Two global fields $K$ and $L$ are said to be Hilbert-symbol equivalent when there exists a group isomorphism $t$ of their groups of square classes and a bijection $T$ of the set of all primes of $K$ onto the set of all primes of $L$ and these two maps preserve Hilbert symbols, i.e., $(a,b)_P=(ta,tb)_{TP}$ for all square classes $a,b$ of $K$ and all primes $P$ of $K$.
Hilbert-symbol equivalence has been introduced (and called reciprocity equivalence) in the article of {\it R. Perlis, K. Szymiczek, P. E. Conner} and {\it E. Litherland} [Contemp. Math. 155, 365-387 (1994;

Zbl 0807.11024)] as a sufficient condition for the Witt equivalence of global fields, i.e., for the existence of an isomorphism of their Witt rings. The authors proves that the converse statement also holds, that is, if $K$ and $L$ are Witt equivalent then $K$ and $L$ are Hilbert-symbol equivalent. Another proof of this result one can find in the paper of {\it K. Szymiczek} [Math. Slovaca 41, 315-330 (1991;

Zbl 0766.11023)].
In the present paper, the author gives a new proof that Witt equivalence implies Hilbert-symbol equivalence. The proof is universal in the sense that it does not use quadratic forms. It is based on the properties of Steinberg symbols and takes advantage of the special role played by Hilbert-symbols.