# Deciding which logical proposition is stronger

I am studying propositional logic and I am trying to solve some exercise related to the following definition:

Given two logical propositions $\alpha$ and $\beta$, we say that $\alpha$ is stronger than $\beta$ if and only if $(\alpha \implies \beta)$ is a tautology. Determine which of the two propositions is stronger in the following cases:

1) True, False

2) True, True

3) False, False

I am a bit confused with this problem. I've seen that for True we always assign the value $1$ and for False, the value $0$. So, (True $\implies$ False) has always value $0$ (actually, there is one possibility) and (False $\implies$ True) has always value $1$ (there is also one possibility). This means that (False $\implies$ True) is a tautology, so False is stronger than True.

I am not sure if what I've said above is correct and I am lost in 2) and 3), I would appreciate some help. Thanks in advance.

• You're doing ok, just keep going. What is your question? – Andrej Bauer Aug 15 '16 at 11:26

The notion of strength of a proposition comes from its power to constrain the model. For instance the proposition $P1 : x = 2$ is stronger than $P2 : x \geq 2$ in Arithmetic. And in propositional notation this can be written as $P1 \Rightarrow P2$.

The stronger the proposition the harder it usually is to prove. On the other hand, if proven it provides more corollaries.

For instance for a signature with 2 propositional symbols $P$ and $Q$ we have the following partial diagram for the Strength relation.

Now to your examples. $\textrm{False}$ happens to be an extreme case. It can never be satisfied. If you are using propositions to constrain your search space for an answer, then $\textrm{False}$ says: "Call off the search and start again". The other extreme is the proposition $\textrm{True}$. It is trivial. It gives absolutely no information on where to look for the answer.

Now we shall use this intuition to tackle your questions:

1) True, False

Since $\textrm{True}$ is a trivial constraint, you would not be able to derive anything non-trivial out of it, this certainly means that you would not be able to derive the extreme constraint $\textrm{False}$.

$\not\vdash \textrm{True} \Rightarrow \textrm{False}$

2) True, True

The propositions have exactly the same strength, they are equivalent

$\vdash \textrm{True} \Rightarrow \textrm{True}$

3) False, False

The propositions have exactly the same strength, they are equivalent

$\vdash \textrm{False} \Rightarrow \textrm{False}$

And an extra case

4) False, True

$\vdash \textrm{False} \Rightarrow \textrm{True}$

Note: there is an issue with saying that proposition $P$ is stronger than $Q$ if $P \Rightarrow Q$ is a tautology since intuitively the relation of being stronger is strict. It would better align with the intuition if we were to say that proposition $P$ is stronger than $Q$ or of equal strength if and only if $P\Rightarrow Q$.

• I'd just add that False=>True - False is stronger than anything and True is weaker than anything. – Petr Pudlák Aug 15 '16 at 13:36