Defending champion John Higgins will face Matthew Selt in the semi-finals of the Indian Open, while Lyu Haotian will take on Anthony Hamilton.
The semis and the final of the world ranking event in Kochi will be played on Sunday, with the winner to take home the trophy and £50,000 top prize.
Romford’s Selt reached the last four of a ranking event for the first time in his 17-year career by edging out Lu Ning 4-3. From 2-0 up, Selt lost the next three frames, but then won the last two with runs of 66 and 86.
Higgins, chasing his first title of the season, edged out China’s Li Hang 4-3 in the quarter-finals. Breaks of 104, 81 and 61 gave Scotland’s Higgins a 3-2 lead, and though he lost the sixth frame, he took the decider 78-1.
“It was a good match,” said four-time World Champion Higgins after reaching the 70th ranking event semi-final of his career. “Li has played the best snooker of anyone this week so I knew it would be difficult. I just got a bit of rub of the green at the right time. Li is what you might call an old fashioned player, a bit like myself. He knows his way around the table and doesn’t hit the ball too hard. He’s a fabulous all-round player. The Chinese players are here to stay and they will be winning big tournaments.”
China’s up-and-coming Lyu reached his third ranking event semi-final with a 4-2 victory over Mark Davis. The 21-year-old previously got to the last four of the 2017 Northern Ireland Open and 2018 China Championship. He was tied at 2-2 but took the last two frames with runs of 106 and 70.
Nottingham’s 47-year-old Hamilton, who won his first ranking title at the German Masters two years ago, scored a 4-2 win over Scott Donaldson. Break of 127, 74, 69 and 65 put Hamilton into the last four.