38.58/38.23 MAYBE 38.58/38.23 38.58/38.23 DP problem for innermost termination. 38.58/38.23 P = 38.58/38.23 init#(x1, x2) -> f1#(rnd1, rnd2) 38.58/38.23 f3#(I0, I1) -> f2#(I0, I2) [28 <= I0 - 1 /\ I0 <= 39] 38.58/38.23 f3#(I3, I4) -> f2#(I3 + 1, I5) [39 <= I3 - 1] 38.58/38.23 f3#(I6, I7) -> f2#(I6 + 1, I8) [10 <= I6 - 1 /\ I6 <= 28] 38.58/38.23 f3#(I9, I10) -> f2#(I9, I11) [I9 <= 10 /\ I9 <= 29] 38.58/38.23 f2#(I12, I13) -> f3#(I12 - 1, I14) [0 <= I12 - 1 /\ I12 <= 49 /\ I12 <= 19] 38.58/38.23 f2#(I15, I16) -> f3#(I15, I17) [19 <= I15 - 1 /\ I15 <= 49] 38.58/38.23 f1#(I18, I19) -> f2#(I19, I20) [-1 <= I19 - 1 /\ 0 <= I18 - 1] 38.58/38.23 R = 38.58/38.23 init(x1, x2) -> f1(rnd1, rnd2) 38.58/38.23 f3(I0, I1) -> f2(I0, I2) [28 <= I0 - 1 /\ I0 <= 39] 38.58/38.23 f3(I3, I4) -> f2(I3 + 1, I5) [39 <= I3 - 1] 38.58/38.23 f3(I6, I7) -> f2(I6 + 1, I8) [10 <= I6 - 1 /\ I6 <= 28] 38.58/38.23 f3(I9, I10) -> f2(I9, I11) [I9 <= 10 /\ I9 <= 29] 38.58/38.23 f2(I12, I13) -> f3(I12 - 1, I14) [0 <= I12 - 1 /\ I12 <= 49 /\ I12 <= 19] 38.58/38.23 f2(I15, I16) -> f3(I15, I17) [19 <= I15 - 1 /\ I15 <= 49] 38.58/38.23 f1(I18, I19) -> f2(I19, I20) [-1 <= I19 - 1 /\ 0 <= I18 - 1] 38.58/38.23 38.58/38.23 The dependency graph for this problem is: 38.58/38.23 0 -> 7 38.58/38.23 1 -> 6 38.58/38.23 2 -> 6 38.58/38.23 3 -> 5, 6 38.58/38.23 4 -> 5 38.58/38.23 5 -> 3, 4 38.58/38.23 6 -> 1, 2, 3 38.58/38.23 7 -> 5, 6 38.58/38.23 Where: 38.58/38.23 0) init#(x1, x2) -> f1#(rnd1, rnd2) 38.58/38.23 1) f3#(I0, I1) -> f2#(I0, I2) [28 <= I0 - 1 /\ I0 <= 39] 38.58/38.23 2) f3#(I3, I4) -> f2#(I3 + 1, I5) [39 <= I3 - 1] 38.58/38.23 3) f3#(I6, I7) -> f2#(I6 + 1, I8) [10 <= I6 - 1 /\ I6 <= 28] 38.58/38.23 4) f3#(I9, I10) -> f2#(I9, I11) [I9 <= 10 /\ I9 <= 29] 38.58/38.23 5) f2#(I12, I13) -> f3#(I12 - 1, I14) [0 <= I12 - 1 /\ I12 <= 49 /\ I12 <= 19] 38.58/38.23 6) f2#(I15, I16) -> f3#(I15, I17) [19 <= I15 - 1 /\ I15 <= 49] 38.58/38.23 7) f1#(I18, I19) -> f2#(I19, I20) [-1 <= I19 - 1 /\ 0 <= I18 - 1] 38.58/38.23 38.58/38.23 We have the following SCCs. 38.58/38.23 { 1, 2, 3, 4, 5, 6 } 38.58/38.23 38.58/38.23 DP problem for innermost termination. 38.58/38.23 P = 38.58/38.23 f3#(I0, I1) -> f2#(I0, I2) [28 <= I0 - 1 /\ I0 <= 39] 38.58/38.23 f3#(I3, I4) -> f2#(I3 + 1, I5) [39 <= I3 - 1] 38.58/38.23 f3#(I6, I7) -> f2#(I6 + 1, I8) [10 <= I6 - 1 /\ I6 <= 28] 38.58/38.23 f3#(I9, I10) -> f2#(I9, I11) [I9 <= 10 /\ I9 <= 29] 38.58/38.23 f2#(I12, I13) -> f3#(I12 - 1, I14) [0 <= I12 - 1 /\ I12 <= 49 /\ I12 <= 19] 38.58/38.23 f2#(I15, I16) -> f3#(I15, I17) [19 <= I15 - 1 /\ I15 <= 49] 38.58/38.23 R = 38.58/38.23 init(x1, x2) -> f1(rnd1, rnd2) 38.58/38.23 f3(I0, I1) -> f2(I0, I2) [28 <= I0 - 1 /\ I0 <= 39] 38.58/38.23 f3(I3, I4) -> f2(I3 + 1, I5) [39 <= I3 - 1] 38.58/38.23 f3(I6, I7) -> f2(I6 + 1, I8) [10 <= I6 - 1 /\ I6 <= 28] 38.58/38.23 f3(I9, I10) -> f2(I9, I11) [I9 <= 10 /\ I9 <= 29] 38.58/38.23 f2(I12, I13) -> f3(I12 - 1, I14) [0 <= I12 - 1 /\ I12 <= 49 /\ I12 <= 19] 38.58/38.23 f2(I15, I16) -> f3(I15, I17) [19 <= I15 - 1 /\ I15 <= 49] 38.58/38.23 f1(I18, I19) -> f2(I19, I20) [-1 <= I19 - 1 /\ 0 <= I18 - 1] 38.58/38.23 38.58/41.21 EOF