1.55/1.92 MAYBE 1.55/1.92 1.55/1.92 DP problem for innermost termination. 1.55/1.92 P = 1.55/1.92 f5#(x1, x2, x3) -> f1#(x1, x2, x3) 1.55/1.92 f4#(I0, I1, I2) -> f2#(I0, I1, I2) 1.55/1.92 f2#(I3, I4, I5) -> f4#(I3, I4 + I5, I5) [-1 * (I4 + I5) <= 0] 1.55/1.92 f1#(I9, I10, I11) -> f2#(I9, I10, I11) 1.55/1.92 R = 1.55/1.92 f5(x1, x2, x3) -> f1(x1, x2, x3) 1.55/1.92 f4(I0, I1, I2) -> f2(I0, I1, I2) 1.55/1.92 f2(I3, I4, I5) -> f4(I3, I4 + I5, I5) [-1 * (I4 + I5) <= 0] 1.55/1.92 f2(I6, I7, I8) -> f3(rnd1, I7 + I8, I8) [rnd1 = rnd1 /\ 0 <= -1 - (I7 + I8)] 1.55/1.92 f1(I9, I10, I11) -> f2(I9, I10, I11) 1.55/1.92 1.55/1.92 The dependency graph for this problem is: 1.55/1.92 0 -> 3 1.55/1.92 1 -> 2 1.55/1.92 2 -> 1 1.55/1.92 3 -> 2 1.55/1.92 Where: 1.55/1.92 0) f5#(x1, x2, x3) -> f1#(x1, x2, x3) 1.55/1.92 1) f4#(I0, I1, I2) -> f2#(I0, I1, I2) 1.55/1.92 2) f2#(I3, I4, I5) -> f4#(I3, I4 + I5, I5) [-1 * (I4 + I5) <= 0] 1.55/1.92 3) f1#(I9, I10, I11) -> f2#(I9, I10, I11) 1.55/1.92 1.55/1.92 We have the following SCCs. 1.55/1.92 { 1, 2 } 1.55/1.92 1.55/1.92 DP problem for innermost termination. 1.55/1.92 P = 1.55/1.92 f4#(I0, I1, I2) -> f2#(I0, I1, I2) 1.55/1.92 f2#(I3, I4, I5) -> f4#(I3, I4 + I5, I5) [-1 * (I4 + I5) <= 0] 1.55/1.92 R = 1.55/1.92 f5(x1, x2, x3) -> f1(x1, x2, x3) 1.55/1.92 f4(I0, I1, I2) -> f2(I0, I1, I2) 1.55/1.92 f2(I3, I4, I5) -> f4(I3, I4 + I5, I5) [-1 * (I4 + I5) <= 0] 1.55/1.92 f2(I6, I7, I8) -> f3(rnd1, I7 + I8, I8) [rnd1 = rnd1 /\ 0 <= -1 - (I7 + I8)] 1.55/1.92 f1(I9, I10, I11) -> f2(I9, I10, I11) 1.55/1.92 1.55/4.90 EOF