0.58/0.70 MAYBE 0.58/0.70 0.58/0.70 DP problem for innermost termination. 0.58/0.70 P = 0.58/0.70 init#(x1, x2) -> f1#(rnd1, rnd2) 0.58/0.70 f2#(I0, I1) -> f2#(I0 + 1, I2) [I0 <= 9 /\ 3 <= I0 - 1] 0.58/0.70 f2#(I3, I4) -> f2#(I3 + 1, I5) [I3 <= 9 /\ I3 <= 2] 0.58/0.70 f2#(I6, I7) -> f2#(3, I8) [3 = I6] 0.58/0.70 f1#(I9, I10) -> f2#(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1] 0.58/0.70 R = 0.58/0.70 init(x1, x2) -> f1(rnd1, rnd2) 0.58/0.70 f2(I0, I1) -> f2(I0 + 1, I2) [I0 <= 9 /\ 3 <= I0 - 1] 0.58/0.70 f2(I3, I4) -> f2(I3 + 1, I5) [I3 <= 9 /\ I3 <= 2] 0.58/0.70 f2(I6, I7) -> f2(3, I8) [3 = I6] 0.58/0.70 f1(I9, I10) -> f2(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1] 0.58/0.70 0.58/0.70 The dependency graph for this problem is: 0.58/0.70 0 -> 4 0.58/0.70 1 -> 1 0.58/0.70 2 -> 2, 3 0.58/0.70 3 -> 3 0.58/0.70 4 -> 1, 2, 3 0.58/0.70 Where: 0.58/0.70 0) init#(x1, x2) -> f1#(rnd1, rnd2) 0.58/0.70 1) f2#(I0, I1) -> f2#(I0 + 1, I2) [I0 <= 9 /\ 3 <= I0 - 1] 0.58/0.70 2) f2#(I3, I4) -> f2#(I3 + 1, I5) [I3 <= 9 /\ I3 <= 2] 0.58/0.70 3) f2#(I6, I7) -> f2#(3, I8) [3 = I6] 0.58/0.70 4) f1#(I9, I10) -> f2#(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1] 0.58/0.70 0.58/0.70 We have the following SCCs. 0.58/0.70 { 2 } 0.58/0.70 { 3 } 0.58/0.70 { 1 } 0.58/0.70 0.58/0.70 DP problem for innermost termination. 0.58/0.70 P = 0.58/0.70 f2#(I0, I1) -> f2#(I0 + 1, I2) [I0 <= 9 /\ 3 <= I0 - 1] 0.58/0.70 R = 0.58/0.70 init(x1, x2) -> f1(rnd1, rnd2) 0.58/0.70 f2(I0, I1) -> f2(I0 + 1, I2) [I0 <= 9 /\ 3 <= I0 - 1] 0.58/0.70 f2(I3, I4) -> f2(I3 + 1, I5) [I3 <= 9 /\ I3 <= 2] 0.58/0.70 f2(I6, I7) -> f2(3, I8) [3 = I6] 0.58/0.70 f1(I9, I10) -> f2(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1] 0.58/0.70 0.58/0.70 We use the reverse value criterion with the projection function NU: 0.58/0.70 NU[f2#(z1,z2)] = 9 + -1 * z1 0.58/0.70 0.58/0.70 This gives the following inequalities: 0.58/0.70 I0 <= 9 /\ 3 <= I0 - 1 ==> 9 + -1 * I0 > 9 + -1 * (I0 + 1) with 9 + -1 * I0 >= 0 0.58/0.70 0.58/0.70 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. 0.58/0.70 0.58/0.70 DP problem for innermost termination. 0.58/0.70 P = 0.58/0.70 f2#(I6, I7) -> f2#(3, I8) [3 = I6] 0.58/0.70 R = 0.58/0.70 init(x1, x2) -> f1(rnd1, rnd2) 0.58/0.70 f2(I0, I1) -> f2(I0 + 1, I2) [I0 <= 9 /\ 3 <= I0 - 1] 0.58/0.70 f2(I3, I4) -> f2(I3 + 1, I5) [I3 <= 9 /\ I3 <= 2] 0.58/0.70 f2(I6, I7) -> f2(3, I8) [3 = I6] 0.58/0.70 f1(I9, I10) -> f2(I10, I11) [-1 <= I10 - 1 /\ 0 <= I9 - 1] 0.58/0.70 0.58/0.70 EOF