3.34/3.38 MAYBE 3.34/3.38 3.34/3.38 DP problem for innermost termination. 3.34/3.38 P = 3.34/3.38 f5#(x1, x2, x3) -> f4#(x1, x2, x3) 3.34/3.38 f4#(I0, I1, I2) -> f1#(I0, I1, I2) 3.34/3.38 f3#(I3, I4, I5) -> f1#(I3, I4, I5) 3.34/3.38 f1#(I6, I7, I8) -> f3#(I6 + I7, -1 + I7, I8) [1 <= I6] 3.34/3.38 f2#(I9, I10, I11) -> f1#(I9, I10, I11) 3.34/3.38 f1#(I12, I13, I14) -> f2#(I12 + I14, I13, -1 + I14) [1 <= I12] 3.34/3.38 R = 3.34/3.38 f5(x1, x2, x3) -> f4(x1, x2, x3) 3.34/3.38 f4(I0, I1, I2) -> f1(I0, I1, I2) 3.34/3.38 f3(I3, I4, I5) -> f1(I3, I4, I5) 3.34/3.38 f1(I6, I7, I8) -> f3(I6 + I7, -1 + I7, I8) [1 <= I6] 3.34/3.38 f2(I9, I10, I11) -> f1(I9, I10, I11) 3.34/3.38 f1(I12, I13, I14) -> f2(I12 + I14, I13, -1 + I14) [1 <= I12] 3.34/3.38 3.34/3.38 The dependency graph for this problem is: 3.34/3.38 0 -> 1 3.34/3.38 1 -> 3, 5 3.34/3.38 2 -> 3, 5 3.34/3.38 3 -> 2 3.34/3.38 4 -> 3, 5 3.34/3.38 5 -> 4 3.34/3.38 Where: 3.34/3.38 0) f5#(x1, x2, x3) -> f4#(x1, x2, x3) 3.34/3.38 1) f4#(I0, I1, I2) -> f1#(I0, I1, I2) 3.34/3.38 2) f3#(I3, I4, I5) -> f1#(I3, I4, I5) 3.34/3.38 3) f1#(I6, I7, I8) -> f3#(I6 + I7, -1 + I7, I8) [1 <= I6] 3.34/3.38 4) f2#(I9, I10, I11) -> f1#(I9, I10, I11) 3.34/3.38 5) f1#(I12, I13, I14) -> f2#(I12 + I14, I13, -1 + I14) [1 <= I12] 3.34/3.38 3.34/3.38 We have the following SCCs. 3.34/3.38 { 2, 3, 4, 5 } 3.34/3.38 3.34/3.38 DP problem for innermost termination. 3.34/3.38 P = 3.34/3.38 f3#(I3, I4, I5) -> f1#(I3, I4, I5) 3.34/3.38 f1#(I6, I7, I8) -> f3#(I6 + I7, -1 + I7, I8) [1 <= I6] 3.34/3.38 f2#(I9, I10, I11) -> f1#(I9, I10, I11) 3.34/3.38 f1#(I12, I13, I14) -> f2#(I12 + I14, I13, -1 + I14) [1 <= I12] 3.34/3.38 R = 3.34/3.38 f5(x1, x2, x3) -> f4(x1, x2, x3) 3.34/3.38 f4(I0, I1, I2) -> f1(I0, I1, I2) 3.34/3.38 f3(I3, I4, I5) -> f1(I3, I4, I5) 3.34/3.38 f1(I6, I7, I8) -> f3(I6 + I7, -1 + I7, I8) [1 <= I6] 3.34/3.38 f2(I9, I10, I11) -> f1(I9, I10, I11) 3.34/3.38 f1(I12, I13, I14) -> f2(I12 + I14, I13, -1 + I14) [1 <= I12] 3.34/3.38 3.43/6.36 EOF