Say a,ba,ba,b are two elements of the ring. Let us denote ab=c,ba=dab=c, ba=dab=c,ba=d , belonging to the ring, since it is closed under multiplication.
Now, we have aba=aba ⟹ ca=ad ⟹ c=daba=aba\implies ca=ad\implies c=daba=aba⟹ca=ad⟹c=d , from the property of the ring.
Thus, ab=baab=baab=ba for arbitrary a,ba,ba,b in the ring. Hence, the ring is commutative.
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