0.61/0.70 MAYBE 0.61/0.70 0.61/0.70 DP problem for innermost termination. 0.61/0.70 P = 0.61/0.70 f4#(x1, x2) -> f3#(x1, x2) 0.61/0.70 f3#(I0, I1) -> f1#(I0, I1) 0.61/0.70 f2#(I2, I3) -> f1#(I2, I3) 0.61/0.70 f1#(I4, I5) -> f2#(rnd1, rnd2) [y2 = y2 /\ y1 = y2 /\ rnd2 = rnd2 /\ 1 + y1 <= rnd2 /\ rnd2 <= 1 + y1 /\ rnd1 = rnd2] 0.61/0.70 R = 0.61/0.70 f4(x1, x2) -> f3(x1, x2) 0.61/0.70 f3(I0, I1) -> f1(I0, I1) 0.61/0.70 f2(I2, I3) -> f1(I2, I3) 0.61/0.70 f1(I4, I5) -> f2(rnd1, rnd2) [y2 = y2 /\ y1 = y2 /\ rnd2 = rnd2 /\ 1 + y1 <= rnd2 /\ rnd2 <= 1 + y1 /\ rnd1 = rnd2] 0.61/0.70 0.61/0.70 The dependency graph for this problem is: 0.61/0.70 0 -> 1 0.61/0.70 1 -> 3 0.61/0.70 2 -> 3 0.61/0.70 3 -> 2 0.61/0.70 Where: 0.61/0.70 0) f4#(x1, x2) -> f3#(x1, x2) 0.61/0.70 1) f3#(I0, I1) -> f1#(I0, I1) 0.61/0.70 2) f2#(I2, I3) -> f1#(I2, I3) 0.61/0.70 3) f1#(I4, I5) -> f2#(rnd1, rnd2) [y2 = y2 /\ y1 = y2 /\ rnd2 = rnd2 /\ 1 + y1 <= rnd2 /\ rnd2 <= 1 + y1 /\ rnd1 = rnd2] 0.61/0.70 0.61/0.70 We have the following SCCs. 0.61/0.70 { 2, 3 } 0.61/0.70 0.61/0.70 DP problem for innermost termination. 0.61/0.70 P = 0.61/0.70 f2#(I2, I3) -> f1#(I2, I3) 0.61/0.70 f1#(I4, I5) -> f2#(rnd1, rnd2) [y2 = y2 /\ y1 = y2 /\ rnd2 = rnd2 /\ 1 + y1 <= rnd2 /\ rnd2 <= 1 + y1 /\ rnd1 = rnd2] 0.61/0.70 R = 0.61/0.70 f4(x1, x2) -> f3(x1, x2) 0.61/0.70 f3(I0, I1) -> f1(I0, I1) 0.61/0.70 f2(I2, I3) -> f1(I2, I3) 0.61/0.70 f1(I4, I5) -> f2(rnd1, rnd2) [y2 = y2 /\ y1 = y2 /\ rnd2 = rnd2 /\ 1 + y1 <= rnd2 /\ rnd2 <= 1 + y1 /\ rnd1 = rnd2] 0.61/0.70 0.61/3.68 EOF