one is told about Q rather than P, it is the assertion that Q is false that tells something new. In this case, PQ¬Q says that P must be false.

Finding a contradiction can be a "good thing" at times; finding a contradiction is after all the goal of proof by contradiction (which BTW some mathematicians still do not accept as a valid form of proof as such proofs are typically non-constructive). At other times, finding a contradiction where no contradiction is expected most like means that someone (usually the investigator) has made a mistake.
 
  • #6
cianfa72 said:
Here your "it" in bold refers to the statement P⟹Q, right?
As @D H said: No, "it" refers to P→Q.
P⟹Q is a statement that P→Q is true. You would make the statement "P⟹Q" when you want to start out with P→Q as a presumption or when you have proven that P→Q is true.
 
  • #7
.Scott said:
As @D H said: No, "it" refers to P→Q.
P⟹Q is a statement that P→Q is true. You would make the statement "P⟹Q" when you want to start out with P→Q as a presumption or when you have proven that P→Q is true.
Ok, so when one writes down P⟹Q that is a true meta-logical statement in itself, let me say.

About proving that the material implication P→Q is true. As I said before, one has to pick a model/interpretation. Take for instance: P: x>6 and Q: x>2 and the compound proposition:

P→Q: "if x>6 then x>2"

in order to prove "P→Q is true" one needs an interpretation for both the predicates "x>6" and "x>2". Using the model/interpretation of mathematical theory of natural numbers N, one is able to assign a truth value to the propositions above for any determination of x (for any given value of the variable x a predicate becomes a proposition). Then, within that model/interpretation, one can show that P true and Q false is never the case, hence P→Q is proven as (always) true and therefore one is entitled to write down P⟹Q.

Does the above make sense?
 
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  • #8
cianfa72 said:
Ok, so when one writes down P⟹Q that is a true meta-logical statement in itself, let me say.

About proving that the material implication P→Q is true. As I said before, one has to pick a model/interpretation. Take for instance: P: x>6 and Q: x>2 and the compound proposition:

P→Q: "if x>6 then x>2"

in order to prove "P→Q is true" one needs an interpretation for both the predicates "x>6" and "x>2". Using the model/interpretation of mathematical theory of natural numbers N, one is able to assign a truth value to the propositions above for any determination of x (for any given value of the variable x a predicate becomes a proposition). Then, within that model/interpretation, one can show that P true and Q false is never the case, hence P→Q is proven as (always) true and therefore one is entitled to write down P⟹Q.

Does the above make sense?
Yes. In the case you described, you might say that for any x, P⟹Q.
 
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  • #9
.Scott said:
Yes. In the case you described, you might say that for any x, P⟹Q.
Ok.

Another source of confusion to me: "P⟹Q if and only if P→Q is a tautology" is a meta-logic theorem.

What does it say? Suppose to fix a model/interpretation (an interpretation in a model is actually a truth-valued function that assigns either True of False to any proposition within the model). P⟹Q says P→Q is true hence "plugging in" the specific truth-values assigned by the interpretation to P and Q into the material implication P→Q it results as true. In this context the theorem tells us nothing.

I believe the real "scope" of the above theorem is at formal object level. Here let me say model/interpretation/semantic is not involved at all (i.e. just the syntax that allows to build new well-formed propositions from old ones).

In this context, however, propositions P and Q entering the theorem are not supposed to be both proposition variables (aka atomic propositions) otherwise one could set P as True and Q as False getting P→Q false. Therefore at least one of them must be a compound proposition.
 
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  • #10
@cianfa72 wrote:
Another source of confusion to me: "P⟹Q if and only if P→Q is a tautology" is a meta-logic theorem.

The statement "P⟹Q if and only if P→Q" is true only in the sense that "true ↔ true" is true. Other than that, I don't like it because it is playing a subtle game with the how English-speaking humans read the statement. It's a kind of a semantic "gotcha".

You can see the "gotcha" by comparing it to the logically equivalent "P→Q if and only if P⟹Q". If interpreted as a Boolean operation, "P⟹Q ↔ P→Q" and "P→Q ↔ P⟹Q" are equivalent and both are true. But P→Q can occur in a proof where its value can be true, but where there is no P⟹Q.

So, what "P⟹Q ↔ (P→Q)" really means is: "Given that 'P⟹Q' has been stated, 'P⟹Q if and only if P→Q' is true.


As a software engineer, let me throw in a third "Boolean voice":
1) In a logical proof, P→Q states a Boolean operation with no expectation about the values or availability of P and Q.
2) P⟹Q asserts that P→Q is true. Thus asserting that either P is false or Q is true.
3) In software, P→Q would be coded as (!P)||Q in C or C++, but it is a command, not a declaration. It instructs the computer to perform the "if then" operation on variables P and Q.

