Politics

Santorum Dismisses Rand Paul’s 2016 Prospects: ‘The Republican Party Is Not A Libertarian Party’

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Brendan Bordelon Contributor
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Former Republican presidential candidate Rick Santorum dismissed Kentucky Republican Sen. Rand Paul’s 2016 presidential ambitions, claiming “the Republican Party is not a libertarian party” and suggesting few GOP voters would be swayed by Paul’s small government message.

Santorum swiped at the senator Monday during an appearance on CNN’s “Crossfire,” where he told left-wing host Van Jones that Republicans “need a positive agenda” to win in 2014 and 2016.

“So positive agenda,” Jones replied. “Rand Paul, who now seems to be everywhere. He’s the frontrunner in your party. Do you agree with his ideas, are you going to support Rand Paul?”

“Well looking, as I said before, there’s diversity in the Republican Party . . . ” Santorum ventured cautiously.

But Van Jones pressed on, asking if Paul is “the new face of the Republican Party? Is this your leader?”

“Well — no he’s not MY leader, I can tell you that for sure,” Santorum laughed. “His father and I had some disagreements during the last campaign.”

“If a libertarian like him becomes the leader of the Republican Party — gets the nomination — would you vote for them?” Jones continued.

“Well first off, I don’t think that will happen,” Santorum claimed, “because the Republican Party is not a libertarian party, it is a conservative party. And it will nominate a conservative, and not a libertarian.”

While running for president in 2012 and during his career as a U.S. senator, Santorum distinguished himself as a strong social conservative who advocated a muscular foreign policy. Paul, in contrast, supports relaxing drug laws, letting states decide gay marriage and a reduced military footprint worldwide.

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