0.00/0.61 MAYBE 0.00/0.61 0.00/0.61 DP problem for innermost termination. 0.00/0.61 P = 0.00/0.61 init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 0.00/0.61 f3#(I0, I1, I2) -> f3#(I3, I4, I5) [-1 <= I3 - 1 /\ 0 <= I0 - 1] 0.00/0.61 f2#(I6, I7, I8) -> f3#(I9, I10, I11) [2 <= I9 - 1 /\ 1 <= I7 - 1 /\ I8 <= 0 /\ I9 - 1 <= I7] 0.00/0.61 f2#(I12, I13, I14) -> f3#(I15, I16, I17) [0 <= I15 - 1 /\ I14 <= 0 /\ 0 <= I13 - 1] 0.00/0.61 f2#(I18, I19, I20) -> f2#(I18 - 1, I21, I18) [2 <= I21 - 1 /\ 0 <= I19 - 1 /\ 0 <= I20 - 1 /\ I21 - 2 <= I19] 0.00/0.61 f1#(I22, I23, I24) -> f2#(I23 - 1, I25, I23) [1 <= I25 - 1 /\ 0 <= I22 - 1 /\ -1 <= I23 - 1 /\ I25 - 1 <= I22] 0.00/0.61 R = 0.00/0.61 init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 0.00/0.61 f3(I0, I1, I2) -> f3(I3, I4, I5) [-1 <= I3 - 1 /\ 0 <= I0 - 1] 0.00/0.61 f2(I6, I7, I8) -> f3(I9, I10, I11) [2 <= I9 - 1 /\ 1 <= I7 - 1 /\ I8 <= 0 /\ I9 - 1 <= I7] 0.00/0.61 f2(I12, I13, I14) -> f3(I15, I16, I17) [0 <= I15 - 1 /\ I14 <= 0 /\ 0 <= I13 - 1] 0.00/0.61 f2(I18, I19, I20) -> f2(I18 - 1, I21, I18) [2 <= I21 - 1 /\ 0 <= I19 - 1 /\ 0 <= I20 - 1 /\ I21 - 2 <= I19] 0.00/0.61 f1(I22, I23, I24) -> f2(I23 - 1, I25, I23) [1 <= I25 - 1 /\ 0 <= I22 - 1 /\ -1 <= I23 - 1 /\ I25 - 1 <= I22] 0.00/0.61 0.00/0.61 The dependency graph for this problem is: 0.00/0.61 0 -> 5 0.00/0.61 1 -> 1 0.00/0.61 2 -> 1 0.00/0.61 3 -> 1 0.00/0.61 4 -> 2, 3, 4 0.00/0.61 5 -> 2, 3, 4 0.00/0.61 Where: 0.00/0.61 0) init#(x1, x2, x3) -> f1#(rnd1, rnd2, rnd3) 0.00/0.61 1) f3#(I0, I1, I2) -> f3#(I3, I4, I5) [-1 <= I3 - 1 /\ 0 <= I0 - 1] 0.00/0.61 2) f2#(I6, I7, I8) -> f3#(I9, I10, I11) [2 <= I9 - 1 /\ 1 <= I7 - 1 /\ I8 <= 0 /\ I9 - 1 <= I7] 0.00/0.61 3) f2#(I12, I13, I14) -> f3#(I15, I16, I17) [0 <= I15 - 1 /\ I14 <= 0 /\ 0 <= I13 - 1] 0.00/0.61 4) f2#(I18, I19, I20) -> f2#(I18 - 1, I21, I18) [2 <= I21 - 1 /\ 0 <= I19 - 1 /\ 0 <= I20 - 1 /\ I21 - 2 <= I19] 0.00/0.61 5) f1#(I22, I23, I24) -> f2#(I23 - 1, I25, I23) [1 <= I25 - 1 /\ 0 <= I22 - 1 /\ -1 <= I23 - 1 /\ I25 - 1 <= I22] 0.00/0.61 0.00/0.61 We have the following SCCs. 0.00/0.61 { 4 } 0.00/0.61 { 1 } 0.00/0.61 0.00/0.61 DP problem for innermost termination. 0.00/0.61 P = 0.00/0.61 f3#(I0, I1, I2) -> f3#(I3, I4, I5) [-1 <= I3 - 1 /\ 0 <= I0 - 1] 0.00/0.61 R = 0.00/0.61 init(x1, x2, x3) -> f1(rnd1, rnd2, rnd3) 0.00/0.61 f3(I0, I1, I2) -> f3(I3, I4, I5) [-1 <= I3 - 1 /\ 0 <= I0 - 1] 0.00/0.61 f2(I6, I7, I8) -> f3(I9, I10, I11) [2 <= I9 - 1 /\ 1 <= I7 - 1 /\ I8 <= 0 /\ I9 - 1 <= I7] 0.00/0.61 f2(I12, I13, I14) -> f3(I15, I16, I17) [0 <= I15 - 1 /\ I14 <= 0 /\ 0 <= I13 - 1] 0.00/0.61 f2(I18, I19, I20) -> f2(I18 - 1, I21, I18) [2 <= I21 - 1 /\ 0 <= I19 - 1 /\ 0 <= I20 - 1 /\ I21 - 2 <= I19] 0.00/0.61 f1(I22, I23, I24) -> f2(I23 - 1, I25, I23) [1 <= I25 - 1 /\ 0 <= I22 - 1 /\ -1 <= I23 - 1 /\ I25 - 1 <= I22] 0.00/0.61 0.00/3.59 EOF