How to explain that ((P-->Q)&P) does not logically imply that " P logically implies Q"? ( On a false interpretation of modus ponens).

Question

Context: reflexion on a possible way to misunderstand " modus ponens" as meaning ( erroneously) : If [(P-->Q) and P] then (P logically implies Q).

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Sure, a truth table shows that [ (P-->Q) & P ] logically implies that " P materially implies Q".

Indeed the formula " [ (P-->Q) & P ] --> ( P-->Q) " is valid,

or ( equivalently " [ (P-->Q) & P ] ==> ( P-->Q) " is correct ( with " ==>"

meaning (tauto)logically implies" )

But I can't manage to explain, using a truth table or another method, that

[ (P-->Q) & P ] ==> ( P==>Q) is not correct,

with " ==> " still meaning (tauto)logically implies .

Which valuation would make [ (P-->Q)&P] true and (P==>Q) false?

I'm not even sure that this way of putting the question is correct.

Answer 1

The expression $(P → Q) ∧ P$ is a formula of the language while $P \vDash Q$ is not.

$\vDash$ is a symbol of the meta-language expressing a relation between formulas, while $(P → Q) ∧ P$ is an expression in the object language.

Every valuation $v$ will assign a truth value to formula; thus, with valuation $v$ such that $v(P)=\text T$ and $v(Q)=\text F$, the formula will be evaluated to $\text F$.

Obviously, with a different valuation the resulting truth value can change (except for tautologies and contradictions).

The situation is different for the relation $\vDash$. In this case we have to consider all valuations in order to assert that it holds or not :

$\varphi \vDash \psi$ iff, for very valuation $v$ such that $v(\varphi)=\text T$ also $v(\psi)=\text T$.

Specifically for propositional variables $P$ and $Q$, we can never have $P \vDash Q$, exactly because we have the valuation $v$ such that :

$v(P)=\text T$ and $v(Q)=\text F$.

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We have to take care in using the expression : $(P → Q) ∧ P \vDash (P \vDash Q)$, because it is not meaningful.

It is not a formula of the language (because $\vDash$ is not part of the object language) and thus we cannot evaluate it with truth tables.

We can write an expression of the meta-language like :

if ..., then $P \vDash Q$.

As said above, if $P$ and $Q$ are propositional variables, then $P \vDash Q$ is false.

Thus, an example of true statement (in the meta-language) involving it will be :

if $\vDash (P → Q) ∧ P$, then $P \vDash Q$.