0.00/0.39 YES 0.00/0.39 0.00/0.39 DP problem for innermost termination. 0.00/0.39 P = 0.00/0.39 init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 0.00/0.39 f2#(I0, I1, I2, I3) -> f2#(I0 - 1, I1 - 1, I2, I2) [I2 = I3 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1] 0.00/0.39 f2#(I4, I5, I6, I7) -> f2#(I4, I6, I6, I6) [I6 = I7 /\ 0 = I5 /\ 0 <= I6 - 1] 0.00/0.39 f1#(I8, I9, I10, I11) -> f2#(I12, I13, I14, I15) [I13 = I15 /\ I13 = I14 /\ 0 <= I8 - 1 /\ -1 <= I12 - 1 /\ -1 <= I9 - 1 /\ -1 <= I13 - 1] 0.00/0.39 R = 0.00/0.39 init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 0.00/0.39 f2(I0, I1, I2, I3) -> f2(I0 - 1, I1 - 1, I2, I2) [I2 = I3 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1] 0.00/0.39 f2(I4, I5, I6, I7) -> f2(I4, I6, I6, I6) [I6 = I7 /\ 0 = I5 /\ 0 <= I6 - 1] 0.00/0.39 f1(I8, I9, I10, I11) -> f2(I12, I13, I14, I15) [I13 = I15 /\ I13 = I14 /\ 0 <= I8 - 1 /\ -1 <= I12 - 1 /\ -1 <= I9 - 1 /\ -1 <= I13 - 1] 0.00/0.39 0.00/0.39 The dependency graph for this problem is: 0.00/0.39 0 -> 3 0.00/0.39 1 -> 1, 2 0.00/0.39 2 -> 1 0.00/0.39 3 -> 1 0.00/0.39 Where: 0.00/0.39 0) init#(x1, x2, x3, x4) -> f1#(rnd1, rnd2, rnd3, rnd4) 0.00/0.39 1) f2#(I0, I1, I2, I3) -> f2#(I0 - 1, I1 - 1, I2, I2) [I2 = I3 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1] 0.00/0.39 2) f2#(I4, I5, I6, I7) -> f2#(I4, I6, I6, I6) [I6 = I7 /\ 0 = I5 /\ 0 <= I6 - 1] 0.00/0.39 3) f1#(I8, I9, I10, I11) -> f2#(I12, I13, I14, I15) [I13 = I15 /\ I13 = I14 /\ 0 <= I8 - 1 /\ -1 <= I12 - 1 /\ -1 <= I9 - 1 /\ -1 <= I13 - 1] 0.00/0.39 0.00/0.39 We have the following SCCs. 0.00/0.39 { 1, 2 } 0.00/0.39 0.00/0.39 DP problem for innermost termination. 0.00/0.39 P = 0.00/0.39 f2#(I0, I1, I2, I3) -> f2#(I0 - 1, I1 - 1, I2, I2) [I2 = I3 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1] 0.00/0.39 f2#(I4, I5, I6, I7) -> f2#(I4, I6, I6, I6) [I6 = I7 /\ 0 = I5 /\ 0 <= I6 - 1] 0.00/0.39 R = 0.00/0.39 init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 0.00/0.39 f2(I0, I1, I2, I3) -> f2(I0 - 1, I1 - 1, I2, I2) [I2 = I3 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1] 0.00/0.39 f2(I4, I5, I6, I7) -> f2(I4, I6, I6, I6) [I6 = I7 /\ 0 = I5 /\ 0 <= I6 - 1] 0.00/0.39 f1(I8, I9, I10, I11) -> f2(I12, I13, I14, I15) [I13 = I15 /\ I13 = I14 /\ 0 <= I8 - 1 /\ -1 <= I12 - 1 /\ -1 <= I9 - 1 /\ -1 <= I13 - 1] 0.00/0.39 0.00/0.39 We use the basic value criterion with the projection function NU: 0.00/0.39 NU[f2#(z1,z2,z3,z4)] = z1 0.00/0.39 0.00/0.39 This gives the following inequalities: 0.00/0.39 I2 = I3 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1 ==> I0 >! I0 - 1 0.00/0.39 I6 = I7 /\ 0 = I5 /\ 0 <= I6 - 1 ==> I4 (>! \union =) I4 0.00/0.39 0.00/0.39 We remove all the strictly oriented dependency pairs. 0.00/0.39 0.00/0.39 DP problem for innermost termination. 0.00/0.39 P = 0.00/0.39 f2#(I4, I5, I6, I7) -> f2#(I4, I6, I6, I6) [I6 = I7 /\ 0 = I5 /\ 0 <= I6 - 1] 0.00/0.39 R = 0.00/0.39 init(x1, x2, x3, x4) -> f1(rnd1, rnd2, rnd3, rnd4) 0.00/0.39 f2(I0, I1, I2, I3) -> f2(I0 - 1, I1 - 1, I2, I2) [I2 = I3 /\ 0 <= I1 - 1 /\ 0 <= I2 - 1 /\ 0 <= I0 - 1] 0.00/0.39 f2(I4, I5, I6, I7) -> f2(I4, I6, I6, I6) [I6 = I7 /\ 0 = I5 /\ 0 <= I6 - 1] 0.00/0.39 f1(I8, I9, I10, I11) -> f2(I12, I13, I14, I15) [I13 = I15 /\ I13 = I14 /\ 0 <= I8 - 1 /\ -1 <= I12 - 1 /\ -1 <= I9 - 1 /\ -1 <= I13 - 1] 0.00/0.39 0.00/0.39 The dependency graph for this problem is: 0.00/0.39 2 -> 0.00/0.39 Where: 0.00/0.39 2) f2#(I4, I5, I6, I7) -> f2#(I4, I6, I6, I6) [I6 = I7 /\ 0 = I5 /\ 0 <= I6 - 1] 0.00/0.39 0.00/0.39 We have the following SCCs. 0.00/0.39 0.00/3.37 EOF