2.50/2.53 MAYBE 2.50/2.53 2.50/2.53 DP problem for innermost termination. 2.50/2.53 P = 2.50/2.53 f5#(x1, x2, x3) -> f4#(x1, x2, x3) 2.50/2.53 f4#(I0, I1, I2) -> f1#(I0, I1, I2) [0 <= I0] 2.50/2.53 f3#(I3, I4, I5) -> f1#(I3, I4, I5) 2.50/2.53 f1#(I6, I7, I8) -> f3#(I6 + I8, 1 + I7, -2 + I8) 2.50/2.53 f2#(I9, I10, I11) -> f1#(I9, I10, I11) 2.50/2.53 f1#(I12, I13, I14) -> f2#(I12 + I13, -2 + I13, I14) 2.50/2.53 R = 2.50/2.53 f5(x1, x2, x3) -> f4(x1, x2, x3) 2.50/2.53 f4(I0, I1, I2) -> f1(I0, I1, I2) [0 <= I0] 2.50/2.53 f3(I3, I4, I5) -> f1(I3, I4, I5) 2.50/2.53 f1(I6, I7, I8) -> f3(I6 + I8, 1 + I7, -2 + I8) 2.50/2.53 f2(I9, I10, I11) -> f1(I9, I10, I11) 2.50/2.53 f1(I12, I13, I14) -> f2(I12 + I13, -2 + I13, I14) 2.50/2.53 2.50/2.53 The dependency graph for this problem is: 2.50/2.53 0 -> 1 2.50/2.53 1 -> 3, 5 2.50/2.53 2 -> 3, 5 2.50/2.53 3 -> 2 2.50/2.53 4 -> 3, 5 2.50/2.53 5 -> 4 2.50/2.53 Where: 2.50/2.53 0) f5#(x1, x2, x3) -> f4#(x1, x2, x3) 2.50/2.53 1) f4#(I0, I1, I2) -> f1#(I0, I1, I2) [0 <= I0] 2.50/2.53 2) f3#(I3, I4, I5) -> f1#(I3, I4, I5) 2.50/2.53 3) f1#(I6, I7, I8) -> f3#(I6 + I8, 1 + I7, -2 + I8) 2.50/2.53 4) f2#(I9, I10, I11) -> f1#(I9, I10, I11) 2.50/2.53 5) f1#(I12, I13, I14) -> f2#(I12 + I13, -2 + I13, I14) 2.50/2.53 2.50/2.53 We have the following SCCs. 2.50/2.53 { 2, 3, 4, 5 } 2.50/2.53 2.50/2.53 DP problem for innermost termination. 2.50/2.53 P = 2.50/2.53 f3#(I3, I4, I5) -> f1#(I3, I4, I5) 2.50/2.53 f1#(I6, I7, I8) -> f3#(I6 + I8, 1 + I7, -2 + I8) 2.50/2.53 f2#(I9, I10, I11) -> f1#(I9, I10, I11) 2.50/2.53 f1#(I12, I13, I14) -> f2#(I12 + I13, -2 + I13, I14) 2.50/2.53 R = 2.50/2.53 f5(x1, x2, x3) -> f4(x1, x2, x3) 2.50/2.53 f4(I0, I1, I2) -> f1(I0, I1, I2) [0 <= I0] 2.50/2.53 f3(I3, I4, I5) -> f1(I3, I4, I5) 2.50/2.53 f1(I6, I7, I8) -> f3(I6 + I8, 1 + I7, -2 + I8) 2.50/2.53 f2(I9, I10, I11) -> f1(I9, I10, I11) 2.50/2.53 f1(I12, I13, I14) -> f2(I12 + I13, -2 + I13, I14) 2.50/2.53 2.50/5.51 EOF