AEW world champion Jon Moxley was the winner of the annual PWI 500.
Pro Wrestling Illustrated announced that Moxley ranks first on the world’s 500 most productive professional wrestlers list.
Seth Rollins ranked No. 1 in 2019.
The PWI 500 2020 is a collection number, while the list becomes 30. Printed copies can now be ordered on the PWI website.
Stay tuned for updates to the PWI 500 for this year. Below is the complete policy of the factor with Moxley, in reaction from MJF, who will challenge the name AEW All Out on September 5:
Since politics is already making its way online, I don’t want to wait any longer: congratulations to @JonMoxley of AEWrestling for being the number one wrestler in this year’s PWI 500! Pre-order your copy now published in https://t.co/EuUXs75XJB pic.twitter.com/O0JrkyvJ6U
What a joke. https://t.co/36Jrocr0u9
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