That third voice might be easier to understand with simple addition - since the '+' operator is the same for all voices.
For T:Total Energy, P:Potential Energy, and K:Kinetic Energy -
1 or 2) T=P+K could be part of the definition of T, P, or K. Or it could be a declaration in a model to demonstrate how energy moves within a system.
3) In software, T=P+K is a command to retrieve the current values of P and K, add them, and store the sum in the variable T.
 
  • #11
.Scott said:
Another source of confusion to me: "P⟹Q if and only if P→Q is a tautology" is a meta-logic theorem.

The statement "P⟹Q if and only if P→Q" is true only in the sense that "true ↔ true" is true.
Sorry, not sure about it, but I believe that in the statement of the theorem above "is a tautology" actually applies to "P→Q" only, not to the entire proposition/statement "P⟹Q if and only if P→Q".

Btw, does the english statement "either P false or Q true" actually rule out the case P false and Q true ?
 
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  • #12
cianfa72 said:
Sorry, not sure about it, but I believe that in the statement of the theorem above "is a tautology" actually applies to "P→Q" only, not to the entire proposition/statement "P⟹Q if and only if P→Q".

Btw, does the English statement "either P false or Q true" actually rule out the case P false and Q true ?
Ahhh, so "P⟹Q if and only if P→Q is a tautology" means "P⟹Q if and only if (P→Q is a tautology)"
not "(P⟹Q if and only if P→Q) is a tautology".

As far as your second question is concerned, depending on the speaker and the context, the speaker could be trying to communicate any of these:
1) "either P false or Q true" and "I am not even considering the possibility of both P false and Q true".
2) "either P false or Q true" and "That means it could be both P false and Q true".
3) "either P false or Q true" and "That means it could be both P false and Q true and isn't that obvious!".
4) "either P false or Q true" and "No it can't be both P false and Q true".
5) "either P false or Q true" and "No it can't be both P false and Q true and isn't that obvious!".
6) "either P false or Q true or wait ... maybe I've got that wrong."
 
  • #13
Ok, so is the statement "P→Q is true when/iff either P is false or Q is true" supposed to include also the case P is false and Q is true ?
 
  • #14
P→Q is true when either P is false or Q is true.
Truth table for P→QQ is TrueQ is False
P is TrueP→Q is TrueP→Q is False
P is falseP→Q is TrueP→Q is True
 
  • #15
.Scott said:
P→Q is true when either P is false or Q is true.
Truth table for P→QQ is TrueQ is False
P is TrueP→Q is TrueP→Q is False
P is falseP→Q is TrueP→Q is True
Yes, however as you can see P→Q is true also when P is false and Q is true.

Just to be clear: my question is about the actual meaning of the english expression "either...or".
 
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  • #16
cianfa72 said:
Yes, however as you can see P→Q is true also when P is false and Q is true.

Just to be clear: my question is about the actual meaning of the english expression "either...or".

When I said "P→Q is true when either P is false or Q is true", I was using the inclusive or.
So, it was equivalent to (~P)∨Q. I included the truth table to avoid any ambiguity.

But you are asking for the "actual meaning of the English expression "either...or"" outside of any
specific context.

The biggest difference between the phrase and P∨Q is that the later has a very specific mathematical definition that is expected to be followed by anyone using the term. In contrast, often when a speaker uses the English phrase, they either don't care about the ambiguity or include something that helps out with the meaning.

Using my last statement (or independent clause) as an example, I was not really considering a situation where the speaker doesn't care about the ambiguity, but includes something to resolve that ambiguity anyway. But, if I did want to address that case, I may have said either:
1) They either don't care about the ambiguity or include something that helps out with the meaning, or both.
or
2) They either don't care about the ambiguity or include something that helps out with the meaning, but not both.
And that last statement demonstrates another very common use of the phrase, when the choice is between two obvious opposites.
A simpler example would be "Either you want a ride home or you don't.". Since that's equivalent to P∨~P, the false→true combination is mute.

All that said, I believe the "inclusive or" (P∨Q; rather than P⊕Q) is the more common interpretation for the "either...or..." phrase.
 
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  • #17
Coming back to what I said in post #9, I believe the use of logical implication is actually two-folded:

1. is a meta-logic proposition (or statement) saying is true in a given (possibly natural/implicit) model/interpretation; the interpretation is the truth function that assign in the model the value either True or False to any proposition within it

2. is a meta-logic proposition (or statement) saying is a tautology; in this context P and Q can't be both proposition variables (aka atomic propositions), i.e. at least one of them must be a compound proposition

In 2. there is no model/interpretation involved; it is simply a matter of evaluating the truth table of the entire proposition for each truth value (either True or False) of any of the proposition variables entering the compound proposition (or statement).

