0.75/0.82 MAYBE 0.75/0.82 0.75/0.82 DP problem for innermost termination. 0.75/0.82 P = 0.75/0.82 init#(x1, x2) -> f1#(rnd1, rnd2) 0.75/0.82 f2#(I0, I1) -> f2#(20, I2) [30 <= I0 - 1] 0.75/0.82 f2#(I3, I4) -> f2#(29, I5) [25 = I3] 0.75/0.82 f2#(I6, I7) -> f2#(I6 - 1, I8) [25 <= I6 - 1 /\ I6 <= 30] 0.75/0.82 f2#(I9, I10) -> f2#(I9 - 1, I11) [10 <= I9 - 1 /\ I9 <= 24 /\ I9 <= 30] 0.75/0.82 f1#(I12, I13) -> f2#(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1] 0.75/0.82 R = 0.75/0.82 init(x1, x2) -> f1(rnd1, rnd2) 0.75/0.82 f2(I0, I1) -> f2(20, I2) [30 <= I0 - 1] 0.75/0.82 f2(I3, I4) -> f2(29, I5) [25 = I3] 0.75/0.82 f2(I6, I7) -> f2(I6 - 1, I8) [25 <= I6 - 1 /\ I6 <= 30] 0.75/0.82 f2(I9, I10) -> f2(I9 - 1, I11) [10 <= I9 - 1 /\ I9 <= 24 /\ I9 <= 30] 0.75/0.82 f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1] 0.75/0.82 0.75/0.82 The dependency graph for this problem is: 0.75/0.82 0 -> 5 0.75/0.82 1 -> 4 0.75/0.82 2 -> 3 0.75/0.82 3 -> 2, 3 0.75/0.82 4 -> 4 0.75/0.82 5 -> 1, 2, 3, 4 0.75/0.82 Where: 0.75/0.82 0) init#(x1, x2) -> f1#(rnd1, rnd2) 0.75/0.82 1) f2#(I0, I1) -> f2#(20, I2) [30 <= I0 - 1] 0.75/0.82 2) f2#(I3, I4) -> f2#(29, I5) [25 = I3] 0.75/0.82 3) f2#(I6, I7) -> f2#(I6 - 1, I8) [25 <= I6 - 1 /\ I6 <= 30] 0.75/0.82 4) f2#(I9, I10) -> f2#(I9 - 1, I11) [10 <= I9 - 1 /\ I9 <= 24 /\ I9 <= 30] 0.75/0.82 5) f1#(I12, I13) -> f2#(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1] 0.75/0.82 0.75/0.82 We have the following SCCs. 0.75/0.82 { 2, 3 } 0.75/0.82 { 4 } 0.75/0.82 0.75/0.82 DP problem for innermost termination. 0.75/0.82 P = 0.75/0.82 f2#(I9, I10) -> f2#(I9 - 1, I11) [10 <= I9 - 1 /\ I9 <= 24 /\ I9 <= 30] 0.75/0.82 R = 0.75/0.82 init(x1, x2) -> f1(rnd1, rnd2) 0.75/0.82 f2(I0, I1) -> f2(20, I2) [30 <= I0 - 1] 0.75/0.82 f2(I3, I4) -> f2(29, I5) [25 = I3] 0.75/0.82 f2(I6, I7) -> f2(I6 - 1, I8) [25 <= I6 - 1 /\ I6 <= 30] 0.75/0.82 f2(I9, I10) -> f2(I9 - 1, I11) [10 <= I9 - 1 /\ I9 <= 24 /\ I9 <= 30] 0.75/0.82 f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1] 0.75/0.82 0.75/0.82 We use the basic value criterion with the projection function NU: 0.75/0.82 NU[f2#(z1,z2)] = z1 0.75/0.82 0.75/0.82 This gives the following inequalities: 0.75/0.82 10 <= I9 - 1 /\ I9 <= 24 /\ I9 <= 30 ==> I9 >! I9 - 1 0.75/0.82 0.75/0.82 All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed. 0.75/0.82 0.75/0.82 DP problem for innermost termination. 0.75/0.82 P = 0.75/0.82 f2#(I3, I4) -> f2#(29, I5) [25 = I3] 0.75/0.82 f2#(I6, I7) -> f2#(I6 - 1, I8) [25 <= I6 - 1 /\ I6 <= 30] 0.75/0.82 R = 0.75/0.82 init(x1, x2) -> f1(rnd1, rnd2) 0.75/0.82 f2(I0, I1) -> f2(20, I2) [30 <= I0 - 1] 0.75/0.82 f2(I3, I4) -> f2(29, I5) [25 = I3] 0.75/0.82 f2(I6, I7) -> f2(I6 - 1, I8) [25 <= I6 - 1 /\ I6 <= 30] 0.75/0.82 f2(I9, I10) -> f2(I9 - 1, I11) [10 <= I9 - 1 /\ I9 <= 24 /\ I9 <= 30] 0.75/0.82 f1(I12, I13) -> f2(I13, I14) [-1 <= I13 - 1 /\ 0 <= I12 - 1] 0.75/0.82 0.75/3.80 EOF