India will release Pakistani prisoner Sikander Azam on Monday (today) after 15 months of detention, said Bhim Singh, the prisoner’s Indian counsel.
Azam had completed his sentence in July 2011. Talking to The Express Tribune, Singh said the Pakistani prisoner will cross Wagha Border on October 9 with two other prisoners.
Sikander Azam was arrested from Jammu railway station on July 18, 2008 under Public Safety Act, and was sentenced to three years’ imprisonment and fine. Despite completing the sentence on July 18, 2011 the prisoner was kept in confinement at Jammu jail due to pending legal formalities.
Singh said that the Indian Supreme Court directed that he should be shifted to a guest house since he was no more a prisoner. In compliance with the order, Sikander Azam was shifted to a guest house at Delhi.
Division Bench of the Supreme Court of India comprising Mr RM Lodha and Mr Anil R Dave in August this year while hearing a petition filed by Bhim Singh said, “Sikander Azam was arrested on July 19, 2008 in connection with FIR No. 58 of 2008.
The maximum sentence awarded to Sikander Azam came to an end on July 19, 2011. We, accordingly, direct the release of Sikander Azam forthwith from Jammu jail where he is presently lodged. On his release, he shall be brought to the Sewa Sadan, Lampur, Delhi by the authorities of the State of Jammu and Kashmir and he will be lodged in Sewa Sadan, Lampur, Delhi, until his repatriation.”
Published in The Express Tribune, October 8th, 2012.
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