See also here MSE.
 
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  • #18
I don't think I addressed that "tautology" part very well.

A logical tautology is something that is true on the face of it. Examples would be P→P, P↔P, and P∨~P.
In other words, "tautology" is not used for claiming something is true, it is used for claiming that something that is true is true by its very form.

So, it would be a mis-use of the term "tautology" to say that P⟹Q states that "P→Q is a tautology".
P⟹Q simply states that "P→Q is true".

So, in post 10, my parsing of "P⟹Q if and only if P→Q is a tautology" was basically correct.
A "P⟹Q" would make "P→Q" a tautology. So, the combination the two statements "P⟹Q" and "P→Q" creates a tautology. But, saying it with logical operators instead of English words imposes a greater expectation of precision. Strictly speaking "(P⟹Q) ↔ (P→Q)" is just wrong because the term "P⟹Q" doesn't belong at the logical operations level. The tautology would be "(P→Q) ↔ (P→Q)".

You haven't asked, but let's consider whether "P⟹Q" is a tautology. When we write "P⟹Q", we are claiming that "P→Q" is true. But isn't "P⟹Q" also true? I would say, not really. It is more of a construction.
 
  • #19
.Scott said:
So, in post 10, my parsing of "P⟹Q if and only if P→Q is a tautology" was basically correct.
No, I don't think so (the correct way is "P⟹Q if and only if (P→Q is a tautology)").

As I said before, there are two ways the symbol is understood, let's call them a) and b). In both cases it is not a symbol at object language level (i.e. at proposition logic aka sentential or first-order logic level), rather it is employed in meta-logic discourses to state/assert something about the relation between propositions/sentences that live at the object language level.

In a) the writing P⟹Q asserts/states that, within a given model interpretation (possibly natural or implicit), the material implication P→Q is true.

In b) P⟹Q asserts that, regardless of specific interpretations, P→Q is always true. Here P and Q are proposition formulas and at least one of them is a compound proposition (i.e. it is not a proposition variable aka atomic proposition). It is straightforward to prove the meta-logical theorem that P→Q is a tautology and the other way around (if and only if).

Note that to distinguish them, in b) the appropriate/relevant symbol to use is .
 
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  • #20
cianfa72 said:
No, I don't think so (the correct way is "P⟹Q if and only if (P→Q is a tautology)").

As I said before, there are two ways the symbol is understood, let's call them a) and b). In both cases it is not a symbol at object language level (i.e. at proposition logic aka sentential or first-order logic level), rather it is employed in meta-logic discourses to state/assert something about the relation between propositions/sentences that live at the object language level.

In a) the writing P⟹Q asserts/states that, within a given model interpretation (possibly natural or implicit), the material implication P→Q is true.

In b) P⟹Q asserts that, regardless of specific interpretations, P→Q is always true. Here P and Q are proposition formulas and at least one of them is a compound proposition (i.e. it is not a proposition variable aka atomic proposition). It is straightforward to prove the meta-logical theorem that P→Q is a tautology and the other way around (if and only if).

Note that to distinguish them, in b) the appropriate/relevant symbol to use is .
P→Q is not a tautology, so (P→Q is a tautology) would be false.
 
  • #21
.Scott said:
P→Q is not a tautology, so (P→Q is a tautology) would be false.
That's not the point. As explained here (see in particular chapter 2 & 3) in sentential/proposition logic taken as object formal language, P and Q are well-formed formulas (wff).

A wff is a sentence/proposition composed from atomic or compound propositions by means of logical connectives that's grammatically acceptable (material implication/conditional is just one of these connectives).

The following is a meta-logic theorem as shown there:

implication.PNG
 
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  • #22
cianfa72 said:
That's not the point. As explained here (see in particular chapter 2 & 3) in sentential/proposition logic taken as object formal language, P and Q are well-formed formulas (wff).

A wff is a sentence/proposition composed from atomic or compound propositions by means of logical connectives that's grammatically acceptable (material implication/conditional is just one of these connectives).

The following is a meta-logic theorem as shown there:

View attachment 356714
You need to parse the sentential statement in the first block as follows:
(Formula A logically implies formula B if and only if the conditional formula A→B) is a tautology.
 
  • #23
.Scott said:
You need to parse the sentential statement in the first block as follows:
(Formula A logically implies formula B if and only if the conditional formula A→B) is a tautology.
No, it isn't. See the example in the page I attached to about "~P logically implies ~(P & Q)".
 

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