0.00/0.14 YES 0.00/0.14 0.00/0.14 DP problem for innermost termination. 0.00/0.14 P = 0.00/0.14 div#(x, y) -> div#(x - y, y) [x > y && y > 0] 0.00/0.14 R = 0.00/0.14 div(x, y) -> div(x - y, y) + 1 [x > y && y > 0] 0.00/0.14 div(I0, I1) -> 0 [I1 >= I0] 0.00/0.14 div(I2, I3) -> 0 [0 >= I3] 0.00/0.14 0.00/0.14 We use the reverse value criterion with the projection function NU: 0.00/0.14 NU[div#(z1,z2)] = z1 0.00/0.14 0.00/0.14 This gives the following inequalities: 0.00/0.14 x > y && y > 0 ==> x > x - y with x >= 0 0.00/0.14 0.00/0.14 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. 0.00/3.12 EOF