I Modus Ponens deduction rule

[cianfa72]

TL;DR Summary

Argument behind the validity of Modus Ponens deduction rule

A question about the Modus Ponens deduction rule used in theory of proof within an axiomatic system. $$[(p \to q) \land p ] \to q$$
One can easily check the above is a tautology (i.e. it evaluates always True for any truth value of propositions $p$ and $q$).

Now what is the logic argument that allows to conclude/infer the truth of $q$ assuming True the truth value of propositions $p$ and $(p \to q)$ ?

 

[andrewkirk]

There can be no logical argument for that. What logical system could it use? It would have to be a logical system without modus ponens. Of course we could make up a logic system without MP but then we're just left asking what is the justification for the rules of deduction in that system? You end up with an infinite regress.

You have to just start by asserting a set of rules. You can't prove them.

You might worry that that makes the rules arbitrary. In practice it doesn't. We select rules that reflect our observations of how the world seems to work, and how humans think.

 

[cianfa72]

Just to check what I understood. Consider the following:

"If Sam earns $200 then Sam buys a new smartphone"

Define the two propositions p := "Sam earns $200" and q:="Sam buys a new smartphone". The above is in the form of material implication $p \to q$.

In order to use Modus Ponens to infer the truth of the conclusion q, what one needs to do is check that in all instances, let me say, Sam in real life happens to earn $200 he actually buys a new smartphone.

Let's assume that in real life the above holds True. Then since the implication $p \to q$ is True, assuming p as True then from Modus Ponens it follows that the conclusion q is also True.

Does it make sense?

Ps. Let me joke: to me it seems a bit weird since when one proves/checks the truth of the implication $p \to q$ they are actually "applying" at the same time an "instance" of Modus Ponens rule.

 

[cianfa72]

Btw, which is the difference between logical vs material implication?

 

[fresh_42]

Btw, which is the difference between logical vs material implication?

Logical: $P\longrightarrow Q$ https://en.wikiversity.org/wiki/Logical_implication
Material: $\lnot P \vee Q$ https://en.wikipedia.org/wiki/Material_implication_(rule_of_inference)

 

[cianfa72]

Logical: $P\longrightarrow Q$ https://en.wikiversity.org/wiki/Logical_implication
Material: $\lnot P \vee Q$ https://en.wikipedia.org/wiki/Material_implication_(rule_of_inference)

Ah..so material implication is just a matter of "valid rule of replacement" of a logical implication?

In other words in any "compound proposition" one may replace all the occurrences of $P \longrightarrow Q$ with $\lnot P \vee Q$ and the other way around since they have the same truth table.

 

[fresh_42]

Yes. I think it is mainly a question about which logical framework / system / calculus you prefer to use. If you click the references in these links, then you see that the material implication is settled in propositional logic. This means you consider functions that have a logical value, here $f(P,Q)=\lnot P \vee Q.$ The logical implication $P\longrightarrow Q$ isn't such a function. It can be wrong even if $Q$ is true, or both.

 

[cianfa72]

The logical implication $P\longrightarrow Q$ isn't such a function. It can be wrong even if $Q$ is true, or both.

Sorry, in the case of logical implication with the term "wrong" do you actually mean false ?

 

[fresh_42]

Sorry, in the case of logical implication with the term "wrong" do you actually mean false ?

Yes.

Take the statements P: "Joe is old." and Q: "Joe has grey hair." Then $\lnot P \vee Q$ reads: "Joe isn't old or has grey hair." which is true if Joe has grey hair. However, the logical implication $P\longrightarrow Q$ is false, since not every old person has grey hair. This example shows that you need to define your logical framework carefully.

 

[cianfa72]

Let me be sloppy in showing what I got

Material implication ($\rightarrow$) is just an operator that returns a new proposition from proposition variables $P$ and $Q$ with a truth-value according its truth table and their truth-values. In principle propositions $P$ and $Q$ are unrelated each other (as you said it is just a function that has a logical value).

In case of Logical implication ($\Rightarrow$), instead, its truth is grounded on an argument showing that when the premise is true the conclusion is also true, i.e. basically premise and conclusion are actually logically related in cause-effect manner.

Basically the proof of theorems are based on logical implications.

 

[cianfa72]

I would like to point out a main difference between material vs. logical implication as far as I understood

Since material implication $P \to Q$ is just a function of two propositions $P$ and $Q$ (i.e. the symbol $\to$ is a formal logical connective/operator), the truth value it returns depends on the actual truth values of proposition variables $P$ and $Q$.

On the other hand, since the logical implication $P \implies Q$ isn't a function (it hasn't a truth table), it has a truth value in itself.

Ps. I found this very informative:

There's a subtle difference between them. $P \to Q$ is the proposition "P implies Q" which may be a true or false statement. $P \implies Q$ is the assertion that $P \to Q$ is true. Therefore they are equivalent whether $P \to Q$ is a tautology.

 

[D H]

TL;DR Summary: Argument behind the validity of Modus Ponens deduction rule

A question about the Modus Ponens deduction rule used in theory of proof within an axiomatic system. $$[(p \to q) \land p ] \to q$$
One can easily check the above is a tautology (i.e. it evaluates always True for any truth value of propositions $p$ and $q$).

That is not the case. Modern ponens is anything but a tautology. There is only one pair of $(p,q)$ values that lead to $[(p \to q) \land p ] \to q$ being true, and that is when both $p \to q$ and $p$ are true. What is a tautology is $$[(p \to q) \land q ] \to q$$.

 

[D H]

Now I see the problem. Modus ponens is not about the truth table of $[(p \to q) \land p ] \to q$. It is instead about what one can deduce about $q$, namely that one can deduce that $q$ must be true if both $p \to q$ and $p$ are true. If either of (or both of) $p \to q$ or $p$ is false, one cannot deduce anything about the veracity of $q$.