0.54/0.55 YES 0.54/0.55 0.54/0.55 DP problem for innermost termination. 0.54/0.55 P = 0.54/0.55 gcd#(x, y) -> gcd#(y - x, x) [y > x && x > 0] 0.54/0.55 gcd#(I0, I1) -> gcd#(I0 - I1, I1) [I0 >= I1 && I1 > 0] 0.54/0.55 R = 0.54/0.55 gcd(x, y) -> gcd(y - x, x) [y > x && x > 0] 0.54/0.55 gcd(I0, I1) -> gcd(I0 - I1, I1) [I0 >= I1 && I1 > 0] 0.54/0.55 gcd(0, I2) -> I2 0.54/0.55 gcd(I3, 0) -> I3 0.54/0.55 0.54/0.55 We use the reverse value criterion with the projection function NU: 0.54/0.55 NU[gcd#(z1,z2)] = z2 0.54/0.55 0.54/0.55 This gives the following inequalities: 0.54/0.55 y > x && x > 0 ==> y > x with y >= 0 0.54/0.55 I0 >= I1 && I1 > 0 ==> I1 >= I1 0.54/0.55 0.54/0.55 We remove all the strictly oriented dependency pairs. 0.54/0.55 0.54/0.55 DP problem for innermost termination. 0.54/0.55 P = 0.54/0.55 gcd#(I0, I1) -> gcd#(I0 - I1, I1) [I0 >= I1 && I1 > 0] 0.54/0.55 R = 0.54/0.55 gcd(x, y) -> gcd(y - x, x) [y > x && x > 0] 0.54/0.55 gcd(I0, I1) -> gcd(I0 - I1, I1) [I0 >= I1 && I1 > 0] 0.54/0.55 gcd(0, I2) -> I2 0.54/0.55 gcd(I3, 0) -> I3 0.54/0.55 0.54/0.55 We use the reverse value criterion with the projection function NU: 0.54/0.55 NU[gcd#(z1,z2)] = z1 0.54/0.55 0.54/0.55 This gives the following inequalities: 0.54/0.55 I0 >= I1 && I1 > 0 ==> I0 > I0 - I1 with I0 >= 0 0.54/0.55 0.54/0.55 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. 0.54/3.53 EOF