0.60/0.67 MAYBE 0.60/0.67 0.60/0.67 DP problem for innermost termination. 0.60/0.67 P = 0.60/0.67 f9#(x1, x2) -> f8#(x1, x2) 0.60/0.67 f8#(I0, I1) -> f2#(I0, 1) 0.60/0.67 f5#(I4, I5) -> f4#(I4, I5) 0.60/0.67 f4#(I6, I7) -> f5#(I6, I7) 0.60/0.67 f2#(I8, I9) -> f3#(rnd1, I9) [rnd1 = rnd1] 0.60/0.67 f3#(I10, I11) -> f1#(I10, I11) [1 + I10 <= 0] 0.60/0.67 f3#(I12, I13) -> f1#(I12, I13) [1 <= I12] 0.60/0.67 f3#(I14, I15) -> f4#(I14, 1) [I14 <= 0 /\ 0 <= I14 /\ y1 = 0] 0.60/0.67 f1#(I16, I17) -> f2#(I16, I17) 0.60/0.67 R = 0.60/0.67 f9(x1, x2) -> f8(x1, x2) 0.60/0.67 f8(I0, I1) -> f2(I0, 1) 0.60/0.67 f6(I2, I3) -> f7(I2, I3) 0.60/0.67 f5(I4, I5) -> f4(I4, I5) 0.60/0.67 f4(I6, I7) -> f5(I6, I7) 0.60/0.67 f2(I8, I9) -> f3(rnd1, I9) [rnd1 = rnd1] 0.60/0.67 f3(I10, I11) -> f1(I10, I11) [1 + I10 <= 0] 0.60/0.67 f3(I12, I13) -> f1(I12, I13) [1 <= I12] 0.60/0.67 f3(I14, I15) -> f4(I14, 1) [I14 <= 0 /\ 0 <= I14 /\ y1 = 0] 0.60/0.67 f1(I16, I17) -> f2(I16, I17) 0.60/0.67 0.60/0.67 The dependency graph for this problem is: 0.60/0.67 0 -> 1 0.60/0.67 1 -> 4 0.60/0.67 2 -> 3 0.60/0.67 3 -> 2 0.60/0.67 4 -> 5, 6, 7 0.60/0.67 5 -> 8 0.60/0.67 6 -> 8 0.60/0.67 7 -> 3 0.60/0.67 8 -> 4 0.60/0.67 Where: 0.60/0.67 0) f9#(x1, x2) -> f8#(x1, x2) 0.60/0.67 1) f8#(I0, I1) -> f2#(I0, 1) 0.60/0.67 2) f5#(I4, I5) -> f4#(I4, I5) 0.60/0.67 3) f4#(I6, I7) -> f5#(I6, I7) 0.60/0.67 4) f2#(I8, I9) -> f3#(rnd1, I9) [rnd1 = rnd1] 0.60/0.67 5) f3#(I10, I11) -> f1#(I10, I11) [1 + I10 <= 0] 0.60/0.67 6) f3#(I12, I13) -> f1#(I12, I13) [1 <= I12] 0.60/0.67 7) f3#(I14, I15) -> f4#(I14, 1) [I14 <= 0 /\ 0 <= I14 /\ y1 = 0] 0.60/0.67 8) f1#(I16, I17) -> f2#(I16, I17) 0.60/0.67 0.60/0.67 We have the following SCCs. 0.60/0.67 { 4, 5, 6, 8 } 0.60/0.67 { 2, 3 } 0.60/0.67 0.60/0.67 DP problem for innermost termination. 0.60/0.67 P = 0.60/0.67 f5#(I4, I5) -> f4#(I4, I5) 0.60/0.67 f4#(I6, I7) -> f5#(I6, I7) 0.60/0.67 R = 0.60/0.67 f9(x1, x2) -> f8(x1, x2) 0.60/0.67 f8(I0, I1) -> f2(I0, 1) 0.60/0.67 f6(I2, I3) -> f7(I2, I3) 0.60/0.67 f5(I4, I5) -> f4(I4, I5) 0.60/0.67 f4(I6, I7) -> f5(I6, I7) 0.60/0.67 f2(I8, I9) -> f3(rnd1, I9) [rnd1 = rnd1] 0.60/0.67 f3(I10, I11) -> f1(I10, I11) [1 + I10 <= 0] 0.60/0.67 f3(I12, I13) -> f1(I12, I13) [1 <= I12] 0.60/0.67 f3(I14, I15) -> f4(I14, 1) [I14 <= 0 /\ 0 <= I14 /\ y1 = 0] 0.60/0.67 f1(I16, I17) -> f2(I16, I17) 0.60/0.67 0.60/3.65 EOF