Mitch McConnell: You can take this bill to protect Robert Mueller, roll it up tight, and cram it up your… Photo: Chip Somodevilla/Getty Images

Republican senator Thom Tillis has taken the risky step of endorsing a bill to safeguard Robert Mueller from being fired by President Trump. But the risk was for naught, as Senate Majority Leader Mitch McConnell told reporters he would not allow any such bill to come to the Senate floor.

McConnell argues, as he has in the past, that such a bill is “unnecessary” because there is “no indication” Trump would fire the special counsel. Of course there are many indications. Innumerable news reports have described Trump raging about Mueller and demanding his firing. Trump actually ordered the firing of Mueller in June, and again in December, and has begun attacking Mueller publicly, as well as attacking the Department of Justice official who oversees and has approved his investigation, both privately and publicly. Trump has also previously fired the FBI director, with whom he closely associates Mueller.

Other than that, there aren’t any indications. But the only indications McConnell will accept as valid will apparently be the actual firing of Mueller, at which point it will be too late to do anything, and McConnell will urge the country to trust Trump’s new attorney general, Sean Hannity, or whoever.

McConnell of course sat in a high-level briefing detailing Russian interference in the election. The Obama administration proposed leaders of both parties close ranks and issue a bipartisan statement warning Russia not to interfere in the election. McConnell reportedly cast doubts on the intelligence and said he would consider any statement about Russian interference to be a partisan effort by Obama to help his party win.

When McConnell refuses to act because he says he doesn’t think a threat is real, it means he is happy to let the threat be carried out.