I recently had this question answered on stack exchange:
if X is in NP but Y is not in NP then can X be reduced to Y?
The answer proposed a counter example using an element of complexity theory I had not given much thought to, specifically it was the problem
$A = \emptyset$
that is, $A$ is the empty problem (or contradiction problem) where $\forall$ inputs $x$, $x \not\in A$
The role of $A = \emptyset$ in my answered question was that it served as problem in $NP$ that could be reduced to any other problem, including those not in $NP$, thereby giving a direct counter example to the statement: $X \in NP \land Y \not\in NP \implies X \not\le^m_p Y$
What interests me was the implication that $A = \emptyset$ could be polynomial-time, many-one reduced to any other problem and that $A \in NP$.
This makes sense to me as these two implication seems vacuously true:
Let $B$ be any other problem such that $B \neq \emptyset$. Now $A \le^m_p B$ if $\exists$ some polynomial reduction $f$ such that $\forall$ inputs $x$, $x \in A \iff f(x) \in B$. But since $\not\exists x, x \in A$ by the definition of A then the reduction of $A$ to any $B$ would be a trivial one in that no matter what the given input $x$, the reduction $f(x)$ could simply produce a same input instance for $B$ such that $f(x) \not\in B$
$A \in NP$ since it seems clear that verifying an input to a problem that always returns false can be trivially accomplished in polynomial time in that we don't really even need to verify anything since there is no input instance to $A$ that would evaluate to true
Using the same logic as above this leads me to reason that there is a tautology problem $C$ such that $\forall$ inputs $x$, $x \in C$ and that $C \in NP$ and $C$ can be polynomial-time, many-one reduced to any other problem just like $A$
Is this true? or have I made an erroneous assumption? And what other useful properties are possessed by the contradiction and tautology problems $A$ and $C$?