Senate rejects Rand Paul’s stand, confirms aid to Libya, Egypt, Pakistan despite terrorism
The New American reports,
Three minutes after midnight Friday leading into Saturday morning, the Senate rejected by a vote of 81 to 10 a proposal offered by Senator Rand Paul (R-Ky.) to hold aid to the governments of Egypt, Libya, and Pakistan pending the surrender to U.S. authorities of those suspected of carrying out the attack on U.S. diplomatic offices in Egypt and Libya. U.S. Ambassador Chris Stevens was murdered in the attack on the Libyan consulate.
Senator Paul’s bill also contained a clause requiring the release of Dr. Shakil Afridi, currently imprisoned by the Pakistani government, before any more money would be sent to Islamabad. Paul observed,
In no way should the United States government be sending money to governments who are not our ally, who blatantly do not respect our country, and who work to compromise the safety of our allies and citizens abroad. I am pleased that the Senate leadership has listened to my pleas for an end to this and have agreed to debate and vote on this pressing issue. . . .
Senate leaders from both major parties were reluctant to take a vote on the measure and only relented after Senator Paul successfully filibustered Friday night, forcing the issue. . . .
The following senators joined Senator Paul in voting to attach these conditions to the delivery of funds to the named governments:
Michael D. Crapo (R-Idaho); Jim DeMint (R-South Car.); Charles E. Grassley (R-Iowa); Mike Lee (R-Utah); Jerry Moran (R-Kan.); Jim Risch (R-Idaho); Pat Roberts (R-Kan.); Richard C. Shelby (R-Ala.); and Patrick Toomey (R-Penn.).
During that same pre-dawn session, the Senate approved 90-1 a bill declaring that the United States will take action to thwart Iran’s purported plan to develop nuclear weapons. Rand Paul was the only senator who voted against the measure.