Show that any PSPACE-hard language is also NP-hard

The NP problems are a class of problems whose solutions are difficult to find but simple to prove, and which are solved in polynomial time by a Non-Deterministic Machine.

A problem is NP-hard if all problems in NP can be reduced to it in polynomial time, even if it isn't in NP.

PSPACE is the collection of all decision problems that a Turing machine can answer using a polynomial space in computational complexity theory.

PSPACE hardness for a language L is defined as all $A\in PSPACE$,$A\le {}_{P}L$. Because SAT is NP complete, it is also NP hard. As a result, $SAT\le {}_{P}L$, and L s NP hard.

That if every NP-hard language is also PSPACE-hard, then PSPACE=NP.

If every NP-hard language is PSPACE-hard , then SAT is PSPACE-hard, and consequently every PSPACE language is polynomial-time reducible to SAT. then

$PSPACE\subseteq NP$ because $SAT\in NP$ and therefore $PSPACE=NP$because we know $NP\subseteq PSPACE$